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4;7< ADVANCED ENGINEERING MATHEMATICS BY Dr. Mohamed Ismail Mohamed Hessein Professor Of Mathematics Faculty Of Engineering (Shoubra) Benha University 2017 CONTENTS Chapter 1 2 3 4 5 6 7 8 CONTENTS Page i Linear Dependence and Independence 1 Second-order linear differential equations 21 Fourier series 54 Fourier integrals 100 Laplace transforms 111 Special functions 149 Partial differential equations 177 Solution of differential equations in series 204 References 229 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Chapter 1 Linear Dependence and Independence Let V be a vector space over a field F. The following defines the notion of linear dependence and independence of vectors over F. (One usually suppresses mentioning F when the field is understood.) This concept plays an essential role in the theory of linear algebra and in mathematics in general. Definition 1.1: We say that the vectors v1, v2, ... , vm in V are linearly dependent if there exist scalars a1, a2, .. . , am in F, not all of them 0, such that a1v1 + a2v2 + amvm = 0 Otherwise, we say that the vectors are linearly independent. The above definition may be restated as follows. Consider the vector equation a1v1 + a2v2 + amvm = 0 (*) where the a’s are unknown scalars. This equation always has the zero solution a1 = 0, a2 = 0, .. . , am = 0 . Suppose this is the only solution, that is, suppose we can show: a1v1 + a2v2 + amvm = 0 implies a1 = 0, a2 = 0, .. . , am = 0. Then the vectors v1, v2, ... , vm are linearly independent, On the other hand, suppose the equation (*) has a nonzero solution, then the vectors are linearly dependent. A set S = {v1, v2, ... , vm} of vectors in V is linearly dependent or independent according to whether the vectors v1, v2, ... , vm are linearly dependent or independent. An infinite set S of vectors is linearly dependent or independent according to whether there do or do not exist vectors v1, v2, ... , vm in S that are linearly dependent. Warning: The set S = {v1, v2, ... , vm} above represents a list or, in other words, a finite sequence of vectors where the vectors are ordered and repetition is permitted. The following remarks follow directly from the above definition. Remark 1: Suppose 0 is one of the vectors v1, v2, ... , vm say v1 = 0. Then the vectors must be linearly dependent, because we have the following linear combination where the coefficient of v1  0: 1v1 + 0 v2++ 0vm = 1. 0 + 0 ++ 0 = 0 Remark 2: Suppose v is a nonzero vector. Then v, by itself, is linearly independent, because k v = 0, v  0 implies k = 0. Remark 3: Suppose two of the vectors v1, v2, ... , vm are equal or one is a scalar multiple of the other, say v1 = k v2. Then the vectors must be linearly dependent, because we have the following linear combination where the coefficient of v1 0: v1 − k v2 + 0 v3 + ··· + 0 vm = 0 Remark 4: Two vectors v1 and v2 are linearly dependent if and only if one of them is a multiple of the other. Remark 5: If the set {v1, v2, ... , vm} is linearly independent, then any rearrangement   of the vectors vi1 , vi 2 ,, vi m is also linearly independent. Remark 6: If a set S of vectors is linearly independent, then any subset of S is linearly independent. Alternatively, if S contains a linearly dependent subset, then S is linearly dependent. 1 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Example 1.1: a) Let u = (0, 1, 1, 1), v = (2, 1, 3, 1), w = ( 5, 4, 9, 4). Then u, v, w are linearly dependent, because 3u + 5v − 2w = 3(0, 1, 1, 1) + 5 (2, 1, 3, 1) − 2 ( 5, 4, 9, 4)= = ( 0+10−10 , 3+5−8, 3+15−18, 3+5−8) = ( 0, 0, 0, 0) b) We show that the vectors u = (1, 2, 3, 3), v = (2, 5, 7, 6), w = ( 1, 3, 5, 3). are linearly independent. We form the vector equation x u + y v + z w = 0, where x, y, z are unknown scalars. This yields x +2 y + z = 0 1  2 1   0  x + 2 y + z = 0 x +2 y + z = 0 2 5  3  0  2x +5 y + 3z = 0 y+ z=0 y+z=0         x y z  3x +7 y + 5z = 0 y + 2z = 0 2z = 0 3 7  5   0  3x +6 y + 3z = 0 0+ 0=0         3 6 3 0  Back-substitution yields x = 0, y = 0, z = 0. We have shown that x u + y v + z w = 0 implies x = 0, y = 0, z = 0 Accordingly u, v, w are linearly independent. c) Let V be the vector space of functions from R into R. We show that the functions t 2 f (t) = sin t, g(t) = e , h(t) = t are linearly independent. We form the vector (function) equation a f + b g + c h = 0, where a, b, c are unknown scalars. This function t 2 equation means that, for every value of t, a sin t + b e + c t = 0. Thus, in this equation, we choose appropriate values of t to easily get a = 0, b = 0, c = 0. For example, (i) Substitute t = 0 to obtain a (0) + b (1) + c (0) = 0 or b = 0.  2 (ii) Substitute t =  to obtain a (0) + 0 (e ) + c ( ) = 0 or c = 0. /2 2 (iii) Substitute t =/2 to obtain a (1) + b (e ) + 0 ( /4) = 0 or a = 0. We have shown a f + b g + c h = 0 implies a = 0, b = 0 , c = 0 . Accordingly, f, g, h are linearly independent. §1.1. Linear Dependence and Linear Combinations: The notions of linear dependence and linear combinations are closely related. Specifically, for more than one vector, we show that the vectors v1, v2, ... , vm are linearly dependent if and only if one of them is a linear combination of the others. Suppose, say, vi is a linear combination of the others, v i  a1v1    a i 1v i 1  a i 1v i 1    a m v m Then by adding −vi to both sides, we obtain a1v1    a i 1v i 1  v i  a i 1v i 1    a m v m  0 where the coefficient of vi is not 0. Hence, the vectors are linearly dependent. Conversely, suppose the vectors are linearly dependent, say, b1v1    b jv j    b m v m  0, where b j1  0 Then we can solve for vj obtaining v j  b 1j b1v1    b 1j b j1v j1  b 1j bi 1vi 1    b 1j b m v m . 2 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence and so vj is a linear combination of the other vectors. We now state a slightly stronger statement than the one above. This result has many important consequences.  Lemma 1: Suppose two or more nonzero vectors v1, v2, ... , vm are linearly dependent. Then one of the vectors is a linear combination of the preceding vectors, that is, there exists k > 1 such that v k  c1v1  c 2 v 2    c k 1v k 1 . § 1.2. Basis and Dimension First we state two equivalent ways to define a basis of a vector space V. Definition 1.2: A set S = {u1, u2, ... , un} of vectors is a basis of V if it has the following two properties: (1) S is linearly independent. (2) S spans V. Definition 1.3: A set S = {u1, u2, ... , un} of vectors is a basis of V if every v V can be written uniquely as a linear combination of the basis vectors. The following is a fundamental result in linear algebra. Theorem 1.1: Let V be a vector space such that one basis has m elements and another basis has n elements. Then m = n. A vector space V is said to be of finite dimension n or n dimensional, written dim V = n. if V has a basis with n elements. Theorem 2 tells us that all bases of V have the same number of elements, so this definition is well defined. The vector space {0} is defined to have dimension 0. Suppose a vector space V does not have a finite basis. Then V is said to be of infinite dimension or to be infinite-dimensional. The above fundamental Theorem 1 is a consequence of the following ‘‘replacement lemma’’ . Lemma 1.2: Suppose {v1, v2, ... , vn} spans V, and suppose {w1, w2, ... , wm} is linearly independent. Then m n, and V is spanned by a set of the form {v1, v2, ... , vn, w1, w2, ... , wm } Thus, in particular, n + 1 or more vectors in V are linearly dependent. Observe in the above lemma that we have replaced m of the vectors in the spanning set of V by the m independent vectors and still retained a spanning set. Examples of Bases This subsection presents important examples of bases of some of the main vector spaces appearing in this text. 3 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence n n a) Vector space R : Consider the following n vectors in R : e1 = (1, 0, 0, 0, ... , 0, 0), e2 = (0, 1, 0, 0, .. . , 0, 0), , en = (0, 0, 0, 0, .. . , 0, 1) These vectors are linearly independent. Furthermore, any vector u = (a1, a2, ... , an) in Rn can be written as a linear combination of the above vectors. Specifically, u = a1e1 + a2e2 + · · · + anen. n n Accordingly, the vectors form a basis of R called the usual or standard basis of R . n n Thus (as one might expect), R has dimension n. In particular, any other basis of R has n elements. b) Vector space M = Mr,s of all r s matrices: The following six matrices form a basis of the vector space M2,3 of all 2 x 3 matrices over R: 1 0 0  0 1 0  0 0 1  0 0 0  0 0 0   0 0 0   0 0 0  ,  0 0 0  ,  0 0 0  , 1 0 0  , 0 1 0  ,  0 0 1  .             More generally, in the vector space M = Mr,s of all r  s matrices, let Eij be the matrix with ij-entry 1 and 0’s elsewhere. Then all such matrices form a basis of Mr,s called the usual or standard basis of Mr,s. Accordingly, dim Mr,s = r s. c) Vector space Pn(t) of all polynomials of degree n: 2 3 n The set S = {1, t, t , t , .. . , t } of n+1 polynomials is a basis of Pn(t). Specifically, any polynomial f (t) of degree n can be expressed as a linear combination of these powers of t, and one can show that these polynomials are linearly independent. Therefore, dim Pn(t) = n + 1. d) Vector space P(t) of all polynomials: Consider any finite set S = {f1(t), f2(t), ... , fm(t)} of polynomials in P(t), and let m denote the largest of the degrees of the polynomials. Then any polynomial g(t) of degree exceeding m cannot be expressed as a linear combination of the elements of S. Thus, S cannot be a basis of P(t). This means that the dimension of P(t) is infinite. We 2 3 note that the infinite set S'= {1, t, t , t , }, consisting of all the powers of t, spans P(t) and is linearly independent. Accordingly, S' is an infinite basis of P(t). § 1.3. Theorems on Bases The following three theorems will be used frequently. Theorem 1.2: Let V be a vector space of finite dimension n. Then: i) Any n + 1 or more vectors in V are linearly dependent. ii) Any linearly independent set S = {u1, u2, ... , un} with n elements is a basis of V. iii) Any spanning set T = {v1, v2, ... , vn} of V with n elements is a basis of V. Theorem 1.3: Suppose S spans a vector space V. Then: (i) Any maximum number of linearly independent vectors in S form a basis of V. (ii) Suppose one deletes from S every vector that is a linear combination of preceding vectors in S. Then the remaining vectors form a basis of V. 4 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Theorem 1.4: Let V be a vector space of finite dimension and let S = {u1, u2, ... , un} be a set of linearly independent vectors in V. Then S is part of a basis of V; that is, S may be extended to a basis of V. Example 1.2: 4 a) The following four vectors in R form a matrix in echelon form: (1, 1, 1, 1), (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0, 1) 1 0 0 0  0 1 0 0    0 0 1 0    0 0 0 1  4 Thus, the vectors are linearly independent, and, because dim R = 4, the four vectors 4 form a basis of R . (1, 2, 3, 4) = 1(1, 1, 1, 1) + 1(0, 1, 1, 1) + 1(0, 0, 1, 1) +1 (0, 0, 0, 1) (3, 6, 5, −3) = 3(1, 1, 1, 1) + 3(0, 1, 1, 1) − 1(0, 0, 1, 1) −8 (0, 0, 0, 1) b) The following n + 1 polynomials in Pn(t) are of increasing degree: 2 n 1, t − 1, (t − 1) , ... , (t − 1) Therefore, no polynomial is a linear combination of preceding polynomials, hence, the polynomials are linear independent. Furthermore, they form a basis of Pn(t), because dim Pn(t) = n + 1. 3 c) Consider any four vectors in R , say (1, 1, 1), (1, 1, 0), (1, 0, 0), (3, −5, −7) By Theorem 2(i), the four vectors must be linearly dependent, because they come 3 from the three-dimensional vector space R . § 1.4. Dimension and Subspaces: The following theorem gives the basic relationship between the dimension of a vector space and the dimension of a subspace. Theorem 1.5: Let W be a subspace of an n-dimensional vector space V. Then dim W n. In particular, if dim W = n, then W = V. Proof: Because V is of dimension n, any n + 1 or more vectors are linearly dependent. Furthermore, because a basis of W consists of linearly independent vectors, it cannot contain more than n elements. Accordingly, dim Wn. In particular, if {w1, w2, ... , wn} is a basis of W, then, because it is an independent 5 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence set with n elements, it is also a basis of V. Thus, W = V when dim W = n. 3 3 Example 1.3: Let W be a subspace of the real space R . Note that dim R = 3. Theorem 2 tells us that the dimension of W can only be 0, 1, 2, or 3. The following cases apply: If dim W = 0, then W = {0}, a point. If dim W = 1, then W is a line through the origin 0. If dim W = 2, then W is a plane through the origin 0. 3 If dim W = 3, then W is the entire space R . § 1.5. Linear Independent Functions Two functions f(x) and g(x) are said to be linearly independent over an interval  ≤ x ≤  if the equation a f(x) + b g(x) = 0 (*) is only true for all x in the interval if a = b = 0. The functions are said to be linearly dependent if (*) is true for some non vanishing constants a and b. When the functions are linearly dependent, provided a  0, equation (*) can be written b a f (x)   g(x) , with a corresponding result g(x)   f (x) , if b  0, a b showing that in each case the linear dependence of the functions means they are proportional. We have established the following simple test. Test for linear independence of f(x) and g(x) over  ≤ x ≤  The two functions f(x) and g(x) will be linearly independent over  ≤ x ≤  if they are not proportional over the interval; otherwise, they will be linearly dependent. Example 1.4: Apply the test for linear independence to the following pairs of functions. (a) ex and e2x are linearly independent for all x because e2x/ex = ex is defined for all x and ex is not a constant. (b) ℓn x2 and ℓn x3 are linearly dependent for x > 0, because ℓn x2 = 2 ℓn x and ℓn x3 = 3 ℓn x, so ℓn x2/ℓn x3 = 2/3 is a constant, and the logarithmic function is defined for x > 0. (c) sinh 2x and sinh x cosh x are linearly dependent for all x because sinh 2x = 2 sinh x cosh x. The notion of the linear independence of functions is of special significance when the functions are solutions of homogeneous differential equations. This is because it will be seen later that all particular solutions of such differential equations can be represented in the form of suitable linear combinations of as many linearly independent solutions as the equation allows. In fact, the number of linearly independent solutions is equal to the order of the differential equation, so the second order differential equation (1) has two linearly independent solutions. So, if f(x) and g(x) are linearly independent solutions of (1), and a and b are general solution arbitrary constants, the general solution of (1) from which all particular solutions can be obtained can be written y(x) = a f(x) + b g(x). § 1.6. Linear Independence and the Wronskian Recall from linear algebra that two vectors v and w are called linearly dependent if there are nonzero constants a and b with a v + b w = 0 6 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence We can think of differentiable functions f(x) and g(x) as being vectors in the vector space of differentiable functions. The analogous definition is Let f(x) and g(x) be differentiable functions. Then they are called linearly dependent if there are nonzero constants a and b with a f(x) + b g(x) = 0 for all x. Otherwise they are called linearly independent. Example 1.5: The functions f(t) = 2sin2 t and g(t) = 1 − cos2 t are linearly dependent since (1)(2sin2 t)+(−2)(1−cos2 t)= 0 Example 1.6: The functions f(t) = t and g(t) = t2 are linearly independent since otherwise there would be nonzero constants a and b such that a t + b t2 = 0 for all t. First let t = 1. Then a + b = 0 . Now let t = 2. Then 2 a + 4 b = 0. This is a system of 2 equations and two unknowns. The determinant of the corresponding matrix is 4−2=2 Since the determinant is nonzero, the only solution is the trivial solution. That is a = b = 0, The two functions are linearly independent. In this example, we arbitrarily selected two values for t. It turns out that there is a systematic way to check for linear dependence. The following theorem states this way. Theorem 1.6: Let f and g be differentiable on [,]. If Wronskian W(f, g)(to) is nonzero for some to in [, ] then f and g are linearly independent on [, ]. If f and g are linearly dependent then the Wronskian is zero for all t in [, ]. Proof If a f(t) + b g(t) = 0. Then we can take derivatives of both sides to get a f'(t) + b g'(t) = 0 This is a system of two equations with two unknowns. The determinant of the f (t) corresponding matrix is the Wronskian. W(f, g)(t) = g(t) f '(t) g '(t) . Hence, if the Wronskian is nonzero at some to, only the trivial solution exists. Hence they are linearly independent. Example 1.7: Show that the functions f(t) = t and g(t) = e2t are linearly independent. Solution: We compute the Wronskian f '(t) = 1, g '(t) = 2e2t 2t f(t) g(t) The Wronskian is f '(t) g'(t) = t e 1 2e 2t 2t 2t 2 t = (t)(2e ) − (e )(1) = (e )(2t − 1). Now plug in 0 to get W(f, g)(0) = −1. which is nonzero. We can conclude that f and g are linearly independent. 7 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Example 1.8: 1. The vectors e1 = (1 ,0 ,0, , 0), e2 = (0 ,1 ,0, , 0), e3 = (0 ,0 ,1, , 0),, en = (0 ,0 ,0, , 1) in Rn are linearly independent. To show this, consider any scalars a1, a2,ak such that a1 e1+ a2e2 + a3e3+ ... + am em = (a1, a2, a3,,an ) = (0 ,0 ,0, , 0) The last equation implies that a1= a2 = a3 == ak = 0. The vector space Rn is spanned by the unit n-vectors e1 = (1, 0, 0, 0, . . . , 0), e2 = (0, 1, 0, 0, . . . , 0), . . . , en = (0, 0, 0, 0, . . . , 1). These vectors can be put in identity matrix in (echelon) form.  e1  1 0  0 0   e  0 1  0 0   2    A =        =  e1 e2  en 1 en          e  0 0 1 0 n  1      en  0 0  0 1  2. The vectors u = (1, 2, 3), v = (4, 5, 6), and w = (7, 8, 9), in R3 are linearly dependent, since u−2v+ w = 0. How did we come up with this? Start with the equation in the definition of linear dependence au +bv +c w = (0, 0, 0), and solve this system for possible values of a, b , c. 2. We can write this system in matrix form as 1 2   3 4 7 8  9  5 6 a  0  b   0       c   0  The coefficient matrix A is simply the matrix with columns u, v, w which we can write as A = [u, v, w]. If we reduce A to row echelon form we get 1 0   0 4 3 6 7  1 6    0    12   0 4 7  1  6   0   0   0 3 0 4 1 0 7 2  0  We have found that rank A = 2 is not maximal, and so the system is singular and has a nonzero solution (many in fact). i.e. a − c = 0 , b + 2c = 0  a = −b/2 = c. If we take c = 1  a = 1 , b = −2 , c = 1. If we take c = 2  a = 2, b = −4 and so on. u − 2v + w = 0 . Theorem 1.7: Let v1, v2, v3, ..., vm Rn be a set of vectors and let A = (v1v2v3...vm) Then the following statements are equivalent. 1. v1, v2, v3, ..., vm are linearly independent. 2. v1, v2, v3, ..., vm span Rn. 3. v1, v2, v3, ..., vm form a basis of Rn. 4. A  0. 5. rank A = n. 8 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence The fact that any basis of Rn contains exactly n vectors tells us that there are n independent directions in Rn. This is why we say that Rn is n-dimensional. Definition 1.4: The number of vectors in any basis of a subspace V is called the dimension of V. This definition only makes sense if any basis of V has the same number of vectors. Fortunately, this can be shown to be true. Theorem 1.8: Let S = {v1, v2, v3, ..., vm} be a set of m vectors in Rm and let A be the square matrix of vectors of S. Then S is linearly independent if and only if det A  0. 1) An example of a linear combination of vectors in R5 is provided by the vector sum (m = 3, n = 5) ; A = 2x + y + 3z, B = x + 2y − z z = (6, 0, 2, 2,−1). The vector where x = (1, 2, 3, 0, 4), y = (2, 1, 4, 1,−3), and in R5 formed by this linear combination is A = 2x+ y+3z = 2(1, 2, 3, 0, 4) + (2, 1, 4, 1,−3) + 3(6, 0, 2, 2,−1) = (22, 5, 16, 7, 2). B = x + 2y2 − z = (1, 2, 3, 0, 4) + 2(2, 1, 4, 1,−3)í (6, 0, 2, 2,−1) = (−1, 4, 9, 0, −1). 2) Let u = (1, 4, 2, 3) , v = (7, 10, −4, −1) , w = (−2, 1, 5, −4), These are dependent vectors since a (1, 4, 2, 3) +b (7, 10, −4, −1) + c (−2, 1, 5, −4) = (0, 0, 0, 0) i.e. −3 (1, 4, 2, 3) +1 (7, 10, −4, −1) + 2 (−2, 1, 5, −4) a = −3 , b = 1 , c = 2 We can calculate the values of a, b, and c by solving the system of linear equations: a + 7b − 2c = 0, 4 a +10 b + c = 0, 2 a − 4 b + 5 c = 0, 3 a − b − 4 c = 0. 3) An example of linear dependence in R4 is provided by the vectors x =(1, 0, 2, 5), y = (2, 1, 2, 1), z = (3, 2, 1, 0), and t = (−1,−1,−1, 7), because t = 2x − 3y + z. 2(1, 0, 2, 5) −3 (2, 1, 2, 1)+(3, 2, 1, 0) = (−1,−1,−1, 7). As example of linear independence in R4 is provided by the vectors x1 =(1, 0, 2, 5), x2 = (2, 1, 2, 1), x3 = (3, 2, 1, 0), and x4 = (−1,−1,−1, 8), a(1, 0, 2, 5) +b (2, 1, 2, 1) + c(3, 2, 1, 0) +G ííí     with a = b = c = d = 0. Example 1.9: The vectors (1,1,0) , (1,0,1), (0,1,1) are independent in R3 . Since a (1,1,0) + b (1,0,1) + c (0,1,1) = (0, 0, 0) This is equivalent to the following set of linear equations: a + b = 0 , a + c = 0, b + c = 0. You can verify that the solution is a = b = c = 0. Hence, the vectors are independent. Example 1.10: We know that the set of vectors {(1, −2) , (ℓn 3 , ℓn 2) . ( e4 , e5)}in R2 is dependent without doing any computation. Any set of three (or more) vectors in R2 is dependent. Example 1.11: R[x] is a vector space over the reals. The set 1 , x , x2 , x3 ,  is independent. For if ao + a1 x + a2 x2 +  + an xn = 0, it follows that ai = 0 for all  i. 9 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Example 1.12: In order to determine if the three vectors in u, v , w  R4, are linearly dependent or independent, where u = (1, 4, 2, 3) , v = (7, 10, −4, −1) , w = (−2, 1, 5, −4) form the matrix, 3 u  1 4 2 3 u  1 4 2  7 10 4 1 v   0 18 18 22 v  7u        2 1 5 4 w  0 9 9 2 w  2u  1 0   0 4 2 3 9  18 9  18 2  22 u  1 w  2u  0   v  7 u   0 4 2 3 9 0 9 0 2  18 u  w  2u   v  3u  2 w  It is clear that these vectors are linearly independent.(Since the last term in the third row is not zero −1  0. In order to determine if the Four vectors in  , ,  ,  R3,  = (1, 7, −2) ,  = (4, 10, 1) ,  = (2, −4, 5) and  = (3, −1, −4) are linearly dependent or independent. We have a set of n = 4 vectors in R3 (i.e., Rm with m = 3). Since n = 4 > m = 3 (i.e., n > m), then S is linearly dependent. In order to find one of them expressed as a linear combination of the other vectors, form the matrix 2 α   1 7 2 α   1 7  4 10 1 β   0 18 9 β  4α      2 4 5 γ   0 18 9 γ  2α       3 1 4 δ   0 22 2 δ  3  α  1 7 2  1  0 2 1 (β  4α)/(-9)   0   0 0 0 γ+2     0     0 11 1 (δ  3 ) / ( 2)   0 7 2 α  1 0.5 (β  4α)/(  18)   0 4.5 (11β  17α  9γ)/18  0 0 2α  β  δ  It is clear that these vectors are linearly dependent and 0 = 2 −  +    = 2 +  or  =  − 2 or 2 =  −   = (4, 10, 1) = 2 (1, 7, −2) + (2, −4, 5) = (2, 14, −4) + (2, −4, 5) Determine whether the set of vectors {(1, 0, 1, 2), (1, 2, 2, 1), (0, 1, 2, 1)} is Linear dependence or Linear Independence in R4. Example 1.13: The vectors (2, − 3) and (−10, 15) are dependent in R2 . To show this, we have to find numbers a and b, not both 0, such that a (2,− 3) + b(−10, 15) = (0, 0). There are many pairs of numbers that work. For example, 5 (2, − 3) + 1 (−10, 15) = (0, 0). Likewise, 10 (2, − 3) + 2 (− 10, 15) = (0, 0). More generally, vectors (a , b) and (c, d) in R2 are dependent if and only if they are multiples of one another. 10 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence Example 1.15: Is S = {(1, 0, 1), (2, 1, −1), (5,−1, 0), (−2, 1, 1)} linearly dependent or linearly independent? Solution: We have a set of n = 4 vectors in R3 (i.e., Rm with m = 3). Since n = 4 > m = 3 (i.e., n > m), then S is linearly dependent. Example 1.16: Is S = {(1, 1, 1), (1, 0, 1), (0, 2, 0)} linearly dependent or linearly independent? 1 1 0  Solution: We form the matrix A = 1 0 2  .   1 1 0  Then expanding along column 3 gives et A = 2(−1)2+3 1 1 1 1  0. Since det A = 0, S is linearly dependent. Example 1.17: Show that the following vectors are linearly independent, u = (1, 2, 0, 2), v = (1, 1, 1, 0) , w = (2, 0, 1, 3) Find a fourth vector x so that the set {u, v, w, x} is a basis for R4? Solution: To find the one vector that we need to complete a basis (the space has dimension four), we could start with the matrix 1 1 2 a  2 1 0 b   0 1 1 c    2 0 3 d  Row reduce this and in the end select a, b, c, d so that the reduced form has independent columns (four pivots). Apply row-reduction suggested above and obtained a  1 1 2 a   1 1 2 a  1 1 2  2 1 0 b  0 1 4 2a  b  0 1 4 2a  b        0 1 1 c  0 1 1 c  0 0 3 c  2a  b         2 0 3 d  0 2 1 d  2a  0 0 7 d  2a  2b  1 1 2  0 1 4   is 3, It is easy to see that the rank of the matrix  0 0 3   0 0 7  11 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence which means that the vectors u, v, w are linearly independent. complete the reduction, we get a a 1 1 2  1 1 2    0 1 4  b  2a 0 1 4 b  2a      (1)  0 0 21 0 0 21 7c  14a  7b  7c  14a  7b      0 0 0 8a  b  7c  3d  0 0 21 3d  6a  6b  Taking a = 1 and b = c = d = 0 in (1) will give us a matrix with independent columns (we want −8a +b + 7c + 3d = −8  0). So we can take (1, 0, 0, 0) to be the forth vector needed to complete a basis. We can see also that 1 1 2 1 1 1 2 0 1 1 2 0 1 1 2 0 2 1 0 0 2 1 0 1 2 1 0 0 2 1 0 0  7,  8 ,  1,  3. 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 0 2 0 3 0 2 0 3 0 2 0 3 0 2 0 3 1 Example 1.18: Determine whether {u, v, w} is linearly independent, where u = (1, 1, 2, 1), v = (0, 2, 1, 1), and w = (3, 1, 2, 0). Solution: Using the definitions, we wish to know whether a u + b v + c w = 0 has only the trivial solution. Suppose that a u + b v + c w = 0 . Then a(1, 1, 2, 1) + b(0, 2, 1, 1) + c(3, 1, 2, 0) = (0, 0, 0, 0) (a, a , 2a, a) + (0, 2b, b, b) + (3c, c, 2c, 0) = (0, 0, 0, 0) (a + 3c, a + 2b + c, 2a + b + 2c, a + b) = (0, 0, 0, 0) Equating corresponding components, we have a+ 3c = 0 a + 2b + c = 0 2a + b + 2c = 0 a+ b =0 1 0 3 0  1 0 0 0  1 2 1 0  0 1 0 0     Forming the augmented matrix, we get 2 1 2 0 0 0 1 0     1 1 0 0  0 0 0 0 We see that the system has the unique solution a = b = c = 0. Thus, the homogeneous system has only the trivial solution. Since there is no solution in which any of the constant values is non-zero, we see that the set {u, v, w} is linearly independent. Example 1.19: Given the set S = {u, v, w, x, y}, with u = (1, 2, 3), v = (1, 1, 2), w = (0, 1, 1), x = (2, 3, 1), and y = (1, 4, 1). Show that S is linearly dependent and express (at least) one of the vectors as a linear combination of some of the others. Solution: We construct the matrix as in the test for linear independence, and transform it to echelon form: 12 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence 1 1 0 2 1  1 0 1 0 2   2 1 1 3 4    0 1 1 0  3       3 2 1 1 1  0 0 0 1 1  Notice that column 3 of the row reduced matrix, (1,−1, 0), is not an elementary column. Notice also that in this matrix, column 3 is equal to one times column 1 plus (−1) times column 2, i.e., (1,−1, 0) = 1(1, 0, 0)+(−1)(0, 1, 0). We see that the entries in the non-elementary column 3 are the coefficients of dependence with which this vector can be expressed as a linear combination of the vectors preceding it in the original matrix. What the theorem is telling us is that the third column of the original matrix is the same linear combination of the vectors which were transformed into these elementary vectors. That is, just as it is true in the last matrix that column 3 is equal to 1 times the first column plus −1 times the second column (using columns of the last matrix), in the original matrix we also have C3 = C1 − C2. We can, of course, verify this: u − v = (1, 2, 3) − (1, 1, 2) = (1 − 1, 2 − 1, 3 − 2) = (0, 1, 1) = w Thus, we see that w is dependent on u and v, and the entries in column 3 of the last matrix give the coefficients of dependency. And since we can express w as a linear combination of u and v, we see that the set S is linearly dependent. Look also at column 5 of the matrix. This column is not an elementary column, so the vector y can be written as a linear combination of some of the other vectors. Which ones? We look for the vectors (i.e., columns) containing leading 1’s in the last matrix, because these are elementary columns. We see that row 1’s leading 1 is in column 1, row 2’s leading 1 is in column 2 and row 3’s leading 1 is in column 4. In the last matrix we have: (2,−3, 1) = 2(1, 0, 0) − 3(0, 1, 0) + 1(0, 0, 1) so that column 5 = 2× (column 1) + (−3)× (column) 2 + 1× (column 4). This tells us that this relationship also holds true in the original matrix. That is, we have y = 2u − 3v + x. (Again, we can easily confirm that this is true.) Thus, we see that y is dependent on u, v and x, and that the entries in column 5 of the last matrix give us the coefficients of dependence, i.e., tell us how to write y as a linear combination of these vectors. Example 1.20: consider the functions x2, x, and 1. Take the Wronskian: x2 W = 2x 2 x 1 1 0  2 0 0 Note that W is always non-zero, so these functions are independent everywhere. Consider, however, x2 and x: x2 x  x 2  2x 2   x 2 2x 1 Here W = 0 only when x = 0. Therefore x2 and x are independent except at x = 0. Consider 2x2 + 3, x2, and 1: W= 13 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence 2x 2  3 x 2 1 2x 0  8x  8x  0 W = 4x 4 2 0 Here W is always zero, so these functions are always dependent. This is intuitively obvious, of course, since 2x2 + 3 = 2(x2) + 3(1) Example 1.21: Consider the functions u = sin x, v = cos x, and w = sin x + cos x . Then f(x) = 3 sin x + 5 cos x = 3 u + 5 v, unique when expressed with linearly independent functions, but f(x) = 3 sin x + 5 cos x = 5 w − 2 u = 3 w + 2 v not unique when expressed with linearly dependent functions. Example 1.22: Let u = sin x , v = cos x and w = sin x + cos x from vector space V. Find which of the following can be represented in terms of u and v, and w. a) f(x) = 3 sin x + 5 cos x = 3 u + 5 v = 5 w − 2 u = 3 w + 2 v not unique when expressed with linearly dependent functions. b) cos2x can't expressed by u, v and w only. c) sin2 x can't expressed by u, v and w only. d) sin 2x can't expressed by u, v and w only. e) cos 3x can't expressed by u, v and w only. f) 5 can't expressed by u, v and w only. Example 1.23: Let u = cos2 x and v = sin2 x form a basis for a vector space V. Find which of the following can be represented in terms of u and v, and so lie in V. (a) 2. (b) sin 2x. (c) 0. (d) cos 2x. (e) 2 + 3x. (f) 3 − 4 cos 2x. Solution: a) 2 = 2 cos2 x + 2 sin2 x = 2(u + v ), b) sin 2x  V c) 0 = cos2 x − cos2 x = sin2 x − sin2 x = u − u = v − v d) cos 2x = cos2 x − sin2 x = u − v e) 2 + 3x V f) 3 − 4 cos 2x = 3(cos2 x + sin2 x) − 4 cos2 x + 4sin2 x = 3( u + v) -4(u − v) = −u + 7v. Example 1.24: Let u = (2, 0) , v = (−3, 5), and w = (8, 6). Explain (without doing any computations) how you know that the set of vectors {u, v, w } is linearly dependent. Then write a linear dependence relation for this set. (This latter part does require computation.) We know that the set of vectors {u, v, w} in R2 is dependent without doing any computation. Any set of three (or more) vectors in R2 is dependent. a (2, 0) + b (−3, 5) + c(8, 6) = (0, 0) 2a −3b + 8c = 0 , 5b +6c = 0 14 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence  2 3 8   a  0   0 5 6   b   0       This means that w can be expressed as a linear combination of {u, v} or v as a linear combination of {u, w} as follows  2 0 u  1 0 u/2  1 0 u/2           3 5 v 0 5 v 3u / 2  0 5 v 3u / 2         8 6 w   0 6 w  4u   0 0 5w  29u  6v  This system will be consistent if 29u + 6 v − 5w = 0 i.e. 5 w = 29 u + 6 v , 29 u = 5 w − 6v and 6v = 5 w − 29 u. Another Solution: Let u = (2, 0) , v = (−3, 5), and w = (8, 6). w = (8, 6) = a (2, 0) + b (−3, 5) = (2a − 3b , 5b ) solving we get 5b = 6 , 2a −3b = 8 a = 5.8, b = 1.2  w = (8, 6) = 5.8 u + 1.2 v  10 w = 58 u + 12 v Example 1.25: Let u = (0, 0, 2) , v = (0, 5, −8) and w = (−3, 4, 1) 1. Determine whether the set {u, v, w} is linearly independent or linearly dependent. 2. If the set {u, v, w} is linearly dependent, then write a linear dependence relation for 0 0 3 this set. = 0 5 4  30 , then the set is linearly independent. 2 8 1 Example 1.26: Let u = (0, 0, 2) , v = (−3, 4, 7) and w = (−3, 4, 1). 1. Determine whether the set {u, v, w} is linearly independent or linearly dependent. 2. If the set {u, v, w} is linearly dependent, then write a linear dependence relation for this set. 0 3 3 = 0 4 4 0 , 2 7 1 then the set is linearly dependent. Let u = (0, 0, 2) , v = (−3, 4, 7) and w = (−3, 4, 1) v = (−3, 4, 7) = a (0, 0, 2) + b (−3, 4, 1) = (−3b , 4b, 2a+b ) solving we get −3b = −3, 2a+ b = 7, a = 3, b = 1  v = (−3, 4, 7) = 3 u + w. Example 1.27: Let u = (−1, 4, 0) , v = (−1, 2, 3) , w = (1, 4 ,−12) , h = (−1, 4, 3) Explain why the set of vectors S= {u, v, w, h } is linearly dependent and find one of them expressed as a linear combination of the other vectors. We have a set of n = 4 vectors in R3 (i.e., Rm with m = 3). Since n = 4 > m = 3 (i.e., n > m), then S is linearly dependent. This means that w can be expressed as a linear combination of {u, v, h} or h as a linear combination of {u, v, w} as follows: 15 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence u 0 u   1 4 0 u  1 4   1 4 0  0 2 3   1 2 3 v v  u   0 2 3 vu  =     0 8 12 w  u   0 0 0 w  u  4(v  u)   1 4 12 w        h 3 hu  0 0 3 hu 0 0   1 4 3 u  1 4 0   0 2 3 vu     0 0 0 w  4v  3u    hu  0 0 3 w + 4v − 3u = 0 or w = 3u − 4v , or 3u = w + 4v or 4v = 3u − w w = 3(−1, 4, 0) − 4(−1, 2, 3) = (1, 4, −12) § 1.7: Linear independence of basis functions: Let V be the vector space of all functions of a real variable t. Then the functions et and e2t in V are linearly independent. Proof: Suppose a and b are two real numbers such that a et + b e2t = 0, for all values of t. We need to show that a = 0 and b = 0. In order to do this, we divide through by et (which is never zero) and subtract to obtain b et = − a. In other words, the function b et must be independent of t, which only occurs when b = 0. It follows that a is also zero. Examples 1.28: consider the functions x2, x, and 1. Take the Wronskian: x2 W = 2x 2 x 1 1 0  2 0 0 Note that W is always non-zero, so these functions are independent everywhere. Consider, however, x2 and x: W= x2 x  x 2  2x 2   x 2 2x 1 Here W = 0 only when x = 0. Therefore x2 and x are independent except at x = 0. 2x 2  3 x 2 1 Consider 2x2 + 3, x2, and 1: W = 4x 2x 0  8x  8x  0 4 2 0 Here W is always zero, so these functions are always dependent. This is intuitively obvious, of course, since 2x2 + 3 = 2(x2) + 3(1) Note: Linear dependence or independence is a property of a set of vectors, not of an individual vector. That is, it has to do with the relationship between the vectors in the set. 16 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence 1 0 1 i = (1, 0) , j = (0, 1), f = (1, 1), In the matrix   The column vectors 0 1 1 are linearly dependent while the row vectors u = (1, 0, 1) and v = (0, 1, 1) are linearly independent. Also note: If S is a set of linearly dependent vectors, then at least one of the vectors in S can be written as a linear combination of the others. This does not necessarily mean that each vector in the set can be written as a linear combination of the others. For instance, Although S = {(1, 0), (3, 0), (0, 1)} is a set of linearly dependent vectors in R2, since (3, 0) =3 (1, 0) + 0 (0, 1). However, (0, 1) cannot be written as a linear combination of (1, 0) and (3, 0), but G = {(1, 3, 0) , (0, 0, 1)} has two linearly independent vectors in R3. Some remarks: m Theorem 1.9: Let S = {v1, v2, v3, ..., vm} be a set of m vectors in R and let A be the square matrix of vectors of S. Then S is linearly independent if and only if det A  0. 5 1) An example of a linear combination of vectors in R is provided by the vector sum (m = 3, n = 5) ; A = 2x + y + 3z, B = x + 2y − z where x = (1, 2, 3, 0, 4), y = (2, 1, 4, 1,−3), and z = (6, 0, 2, 2,−1). The vector 5 in R formed by this linear combination is A = 2x+ y+3z = 2(1, 2, 3, 0, 4) + (2, 1, 4, 1,−3) + 3(6, 0, 2, 2,−1) = (22, 5, 16, 7, 2). B = x + 2y2 − z = (1, 2, 3, 0, 4) + 2(2, 1, 4, 1,−3) − (6, 0, 2, 2,−1) = (−1, 4, 9, 0, −1). 2) Let u = (1, 4, 2, 3) , v = (7, 10, −4, −1) , w = (−2, 1, 5, −4), These are dependent vectors since a (1, 4, 2, 3) +b (7, 10, −4, −1) + c (−2, 1, 5, −4) = (0, 0, 0, 0) i.e. −3 (1, 4, 2, 3) +1 (7, 10, −4, −1) + 2 (−2, 1, 5, −4) a = −3 , b = 1 , c = 2 We can calculate the values of a, b, and c by solving the system of linear equations: a + 7b − 2c = 0, 4 a +10 b + c = 0, 2 a − 4 b + 5 c = 0, 3 a − b − 4 c = 0. 4 3) An example of linear dependence in R is provided by the vectors x =(1, 0, 2, 5), y = (2, 1, 2, 1), z = (3, 2, 1, 0), and t = (−1,−1,−1, 7), because t = 2x − 3y + z. 2(1, 0, 2, 5) −3 (2, 1, 2, 1) + (3, 2, 1, 0) = (−1,−1,−1, 7). 4 As example of linear independence in R is provided by the vectors x1 =(1, 0, 2, 5), x2 = (2, 1, 2, 1), x3 = (3, 2, 1, 0), and x4 = (−1,−1,−1, 8), a(1, 0, 2, 5) +b (2, 1, 2, 1) + c(3, 2, 1, 0) + d(−1,−1,−1, 8) = (0, 0, 0, 0). with a = b = c = d = 0. 3 Example 1.29: The vectors (1,1,0) , (1,0,1), (0,1,1) are independent in R . Since a (1,1,0) + b (1,0,1) + c (0,1,1) = (0, 0, 0) This is equivalent to the following set of linear equations: a + b = 0 , a + c = 0, b + c = 0. You can verify that the solution is a = b = c = 0. Hence, the vectors are independent. 17 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence 4 5 2 Example 1.30: We know that the set of vectors {(1, −2) , (ℓn 3 , ℓn 2) . ( e , e )}in R is 2 dependent without doing any computation. Any set of three (or more) vectors in R is dependent. 2 3 Example 1.31: R[x] is a vector space over the reals. The set 1 , x , x , x ,  is 2 n independent. For if ao + a1 x + a2 x +  + an x = 0, it follows that ai = 0 for all  i. Example 1.32: Determine whether the set of vectors {(1, 0, 1, 2), (1, 2, 2,1), (0, 1, 2,1)} 4 is Linear dependence or Linear Independence in R . Solution: In order to determine if the three vectors in u, v , w  R4, are linearly dependent or independent, where u = (1, 4, 2, 3) , v = (7, 10, −4, −1) , w = (−2, 1, 5, −4) form the matrix, u  2 3  1 4 2 3 u  1 4  7 10 4 1 v   0 18 18 22 v  7u        2 1 5 4 w  0 9 9 1 w  2u  u  1 4 2 3 u 2 3 1 4  0 9 w  2u  9 1 w  2u   0 9 9 1     0 18 18 22 v  7u  0 0 0 20 v  3u  2w  It is clear that these vectors are linearly independent.(Since the last term in the third row is not zero −20  0. 3 In order to determine if the Four vectors in  , ,  ,  R ,  = (1, 7, −2) ,  = (4, 10, 1) ,  = (2, −4, 5) and  = (3, −1, −4) are linearly dependent 3 m or independent. We have a set of n = 4 vectors in R (i.e., R with m = 3). Since n = 4 > m = 3 (i.e., n > m), then S is linearly dependent. In order to find one of them expressed as a linear combination of the other vectors, form the matrix α  1 7 2 α  1 7 2  4 10 1 β  0 18 9 β  4α       2 4 5 γ  0 18 9 γ  2α       3 1 4 δ  0 22 2 δ  3  α α 1 7 2  1 7 2  0 2 1 (β  4α)/(-9)  0 1 0.5 (β  4α)/(  18)     0 0 γ+2    0 0 4.5 (11β  17α  9γ)/18  0     2α  β  δ 0 0 11 1 (δ  3) / (2)  0 0  It is clear that these vectors are linearly dependent and 0 = 2 −  +    = 2 +  or  =  − 2 or 2 =  −   = (4, 10, 1) = 2 (1, 7, −2) + (2, −4, 5) = (2, 14, −4) + (2, −4, 5) In order to determine if the three vectors in R4, u = (1, 4, 2, 3) , v = (7, 10, −4, 1) , w = (−2, 1, 5, 4) are linearly dependent or independent, form the matrix 18 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence u  1 4 2 3 u  1 4 2 3   7 10 4 1 v   0 9 9 2 w  2u       2 1 5 4 w  0 0 0 16 v  3u  2w  It is clear that these vectors are linearly independent. Now, let us see the same required for the column vectors of the above matrix. Example 1.33: Determine whether the set of vectors u = (1, 0, 1, 2), v = (0, 1, 2, 1) and w = (1, 2, 2, 1) are linearly independent in R4 . u  1 0 1 2 u 1 0 1 2 u  1 0 1 2   0 1 2 1 v   0 1 2 1 v   0 1 2 1 v       0 2 1 1 w  u  0 0 3 3 w  u  v  1 2 2 1 w  It is clear that these vectors are linearly independent in R4. But the vectors A = (1, 0, 1) , B = (0, 1, 2), C = (1, 2, −3) , D = (2, 1, −1) in R3 are linearly dependent since n = 4 > m = 3. And A  1 0 1 A 1 0 1 A  1 0 1   0 1 2 B  0 1 2 B  0 1 2 B      1 2 2 C  0 2 1 C  A  0 0 3 C  A  2B        2 1 1 D 0 1 1 D  2A  0 0 0 D  A  C  B  Then D − A − C + B = 0  D = A + C − B and B = A + C − D − A and C = D − A + B. The question is whether you can find a, b , c  R, not all 0, such that a u + b v + c w = a (1, 0, 1, 2) + b (1, 2, 2, 1) +c (0, 1, 2, 1) = (0, 0, 0, 0) This amounts to solving the system 1 1 0 0  1 1 0  0  a     0 2 1  0 0 2 1 0      b    The augmented  1 2 2 0  1 2 2   0      c    2 1 1 0 2 1 1 0       However, we can see that row operations will never change the fourth column, so we can omit it. So the row reduction is  1 1 0  1 0 1   0 2 1  0 1 2    1 2 2 0 0 0       2 1 1  0 0 0  Remembering that there is a fourth all zero column that we haven't writing, this says a + c = 0 , b + 2c = 0. The parameterized solution is 2a = −2c = b = 2t a = t, b = 2t, c = − t. So, for example, we can get a nonzero solution by setting c = −t = −1. Then a = 1 and b = −2 . In the row reduced echelon matrix, you can see that we didn't get a copy of the identity matrix. If this had happened, 19 Advanced Engineering Mathematics Chapter 1- Linear Dependence and Independence 1 0 0  0 1 0   0 0 1    0 0 0 says a = 0, b = 0, and c = 0, and the vectors would have been independent. You can just do the algorithm if you wish, but it's always better to understand where it's coming from. 3 m We have a set of n = 4 vectors in R (i.e., R with m = 3). Since n = 4 > m = 3 (i.e., n > m), then S is linearly dependent. 20 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. Chapter 2 SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS § 2.1 A second-order linear differential equation is one of the form d2y dy  q(x)y  f (x) ..........................................................................(1) dx dx 2 or, in alternative notation, y" + p(x)y' + q(x) y = f(x) If f(x) is identically 0, then (1) reduces to  p(x) d2y dy  q(x)y  0 ..................................................(2) dx dx 2 which is called the second-order linear homogeneous differential equation. In order to discuss the solutions to a second-order linear homogeneous differential equation, it will be useful to introduce some terminology. Two functions f and g are said to be linearly dependent if one is a constant multiple of the other. If neither is a constant multiple of the other, then they are called linearly independent. Thus, g(x) = 3 sin x and h(x) = 5 sin x are linearly dependent, but 2 g(x) = x and h(x) = x are linearly independent. The following theorem is central to the study of second-order linear homogeneous differential equations.  p(x) d2y dy  q(x)y  0 dx dx 2 where the functions p(x) and q(x) are continuous on a common open interval. Then there exist linearly independent solutions u(x) and v(x) to (2) on that interval. Furthermore, given any such pair of linearly independent solutions u(x) and v(x), a general solution of (2) on the interval is given by y(x) = A u(x) + B v(x)............ (3) That is, every solution of (2) on the interval can be obtained from (3) by choosing appropriate values of the constants A and B; conversely, (3) is a solution of (2) for all choices of A and B. Theorem 2.1 Consider the homogeneous equation  p(x) Second-order linear homogeneous differential equations with Constant coefficients We will restrict our attention to second-order linear homogeneous equations of the d2y dy  q y  0 ........................................................................................(4) dx dx where p and q are constants. Since the constant functions p(x) = p and q(x) = q are continuous on (− , ), it follows from Theorem 2.1 that to determine a general solution to (4) we need only find two linearly independent solutions u(x) and v(x) on (− , ) . The general solution will then be given by y(x) = A u(x) + B v(x), where A and B are arbitrary constants. mx We will start by looking for solutions to (4) of the form y = e . This is motivated by the fact that the first and second derivatives of this function are multiples of y, suggesting that a solution of (4) might result by choosing m appropriately. To find such an m, we substitute mx mx 2 mx y = e , y' = me , y" = m e .................................................(5) form 2 p 21 Advanced Engineering Mathematics 2 Chapter 2- Second-Order Linear D.E. mx into (4) to obtain (m + p m + q) e = 0 ................................................................. (6) 2 which is satisfied if and only if m + p m + q = 0 ................................................... (7) mx since e  0 for every x. Equation (7), which is called the auxiliary equation for (4), can be obtained from (4) d2y dy 1 0 by m (= m ), and y by 1 (= m ). The solutions, m1 dx dx 2 and m2, of the auxiliary equation can be obtained by factoring or by the quadratic formula. These solutions are by replacing 2 by m ,  p  p 2  4q  p  p 2  4q , ............................(8) m2    2 2 2 Depending on whether p − 4q is positive, zero, or negative, these roots will be distinct and real, equal and real, or complex conjugates. We will consider each of these cases separately. Distinct Real Roots x x If  and  are distinct real roots, then (4) has the two solutions y1 = e , y2 = e x x Neither of the functions e and e is a constant multiple of the other, so the general x x solution of (4) in this case is y = A e + B e ......................................................(9) m1    Example 2.1: Find the general solution of y" + y' − 6 y = 0 2 Solution: The auxiliary equation is m + m − 6 = 0 or equivalently (m − 2)(m +3) = 0 so its roots are  = 2,  = −3. Thus, from (9) the general solution of the differential 2x −3x equation is y = A e + B e , where A and B are arbitrary constants. Equal Real Roots If m1 and m2 are equal real roots, say m1 = m2 (= ), then the auxiliary equation mx yields only one solution of (4): u(x) = e mx However, substitution of the function v(x) = x e into the differential equation shows that it is also a solution. The functions v(x) and u(x) are linearly independent because v(x)/u(x) = x is not a constant, so in this case a basis for the solution space of mx mx (1) is formed by the functions e and x e , with the corresponding general mx solution y = A u(x) + B v(x) = e [ A + B x ]........................................................(10) Example 2.2: Find the general solution of y" − 4 y' + 4 y = 0 2 2 Solution: The auxiliary equation is m − 4m + 4 = 0 or equivalently (m − 2) = 0 so m = 2 is the only root. Thus, the general solution of the differential equation 2x is y = e ( A + B x). Complex Roots If the auxiliary equation has complex roots m1 =  + i and m2 =  − i, then both x x u(x) = e cos x and v(x) = e sin x are linearly independent solutions of (4) and x y = e (A cos  x + B sin x)..................................................................................(12) is the general solution. 22 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. Example 2.3: Find the general solution of y" + y' + y = 0 2 Solution: The auxiliary equation is m + m + 1 = 0 has roots 1  1  4 1  1  4 1 3 1 3 m1 =     i , m2 = i 2 2 2 2 2 2 Thus, the general solution of the differential equation is  x2  3 3   A cos 2 x  Bsin 2 x  .   § 2.2 Linear Equations with Constant Coefficients We have already remarked that the problem of finding the general solution of the complete equation reduces to that of finding n linearly independent particular solutions of the reduced equation, and any (preferably the simplest possible) particular solution of the complete equation. This can be done in only a very few cases. However, if the coefficients c1, c2,, cn, are constants, a complete theory and several methods of solution are available. Because of this fact, and also because linear equations with constant coefficients are of great practical importance as well as theoretical interest we shall devote most of this section to their consideration. Case I : Roots, Real and Different. If  and  are the roots: Let us consider the equation with constant coefficients y"  ( + ) y' +   y = 0 (D2  ( + )D +   )y = 0 (D  )(D  )y = 0 (D  )[(D  )y] = 0 Let y = e mx , then m2  ( + ) m +   = 0  (m  )(m  ) = 0, If   , then we have two different solutions ex and ex , the general solution y = A ex + B ex. Case II : Roots, Real and Equal. If  = , then we have y"  2 y' + 2 y = 0 (D2  2 D + 2 )y = 0 (D  )2 y = 0, we have ex as solution , the second solution u ex, where u is a function of x, to determine it, applying (D  )2 u ex to it and equate to 0. Since D u ex = a ex u + ex u'  (D  ) ex u = ex u', then (D  ) ex u' =  ex u + ex u"   ex u = ex u" , and (D  )2 ex u = ex u" = 0, u" = 0  u' = A , u = A + B x. Then y = (A + B x) ex. Also from the fact that (D  )2 ex u = ex u", we can write (D  )2 ex u = ex D2 u  In general F(D  ) ex u = (D  )n ex u = ex Dn u = ex F(D) u Case III : Roots Imaginary. It must appear in conjugate pairs i.e. if the roots are  ± i, then the solution will be y = C1 e( + i)x + C2 e(  i)x = ex [C1 eix + C2 eix] = = ex [C1{cos x + i sin x} + C2 {cos x  i sin x}] = = ex [(C1 + C2)cos x + i (C1  C2) sin x] = ex [A cos x + B sin x]. y= e 23 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. Summary d2y dy  q(x)y  0 dx dx 2 2 auxiliary equation m + p m + q = 0 Equation  p(x) Case General Solution Distinct real roots  ,  of the auxiliary equation y = A ex + B ex, Equal real roots  =  (= ) of the auxiliary equation y = ex (A + B x), Complex roots  +i,  − i of the auxiliary equation y = ex[A cos x+ B sin x]. Ordinary Differential Equations - Second-Order Equations The simplest of the second-order equations is y" + p y' + q y = f(x) (p, q real), with the initial conditions y(xo) = yo , y'1(xo) = y'o or the boundary conditions y(xo) = yo , y(x1) = y1 . The general solution of the equation is given as follows. y(x) = yh(x) + yC where yH(x), is the solution of Homogenous equation yp the particular solution. § 2.3 The operator D and the fundamental laws of algebra. When a particular integral is not obvious by inspection, it is convenient to employ d certain methods involving the operator D, which stands for . This operator is also dx useful in establishing the form of the complementary function when the auxiliary equation has equal roots. 2 D will be used for d2 3 d3 , D for , and so on. dx 2 dx 3 d2y dy 2 The expression 2  5  2y may then be written 2 D y + 5 D y + 2y, 2 dx dx 2 or (2 D + 5 D + 2)y. We shall even write this in the factorized form (2D + l)(D +2)y, factorizing the expression in D as if it were an ordinary algebraic quantity. Is this justifiable ? The operations performed in ordinary algebra are based upon three laws: I. The Distributive Law: m (a +b) = m a +m b; II. The Commutative Law: a b = b a; m n m+n III. The Index Law: a . a =a . Now D satisfies the first and third of these laws, for m n m+n D (u + v) = D u + D v, and D . D u = D u (m and n positive integers). As for the second law, D(c u) =c (D u) is true if c is a constant, but not if c is a variable. m n n m Also D (D u) =D (D u) (m and n positive integers). Thus D satisfies the fundamental laws of algebra except in that it is not commutative with variables. In what follows we shall write n n−1 F(D)  ao D + a1 D + ... + an−l D + an , 24 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. where the a's are constants and n is a a positive integer. We are justified in factorizing this or performing any other operations depending on the fundamental laws of algebra. For an example of how the commutative law for operators ceases to hold when negative powers of D occur, see Ex. (iii). x x 1. F(D) e =e F(a): x x 2 ax 2 x 3 x 3 x Since D e =  e , D e =  e , D e =  e ,  and so on, x n n−1 x F(D) e = (ao D + a1D + · · · + an−1 D + an) e = n n−1 x x = (ao  + a1 + · · · + an−1  + an) e = e F(). x x 2. F(D){ e V} =e F(D + ) V, where V is any function of x. th By Leibniz's theorem for the n differential coefficient of a product, n x n x n−1 x n−2 x 2 D {e V} = (D e ) V + n(D e )(D V) + ½ n(n − l)(D e )(D V) + ... + x n + e (D V) = n x n−1 x n−2 x 2 x n =  e V + n  e D V + ½ n(n−1) e D V +  + e D V = x n n−1 n−2 2 n = e ( + n  D + ½ n(n−1) D +  +D ) V = x n = e (D + ) V. n−1 x x n−1 Similarly D {e V} = e (D + ) V and so on. Therefore x n n−1 x + ... + an−l D + an ) {e V} = F(D) {e V} = ( ao D + a1 D x n n−1 x = e { ao (D +) + a1 (D +) + ... + an−l (D +) + an } V = e F(D + ) V. 2 2 3. F(D ) cos  x = F( −  ) cos  x: 2 2 4 2 2 Since D cos  x = −  cos  x, D cos  x = (−  ) cos  x , and so on, 2 n 2n−2 F (D ) cos  x = ( ao D2 + a1 D + ... + an−l D2 + an ) cos  x n 2n−2 = ( ao D2 + a1 D + ... + an−l D2 + an ) cos  x = 2 n 2 n−1 2 2 = { ao (−  ) + a1 (−  ) + ... + an−l (−  ) + an } cos  x = F( − ) cos  x. 2 2 Similarly F ( D ) sin  x = F ( −  ) sin  x. § 2.4. Complementary Function when the auxiliary equation has equal roots. When the auxiliary equation has equal roots  and , it may be written 2 2 m − 2 m  +  = 0. d2y dy   2 y  0, 2 dx dx 2 2 2 i.e. (D −2  D +  ) y = 0,  (D − ) y =0  (9) x We have already found that y =A e is one solution. To find a more general one put x y = e V, where V is a function of x. 2 x x 2 x 2 (D − ) {e V} = e (D −  +  ) V = e D V. 2 x Thus equation (9) becomes D V = 0, i.e. V = A + B x, so that y = e (A + B x). p p Similarly the equation ( D −  ) y = 0, reduces to D V = 0, giving 2 p−1 V = (A1 + A2 x + A3 x +  + Ap−1 x ), and 25 The original differential equation will then be  2 Advanced Engineering Mathematics 2 x Chapter 2- Second-Order Linear D.E. p−1 y = e (A1 + A2 x + A3 x +  + Ap−1 x ). When there are several repeated roots, as in p q r (D − ) (D − ) (D − ) y = 0, (10) we note that as the operators are commutative we may rewrite the equation in the q r p form (D − ) (D − ) {(D − ) y} = 0, which is therefore satisfied by any solution of the simpler equation p (D − ) y = 0  (11) Similarly equation (10) is satisfied by any solution of q (D − ) y = 0 . (12) r or of (D − ) y = 0 . (13) The general solution of (10) is the sum of the general solutions of (11), (12), and (13), containing together (p + q + r) arbitrary constants. 4 2 2 2 Example 2.4 Solve (D − 8 D + 16) y = 0, i.e. (D − 4) y = 0. 2 2 The auxiliary equation is (m − 4) = 0, m = 2 (twice) or − 2 (twice). 2x −2x Thus by the rule the solution is y= (A+ B x) e + (E + F x) e . 2 2 2 2 Example 2.5: Solve (D + 1) y = 0. The auxiliary equation is (m + 1) =0, ix −ix m = i (twice) and −i (twice). Thus y = (A + B x) e + (E + F x) e , or better y = (P +Q x) cos x + (R + S x) sin x. Examples for solution: 4 3 2 6 4 2 (1) (D +2D +D ) y = 0. (2) (D + 3D + 3D + 1) y = 0, 4 3 2 5 3 2 (3) (D − 2D + 2D − 2D + l) y = 0. (4) (4D − 3D − D ) y = 0. 2 2 (5) Show that: F (D )(P cosh ax + Q sinh ax) = F (a )(P cosh ax + Q sinh ax). 4n ax 4n ax (6) Show that (D − a) (e sin p x) = p e : sin p x. x § 2.5. Symbolical methods of finding the Particular Integral when f(x) = e . The following methods are a development of the idea of treating the operator D as if it were an ordinary algebraic quantity. We shall proceed tentatively, at first performing any operations that seem plausible, and then, when a result has been obtained in this manner, verifying it by direct differentiation. We shall use the 1 notation f (x) to denote a particular integral of the equation F(D)y  f (x). F(D) x (i) If f(x) = e , F()  0 x F(D) e x =e F() suggests that, as long as F()  0 , 1 x may be a value of e F()  1 x  ex F() 1 x  e x , e . This suggestion is easily verified, for F(D)  e = F(D) F()  F()  (ii) If F() = 0, (D − ) must be a factor of F(D). Suppose that F(D) = (D − )p g(D), where g()  0. Then the result, F(D){ ex V} =ex F(D + ) V, suggests that the following may be true, if V is 1, 26 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E.  ex .1   ex  1 1 1 1 x e x x p x .1 = , e = e =   =  F(D) g() p! (D  ) p  g()   g()  Dp (D  )p g(D) 1 is the operator inverse to D, that is the adopting the very natural suggestion that D 1 integrates p times. Again the operator that integrates with respect to x, while p D result obtained in this tentative manner is easily verified, for   ex x p   ex x p   x x p  p pe   F(D)  (D ) g(D) = =    g(D) (D )      =  g( ) p! g( ) p!   g( ) p!         e x  p   e x  x p x  = g(D)  D    g(D)  .1  e   g( ) p! g( )      In working numerical examples it will not be necessary to repeat the verification of our tentative methods. 2 2x 2 2x Ex. (ii): (D − 2) y = 50 e . Ex. (i): (D + 3) y = 50 e , 1 i) The particular integral is 2 .50e 2x  50e2x 2  2e2x (D  3) (2  3) 2x −3x Adding the complementary function, we get y = 2e + (A + B x) e ii) If we substitute 2 for D in 1 (D  2)2 .50e2x , we get infinity. But using the other method, 1 1 1 .50e2x  50e2x .1  50e 2x . x 2  25x 2e 2x . 2 (D  2)2 D2 2 2x Adding the complementary function, we get y = 25 x e Examples for solution: Solve 2 3x (1) (D + 6 D + 25) y = l04 e , 2 3x (3) (D − 9) y = 54 e , 2 2 (5) (D −  ) y = a sinh  x, Answers: 2 2x + (A + B x) e 2 2 ax (2) (D + 2 D +  +  ) y = e , 3 x −x (4) (D − D) y = e + e , 3 2 −2x (6) (D + 4D + 4D) y = 8e . −x (2) y = e (A cos  x + B sin x) + (1) y=2e +e (A cos 4x+ B sin 4x). ax 2 2 + e /{(a + ) +  }. 3x −3x x −x (3) y = (A + 9x) e + B e . (4) y = A + (B + ½ x)e + (C + ½ x)e . (5) y = (A + ax/2) cosh  x+ B sinh  x, (6) y = A + (B + C x − 2x2) e−2x. 3x −3x 27 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. § 2.6. Particular Integral when f (x) = cos ax. 2 2 g (D ) cos ax = g( − a ) cos ax. 2 2 This suggests that we may obtain the particular integral by writing − a for D wherever it occurs. 2 Example 2.6: Solve (D + 3D + 2) y = cos 2x. 1 1 1 .cos 2x  .cos 2x  .cos 2x 3D  2 4  3D  2 D2  3D  2 1 3D  2 2 To get D in the denominator, try the effect of writing ,  3D  2 9D2  4 suggested by the usual method of dealing with surds. 3D  2 1 This gives cos 2x   (3Dcos 2x  2cos 2x) = 40 9D2  4 1 1 =  (6sin 2x  2cos 2x)  (3sin 2x  2cos 2x) . 40 20 3 2 Example 2.7: Solve (D + 6 D + 11D + 6) y = 2 sin 3x. 1 1 1 sin 3x  2sin 3x  2 sin 3x  3 2 D 24 9D 54 11D 6      D  6D  11D  6 D  24 D  24 1 D  24 sin 3x  sin 3x = (D  24)sin 3x = sin 3x   585 9  576 585 D 2  576 1 1 [ 3 cos 3x + 24 sin 3x] =  [cos 3x + 8 sin 3x]. = 585 195 We may now show, by direct differentiation, that the results obtained are correct. Examples for solution: Solve 2 (1) (D + 1) y = 10 sin 2x. (2) (D − 5D + 6) y = 100 sin 4x. (3) (D2+ 8D + 25) y = 48 cos x−16 sin x . (4) (D2 +2D+401) y = sin 20x+40 cos 20x. Answers: (1) y =2 sin 2x − 4 cos 2x + A e−x. (2) y = 4 cos 4x − 2 sin 4x + A e2x + B e3x. (3) y = 2 cos x+e−4x (A cos 3x + B sin 3x). (4) y = sin 20x+e−x (A cos 20x+ B sin 20x). m §2.7. Particular integral when f(x) = x , where m is a positive integer. 1 In this case the tentative method is to expand in a series of ascending powers F(D) of D. 1 Example 2.8: x 2 = 1 (1  1 D 2 )1 x 2 = 4 4 D2  4 1 (1  1 D 2  1 D 4  )x 2  1 (x 2  1 ). 4 4 16 4 2 Hence, adding the complementary function, the solution suggested for 2 2 (D + 4) y = x , is y = 1 (x 2  1 )  A cos 2x  Bsin 2x. 4 2 28 Advanced Engineering Mathematics Example 2.9:  Chapter 2- Second-Order Linear D.E. 1 1 1 1  3 x3 =   x = D2  4D  3 2 1 D 3  D  by partial fractions,  = 1 (1  D  D 2  D3  D4  )  1 (1  1 D  1 D2  1 D3  1 D 4  ) x 3 = 2 3 3 4 3 9  27 81 = 1  4 D  13 D2  40 D3  121 D  x 3  1 x 3  4 x 2  26 x  80 . 9 27 81 243 3 3 9 Adding the complementary function, the solution suggested for 2 3 (D − 4D + 3) y = x 27 is y  1 x 3  4 x 2  26 x  80  Ae x  Be3x . 3 3 9 27 Example 2.10: 1 1  1 1 1 2 1 2 96x 4  96 x 96 (x  ),    2 D2 (D2  4) D2  D2  4  D2 4 1  96 ( 1 x 4  1 x 2 )  2 x 4  6 x 2 . 4 4 12 2 2 2 Hence the solution of D ( D + 4) y = 96 x should be 4 2 y = 2x − 6x + A cos 2x + B sin 2x + E + F x. Alternative method . 1 1 96 1 .96x 2  . (1  1 D 2  1 D4 )x 2 = 4 16 D2 (D 2  4) D2 4 from Ex. (i), = (24D2  6  3 D2  )x 2  2x 4  6x 2 + 3. 2 This gives an extra term 3, which is, however, included in the complementary function ( in constant E). −1 Note that if D u denotes the simplest form of the integral of u, without any arbitrary −l −1 −l constant, D (D. l) = D . 0 = 0, while D (D . l) = D. x = 1, −l −l so that D (D . l)  D (D. l) . m −m n −m m n Similarly D (D . x )  D (D . x ), if m is greater than n. So when negative powers of D are concerned, the laws of algebra are not always obeyed. This explains why the two different methods adopted in Example 2.10: give different results. Examples for solution: Solve 3 (1) (D + l) y=x . 2 (3) (D − 6 D + 9) y = 54x + 18. 2 2 (5) (D − D − 2) y = 44 − 76x − 48x . Answers: 3 2 −x (1) y = x − 3x + 6x − 6 + A e . 3x (3) y = 6x + 6 + (A + B x) e . 2 −x 2x (5) y = 24x + 14x − 5 + A e + B e . (2) (D2 + 2D) y = 24x. 4 3 2 (4) (D − 6D + 9 D )y = 54x + l8. 3 2 2 (6) (D − D − 2D) y = 44 − 76x − 48x . 2 −2x (2) y = 6x − 6x + A + Be . 3 2 3x (4) y = x + 3x + Ex + F + (A + B x)e . 3 2 −x 2x (6) y= 8x + 7x − 5x + Ae + Be + C 29 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. § 2.8. Particular integrals in other simple cases: We shall now give some typical examples of the evaluation of particular integrals in simple cases which have not been dealt with in the preceding articles. The work is tentative, as before. For the sake of brevity, the verification is omitted, as it is very similar to the verifications already given. 2 Example 2.11: Solve (D + 4) y = sin 2x. 1 2 2 We cannot evaluate sin 2x by writing −2 for D , for this gives zero in the D2  4 denominator. 2ix But i sin 2x is the imaginary part of e , and 1 1 e2ix  e2ix .1, D2  4 (D  2i) 2  4 e 2ix   D D2 1 D 1 2ix 1  2ix 1 (1  ) .1  e .1  e (1   2 2    (1) 4iD  4i 4 i 4iD 4i D(D  4i)  1 x .1 = e2ix = − 1 ix( cos 2x + i sin 2x); = e2ix 4 4iD 4i hence, picking out the imaginary part, 1 sin 2x = − 1 x cos 2x. 2 4 D 4 Adding the complementary function, we get y = A cos 2x + B sin 2x − 1 x cos 2x. 2 4 3 2x Example 2.12: Solve (D − 5D + 6) y = x e . 1 1  3 2x 1  3  1  1 .x e = e2x     .x 3e2x =    .x = 2 2 D 3 D D 1 D        (D  5D  6)  1  = e2x    1  D  D 2  D3  D4    x 3 = e2x  1 x 4  x 3  3x 2  6x  6 . 4  D  Adding the complementary function, we get     3x y = Ae − e2x  1 x 4  x 3  3x 2  6x  B . 4 2x 2x including the term − 6e in B e . 2 3x Example 2.13: Solve (D − 6 D + 13) y = 8 e sin 2x. 1 1 .8e3x sin 2x = 8e3x .sin 2x = 2 2 (D  6D  13) {(D  3)  6(D  3)  13} 1 1 = 8e3x sin 2x = 8e3x ( 1 x cos 2x) .sin 2x = 8e3x 2 2 4 D 4 {D  6D  9  6D  18  13} 3x = −2 x e cos 2x. (see Ex. (i)) Adding the complementary function, we get 3x y = e (A cos 2x + B sin 2x − 2 x cos 2x). 30 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. These methods are sufficient to evaluate nearly all the particular integrals that the student is likely to meet. All other cases may be dealt with on the lines indicated in (33) and (34) of the miscellaneous examples at the end of this chapter. Examples for solution: Solve 2 2x (2) (D − l) y = (x + 3) e . (l) (D +1) y = 4cosx. 3 3 −x 2 −x (4) (D + 2 D +2)y = 2e sin x . (3) (D − 3D − 2) y = 540 x e . 2 2 5 x (5) (D + 1) y = 24 x cos x. (6) (D − D) y = l2e + 8 sin x −2x . 2 3x 3x (7) (D − 6D + 25) y = 2 e cos 4x + 8e (l − 2x) sin 4x. Answers: x 2x (1) y = A cos x + (B + 2x) sin x. 2) y =A e + (x + 2) e . (3) y = Ae2x + (B + Cx − 20 x2 − 20 x3 − 15x4 − 9x5)e−x. (4) y = {A sin x + (B− x) cos x}e−x. (5) y = (A + B x − x3) cos x + (E + F x + 3 x2) sin x. (6) y = A + (B + 3x)ex + C e−x +x2 + E cos x +(F + 2x) sin x. 2 3x (7) y = {A sin 4x + (B − x + x ) cos 4x}e . § 2.9. The Homogeneous Linear Equation: This is the name given to the form n n n−1 n−1 (ao x D + a1 x D + ... + an−l D + an ) y = f (x). t It reduces to the type considered before if we put x = e . 3 2 dy 3d y 2d y Example 2.14: Solve x  3x x  24x 2 . 3 2 dx dx t Put x = e , t = ℓog x, denote to dx d D dt 2 dy dy dy dy dt dy 2d y 1   Dy  x  D(D  1)y .  . = or x x dt dx dt dx dt dx dx 2 3 Similarly x d3 y dx 3  D(D  1)(D 2)y t D(D−1)(D−2)y +3D(D−1)y + Dy = 24 e2 2t (D3 − 3D2 + 2D + 3D2 − 3D + D) y = 24 e 2t D3 y = 24 e 3 Auxiliary equation is m = 0  m = 0 , three times (trios) 2 or C.F. = A + B t + C t . 1 2t 24 2t P.I. = 24 e  e  3e2t , 3 8 D 2 2t 2 2 giving y = A + B t + C t + 3e = A+ B ℓog x+ C (ℓog x) +3x 31 Advanced Engineering Mathematics Example 2.15: Solve x 3 d3 y dx 3  3x 2 Put x = et, t = ℓog x, denote to Chapter 2- Second-Order Linear D.E. d2y dx 2 x d D dt dy  x 3 og x. dx 2 dy dy dy dy dt dy 2d y 1   Dy  x  D(D  1)y .  . = or x x dt dx dt dx dt dx dx 2 3 3d y  D(D  1)(D 2)y Similarly x 3 dx D(D−1)(D−2)y +3D(D−1)y + Dy = t e3t 3 2 2 (D − 3D + 2D + 3D − 3D + D) y = t e3t D y = t e3t Auxiliary equation is m3 = 0  m = 0 , three times (trios) or C.F. = A + B t + C t2. 3 P.I. = 1 D3 t e3t  e3t 1 (D  3)3 t = e3t e3t 1 D (1  D)t = (1  ) 3 t = 27 27 3 e3t x3 (t  1)  (og x  1) . = 27 27 Complete solution is e3t x3 2 y=A+Bt+Ct + (t  1) = A+ B ℓog x+ C (ℓog x) + (og x  1) . 27 27 2 Example 2.16: Solve x 2 Solution: d2y dx 2 x dy  y  x. dx 2 dy 2d y x  y  x. (1) x 2 dx dx Put x = et, t = ℓog x, denote to d D dt 2 dy dy dy dy dy dt 2d y 1 x   Dy x  D(D  1)y . or   . = 2 x dt dx dt dx dt dx dx 3 3d y  D(D  1)(D 2)y . Similarly x 3 dx The equation (l) becomes t t t D(D−1)y + Dy + y = e  (D2 − D + D + 1) y = e  (D2 + 1) y = e Auxiliary equation is m2 + 1 = 0  m = i or C.F. = (A cos t + B sin t) . 32 Advanced Engineering Mathematics 1 1 t 1 t P.I. = et  e  e . 2  1 1 2 D 1 Chapter 2- Second-Order Linear D.E. 1 The complete solution is y = (A cos t + B sin t) + e t = 2 x y = A cos (ℓog x) + B sin (ℓog x) + = 2 Another method is indicated in (28)-(30) of the miscellaneous examples at the end of this chapter. The equation n n n−1 n−1 ao( + x) D y + a1 ( + x) D y + ... + an−l( + x) D y+ an y = f (x). can be reduced to the homogeneous linear form by putting z =  + x, giving dy dy dz dy   Dy= dx dz dx dz Examples for solution: Solve, 2 dy 2d y (1) x  2x  2y  4x 3 2 dx (3) x (4) x dx 3 3d y dx 3 4 4d y dx 4  3x 2  2x d2y dx 2 3 3d y dx x 2 dy 2d y (2) x  9x  25y  50 2 dx dx dy  8y  65 cos (og x) dx 2 dy 2d y x x  y  og x. 3 2 dx dx 2 dy 2d y (5) (1  2x)  6(1  2x)  16y  8(1  2x) 2 dx dx 2 2 dy 2d y (6) (1  x)  (1  x)  y  4cos og(1  x). 2 dx dx Answers: 2 3 (1) y = Ax + Bx + 2x . −4 −4 (2) y = 2 + A x cos (3 ℓog x) + B x sin (3 ℓog x). −2 (3) y = 8 cos (ℓog x) − sin (ℓog x) + A x + B x cos ( 3 ℓog x − ). 2 3 (4) y = 4 + ℓog x + A x + B x ℓog x + C x (ℓog x) + D x (ℓog x) 2 2 (5) y = (l +2x) [{ℓog (1 + 2x)} +A ℓog (1 + 2x) + B]. (6) y=A cos {ℓog (1 + x) − a}+2 ] ℓog (1 + x) sin ℓog (1 + x). § 2.10. Simultaneous linear equations with constant coefficients: The method will be illustrated by an example. We have two dependent variables, y and z, and one independent variable x. 33 Advanced Engineering Mathematics D stands for Consider Chapter 2- Second-Order Linear D.E. d , as before. dx −x (5D + 4) y − (2D + 1) z = e ,  (1) −x ( D + 8) y − 3z =5 e (2) Eliminate z, as in simultaneous linear equations of elementary algebra. To do this we multiply equation (1) by 3 and operate on equation (2) by (2D + 1 ) Subtracting the results, we get −x −x {3 (5D + 4) − (2D + l)(D +8)} y = 3e − (2D + 1) 5 e , 2 −x 2 −x i.e. ( −2D − 2D + 4) y = 8 e , or (D + D −2) y= − 4 e . −x x −2x Solving this in the usual way, we get y = 2e + A e + B e . The easiest way to get z in this particular example is to use equation (2), which does not involve any differential coefficients of z. Substituting for y in (2), we get −x x −2x −x l4 e + 9 A e + 6 B e − 3z = 5 e , −x x −2x so that z = 3 e + 3A e + 2B e . However, when the equations do not permit of such a simple method of finding z, we may eliminate y. In our case this gives −x −x { −(D + 8)(2D + 1) + 3(5D +4)} y = (D +8) e − (5D +4)5 e , 2 −x −x x −2x i.e. ( − 2D − 2D + 4) z = 12 e , giving z = 3 e + E e + F e . To find the relation between the four constants A, B, E, and F, substitute in either of the original equations, say (2). This gives −x x −2x −x x −2x −x (D + 8) (2 e + A e + B e ) − 3 (3 e + E e + F e ) = 5 e , x −2x i.e. (9A − 3E) e + (6B −3F) e = 0, whence 9A − 3E = 6B −3F = 0 i.e. E = 3A and F = 2B, −x x −2x −x x −2x so z = 3 e + E e + F e = 3 e + 3A e + 2B e , as before. Examples for solution: Solve, (1) D y − z = 0 , (D − l) y −(D + l)z = 0. (2) (D − 17) y + (2D − 8) z = 0, (13D − 53) y −2z = 0 . 2 2 2 2 (3) (2D − D + 9) y − (D + D + 3) z = 0, (2D + D +7) y − (D − D + 5) z = 0. x x (4) (D + l)y = z + e , (D + l) z = y + e 2 2 (5) (D + 5)y − 4z = −36 cos 7x, y + D z = 99 cos 7x . −x (6) (2D + l) y + (D + 32)z = 91e +147 sin 2x + 135 cos 2x, −x y − (D − 8) z = 29 e + 47 sin 2x + 23 cos 2x. Answers: (1) y = A cos (x − ); z = −A sin (x − ). 5x 3x 5x 3x (2) y = A e + B e ; z = 6 A e − 7 B e x x (3) y = A e + B cos (2x − ); z = 2 A e − B cos (2x −). x −2x x −2x (4) y = e +A + Be ; z = e + A − Be . (5) y = A cos (x − ) + 4 B cos (2x − ) + cos 7x; z = A cos (x − ) + B cos (2x − ) − 2 cos 7x. 34 Advanced Engineering Mathematics 3x 4x Chapter 2- Second-Order Linear D.E. −x (6) y = − 5Ae − 4B e + 2e + cos 2x − sin 2x ; 3x 4x −x z = A e + B e + 3e + 4 cos 2x + 5 sin 2x. MISCELLANEOUS EXAMPLES ON CHAPTER 2. Solve: 3 3x (1) (D − 1) y = 16 e 2 −5x/2 . (2) (4D + 12D + 9) y = l44x e 4 3 2 −2x (3) (D + 6D + 11D + 6D) y = 20e sin x. 3 x (4) (D − D2 + 4D − 4) y = 68 e sin 2x. 4 2 3x (5) (D − 6D −8D −3) y = 256 (x + l) e . 4 2 (6) (D − 8D − 9) y =50 sinh 2x. 4 2 (7) (D − 2D + 1) y = 40 cosh x. 2 2 2x (8) (D − 2) y = 8(x + e + sin2x). 2 2 2x (9) (D − 2) y = 8 x e sin2x . 2 2 3 (10) (D + 1) y = 3 cos x+2 sin x . 4 2 (11) (D + 10D + 9) y= 96 sin 2x cos x . n x (12) (D − n) y = n , where n is a positive integer . (l3) d2y dx 2 d2y  1 dy 12og x .  x dx x2 2 dy d3y 6y (15) (l4)   10.  . dx 2 x dx dx 3 x 3 d2y dy (16) (x  1) 2  (x  1)  (2x  3)(2x  4) . dx dx 2 d2y dy d2x dx (l7)  4  4y  25x  16e t .  4  4x  y, dt dt dt 2 dt 2 dx dy dz  2y ;  2z. ;  2x ; (18) dt dt dt dx dy (19) t  y  0; t  x  0 . dt dt 2 dy 2d y (20) t  t  2x  0 , dt dt 2 2 dx 2d x t  2y  0 . t dt dt 2 (21) Prove that if V is a function of x and F(D) has its usual meaning, n n n−1 (i) D [x V] = x D V + n D V; (ii) F(D) [x V] = x F(D) V + F'(D) V; 1 1 F'(D) (iii) V V. x V  x F(D) F(D) [F(D)]2 n n n−l n (iv) g(D) [x V] = x g(D) V + n x g'(D) V +  + Cr x 1 . terms, where g(D) stands for F(D) 2x (22) Obtain the Particular Integrals of (i) (D −1) y = xe , 35 n−r (r) g (D) V+to (n + 1) Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 2 (ii) (D + 1) y = x cos x, by using the results (iii) and (iv) of the last example. Answers: 2 x 3x 3 −3x/2 (2) y = (A + B x+ 6 x ) e . (1) y = (A+ B x+ C x ) e + 2e . −3x −2x −x −2x (3) y = A e + B e + C e + E + 2e (sin x − 2 cos x). x x (4) y =Ae + B cos (2x− ) − 2e (4 sin 2x + cos 2x). 2 −x 2 3x (5) y = (A + B x + C x ) e + (E + x + 2x )e . (6) y = A sin (x − ) + B sinh (3x − ) − 2 sinh 2x. 2 (7) y = (A + B x + 5 x ) cosh x+ (E + F x) sinh x. 2 2 2x (8) y = 3 + 4 x + 2 x + (A + B x + 4x ) e + cos 2x. 2 2x (9) y = (A+ B x+ 3 sin 2x − 4x cos 2x − 2x sin 2x) e . (10) y = A cos (x − ) + 3 − 1 cos 2x − 3 x cos x + 1 sin 3x. 2 2 4 16 (11) y =A cos (x − ) + B cos (3x − ) − 3x cos x + x cos 3x. 2 n−1 nx x n (12) y = (Ao + A1 x +A2 x +  + An−1 x ) e + n /(ℓog n − n) . 3 −1 2 (13) y =A + B ℓog x + 2(ℓog x) . (14) y =A + B x + 5 x . 8 3 (15) y = A x + B cos ( 2 ℓog x − ). 2 2 (16) y = A + B ℓog (x + 1) +{ℓog (x + 1)} + x + 8 x. 3t −3t t (17) x = Ae + Be + E cos t + F sin t − e ; 3t −3t t y = Ae + 25e + (3E − 4F) cos t + (3F + 4E) sin t − e . 2t −t (18) x = A e + B e cos ( 3 t − ) ; 2t −t y = A e + B e cos ( 3 t −  + 2/3) ; 2t −t z = A e + B e cos ( 3 t −  + 4/3) −1 ; (19) x = A t + B t −1 (20) x = A t cos (ℓog t − ) + B t cos (ℓog t − ) ; −1 y = A t sin (ℓog t − ) − B t sin (ℓog t − ). 2x 2 2 (22) (i) (x − l) e ; (ii) ½ (x − 2x + l) sin x + ½ (x − l) cos x. § 2.11. Rules to find particular integral : (i) 1 ex  1 ex ,f ()  0. f (D) f () If f() = 0 , the above rule fails, then we apply If f() = 0 , then 1 e x  x 1 e x , f (D) f '( ) f()  0, f '()  0. If f() = 0 , f '() = 0 the above rule fails, then we apply If f() = 0 , f '() = 0, then (ii) 1 e x  x 2 1 e x , f (D) f ''( ) 1 x n  f (D) 1 , Expand [f(D)]−1 and then operate.   f (D) 36 y = −A t + B t −1 . Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 1 sin x  1 sin x, 2 f (D ) f (  2 ) 1 cos x  1 cos x, 2 f (D ) f (  2 ) 2 1 sin x  x 1 If f(− ) = 0 , then sin x, f( 2 )  0, 2 2 f (D ) f '(  ) 2 1 cos x  x 1 cos x, f( 2 )  0, If f(− ) = 0 , then 2 2 f (D ) f '(  ) (iii) (iv) (v) f '( 2 )  0. f '( 2 )  0. 1 ex .(x)  ex 1 (x). f (D) f (D  ) 1 (x)  e x ex (x)dx.  D  Solve Example 2.17: dx t t Put x = e , =e =x, dt d dt d 1 d so that D = = = ; dx dx dt x dt 3 3 2 2 (x D + 3x D + x D) y = 24 x 2 1 d 1 d 1  d d2  1 d  D =D   =  2  D  2    2  ;  x dt  x dt x dt x  dt dt   1  d d2  2  d d2  1  d d2  3 D = D    =  3    2  + 2 D    2  =  dt dt  x  dt dt  x  x 2  dt dt 2     2  d d 2  1  d 2 d3  1  d d 2 d3   =    2 3 2  3 ; +  = dt dt  x 3  dt dt 2  x 3  dt 2 dt 3  x 3  dt 2 thus the given differential equation reduces to d3 y 2t = 24e , dt 3 2 2t 2 2 giving y = A + B t + C t + 3e = A+ B ℓog x+ C (ℓog x) +3x Solution of the differential equation, consisting of two parts i.e. ( A u + B v + W). The first part is the solution of the differential equation whose R.H.S. is zero. A u + B v is known as complementary function C.F.. Second parts W is free from any arbitrary constant and is known as particular integral. Complete Solution = Complementary Function + Particular Integral or y = C.F. + P.I. 37 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. § 2.12. Method for finding the complementary function. d2y dy  Qy  F(x) 2 dx dx (l) In finding the complementary function, R.H.S. of the given equation is replaced by Let the given equation is zero. i.e. d2y dx 2 P (2) Let y = A e mx P dy  Qy  0 . dx be the complementary function of d2y dx 2 P dy  Qy  0  (1) dx dy d2y mx 2 Putting the values of y, and in (1) , then A e ( m + P m + Q ) = 0 2 dx dx 2 or m + P m + Q = 0. It is called Auxiliary equation. (3) Solve the auxiliary equation, there are three cases: Case I : Roots, Real and Different. If m and n are the roots, then the C.F. is mx nx y=Ae +Be . Case II : Roots, Real and Equal. If both roots are m = n, then C.F. is mx y = (A + B x)e . Case III : Roots Imaginary. If the roots are   i, then the solution will be ( + i)x (−i)x x y=Ae +Be or y = e [A' cos x + B' sin x], where A' = (A + B) and B' = i(A − B). d2y dy  15y  0 2 dx dx 2 Solution: {D − 8 D + l5}y = 0, 2 Auxiliary equation is m − 8 m + 15 = 0 or m = 3, 5 3x 5x C.F. = A e + B e x, Example 2.18: Solve d2y 8 dy  9y  0 dx dx 2 2 Solution: Given equation can be written as : (D − 6D + 9) y = 0 2 2 A.E. is m − 6 m + 9 = 0 or (m − 3) = 0 or m = 3, 3 3x Hence the required solution is y ={A + B x) e . Example 2.19: Solve 6 d2y dy  5y  0 dx dx 2 2 Solution: Here the auxiliary equation is m + 4m + 5 = 0 −2x Its root are −2 ± i. The complementary function is e (A cos x + B sin x). Example 2.20: Solve Example 2.2: Solve d2y dx 2 4 6 dy  9y  5e3x dx Solution: 2 −3x Auxiliary equation is m + 6 m + 9 = 0 or m = −3 twice C.F. = (A + B x) e 38 Advanced Engineering Mathematics P.I. = 1 D 2  6D  9 5.e3x = 5 Chapter 2- Second-Order Linear D.E. e3x (3)2  6(3)  9 .= 5 −3x The complete solution is y = (A + B x) e Example 2.22: Solve d2y dx 2  2a e3x . 36 e3x +5 . 36 dy  a 2 y  eax . dx Solution: 2 ax Auxiliary equation is m − 2a m + a2 = 0 or m = a twice C.F. = (A + B x) e 1 1 P.I. = eax = eax Since the denominator equal zero, 2 2 2 D  2aD  a (D  a) 1 differentiating the denominator P.I. = x eax Since the denominator equal 2(D  a) 2 zero, differentiating the denominator, again differentiating P.I. = x eax . 2 2 ax Complete solution is y = (A + B x) e + x eax . d2y 2 dy  12y  x dx dx 2 2 Solution: {D −7 D + l2}y = x, 2 Auxiliary equation is m − 7 m + 12 = 0 or m = 3, 4 3x 4x C.F. = A e + B e x, 2  1 1 1 1 1  7 D .x = .x = 1 D  x= P.I. = 12  12 1  7 D  D2 12  12 D 2  7D  12 Example 2.23: Solve = 7 12 12 1  1 1  7 D  D   x  (x  7 ) . 12 12  12 12  12 2 3x 4x Complete solution is y = A e + B e x + 2 1 (x  7 ) . 12 12 Example 2.24: Solve (D + 4) y = sin 3x. 2 Solution: (D + 4) y = sin 3x 2 Auxiliary equation is m + 4 = 0 or m =  2i  C.F. = A cos 2x + B sin 2x 1 1 sin 3x P. I. = =  sin 3x sin 3x = 5 (32 )  4 D2  4 1 Complete solution is y = A cos 2x + B sin 2x +  sin 3x 5 Example 2.25: Solve d2y dx 2  dy  y  cos 2x dx 39 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 2 Solution: Auxiliary equation is m + m + 1 = 0  m = x 2 1  3 1  i 3  2 2  A cos 3 x  Bcos 3 x  2 2   1 1 1 D3 P. I. = cos 2 x = cos 2 x = cos 2 x = cos 2 x = 2 2 2 D  3 D  D 1 (2 )  D  1 D 9 D3 D3 = cos 2 x = cos 2 x =  1 (D  3)cos 2x = 1 (2sin 2 x  3cos 2x) 13 13 2 2 D 9 (2 )  9 or C.F. = e x 2  A cos 3 x  Bcos 3 x  + 1 (2sin 2 x  3cos 2x) . 13 2 2   2 Example 2.26: Solve (D + 4) y = cos 2x. 2 Solution: (D + 4) y = cos 2x 2 Auxiliary equation is m + 4 = 0 or m =  2i C.F. = A cos 2x + B sin 2x 1 1 cos 2x = x ( 1 sin 2x) = x sin 2x P. I. = cos 2x = x 2 2 4 2 2D D 4 Complete solution is y = e Complete solution is y = A cos 2x + B sin 2x + x sin 2x . Example 2.27: Solve Solution: d2y dx 2 4 2 dy  5y  e2x sin x dx 2 Auxiliary equation is m − 2 m + 5 = 0  m= 2  16  1  2i 2 or C.F. = e x  A cos 2x  Bcos 2x  1 1 P. I. = e2x sinx = e2x sinx D2  2D  5 (D  2)2  2(D  2)  5 1 1 1 sinx = = e2x sinx = e2x sinx = e2x 1  2D  5 D2  4D  4  2D  4  5 D2  2D  5 1 D2 1 = e2x sinx = 1 e2x sinx = sinx = 1 e2x 2 2 2 D2 2D  4 D 4 D2 2x = 1 e2x sinx =  1 e2x (D  2)sinx = 1 e (2 sin x − cos x). 10 2 10 1  4 2x Complete solution is y = e x  A cos 2x  Bcos 2x  + 1 e (2 sin x − cos x) . Example 2.28: Solve 2 d y dx 2 10 4 dy  3y  2xe3x  3e x cos x dx Solution: 2 Auxiliary equation is m − 4 m + 3 = 0  m = 1 , 3 40 Advanced Engineering Mathematics x Chapter 2- Second-Order Linear D.E. 3x or C.F. = A e + B e . 1 1 P. I. = 2 xe3x + 2e x cos 2 x = D 2  4D  3 D2  4D  3 1 1 = 2e3x .x + 3e x .cos x 2 2 (D  3)  4(D  3)  3 (D  1)  4(D  1)  3 1 1 = 2e3x .x + 3e x .cos x = D2  6D  9  4D  12  3 D2  2D  1  4D  4  3 1 1 = 2e3x .x + 3e x .cos x = D2  2D D2  2D 1 1 = 2e3x .cos x = .x + 3e x 4  2D 2D(1  D / 2) 1 D 1 = e3x (1  )1.x − 3 e x .cos x = 2 D 2 2D D D2 2D 3x 1 =e (1   ).x − 3 e x .cos x = 2 D 2 4 4  D2 1 1 D 2D = e3x (    ).x − 3 e x .cos x = 2 D 2 4 44 2 x 1 3x x = e (   ) − 3 e x (2cos 2x  2sin 2x) . 2 2 4 16 The complete solution is x 3x 3x x y=Ae +Be + e ( 2 x 1 3 x  ) − e (cos 2x  sin 2x) 2 4 8 x  ) − 3 e x (cos 2x  sin 2x) 2 8  2 2 x 3x 3x x y=Ae +Ce + e ( 2 e3x 3x has been omitted from the P.I. , since B e is present in the C.F. The term 4 d2y dy Example 2.29: Solve  6  13y  8e3x sin 4x  2 x. dx dx 2 Solution: 2 Its auxiliary equation is m − 6 m + 13 = 0  m = 3  2i 3x or C.F. = e (A cos 2x + B sin 2x). 1 8e3x sin 4x  2 x  = P. I. = 2  D  6D  13  1 1 8e3x sin 4x  + e x og2  =  D2  6D  13   D2  6D  13  Replacing D by D+3 in first part and by ℓog 2 in second part we get 41 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 1 1 = P. I. = 8e3x sin 4x + e x og2 2 2 (D  3)  6(D  3)  13 (og2)  6(og2)  13 1 1 = 8e3x sin 4x + e x og2 D 2  6D  9  6D  18  13 (og2)2  6(og2)  13 1 1 = = 8e3x sin 4x + e x og2 D2  4 (og2)2  6(og2)  13 1 1 sin 4x + e x og2 = 8e3x 16  4 (og2) 2  6(og2)  13 2x 2 =  e3x sin 4x + . 2 3 (og2)  6(og2)  13 The complete solution is 2x 2 3x y = e (A cos 2x + B sin 2x)  e sin 4x + . 3 (og2) 2  6(og2)  13 3x Example 2.30: Solve d2y dx 2 2 dy  y  x sin x. dx Solution: 2 Auxiliary equation is m − 2 m + 1 = 0  m = 1 twice x or C.F. = (A + B x) e . 1 P. I. =  x sin x  = 2 D  2D  1 1 = Imaginary part of x  cos x  isin x  = D2  2D  1 1 x.eix = = Imaginary part of 2 D  2D  1 1 = Imaginary part of eix x (D  i)2  2(D  i)  1 1 = Imaginary part of eix x= D2  2iD  1  2D  2i  1 1 x = Imaginary part of eix 2 D  2(1  i)D  2i 1 ix 1  1 2    1 (1 i)D D .x = = Imaginary part of e 2i  2i  i = Imaginary part of (cos x  isin x) 1  (1  i)D .x 2 1 = Imaginary part of (i cos x  sin x)  x  1  i  = 1 x cos x  1 cos x  1 sin x 2 2 2 2 42 Advanced Engineering Mathematics The complete solution is x y = (A + B x) e + 1 x cos x  1 cos x  1 sin x . 2 Example 2.31: Solve 2 d y dx 2 2 2 Chapter 2- Second-Order Linear D.E. 2 dy  y  xe x sin x. dx Solution: 2 Auxiliary equation is m − 2 m + 1 = 0  m = 1 twice x or C.F. = (A + B x) e . 1 1  xe x sin x  =  xe x sin x  = e x 1  x sin x  = P. I. =  (D  1)2   D2  2D  1  D2 1 = e x   x cos  sin x  Note that 1/D =  D x x = e [ −x sin x − x cos x − cos x] = − e [ x sin x + 2 cos x]. x x The complete solution is y = (A + B x) e − e [ x sin x + 2 cos x]. § 2.13 General method of finding the particular integral of any function (x). 1 P.I. = (x) = y Da 1 or (D − a) (x) = (D − a) y  (x) = (D − a) y or (x) = D y − ay. Da dy i.e.  ay  (x) which is the linear differential equation. dx  adx  a dx Its solution is y e  = e  .(x)dx  =  eax .(x)dx  y = e  eax .(x)dx 1 ax Then y = (x) = e  eax .(x)dx Da −ax ye Example 2.32: Solve d2y dx 2 ax  9y  sec3x. Solution: 2 Auxiliary equation is m + 9 = 0  m =  3i or C.F. = (A cos 3x + B sin 3x) . 1 1 1 1 1  P. I. = sec3x =  sec3x  = sec3x  =    (D  3i)(D  3i) 6i  D  3i D  3i  D2  9 1 1  1 1  sec3x −   sec3x =   6i  D  3i  6i  D  3i  Now 1 cos3x  isin 3x sec3x  e3ix  e 3ix sec3xdx = e3ix  dx = e3ix  [1  i tan 3x]dx = D  3i cos3x = e3ix (x  i og cos3x) , 3 43 Advanced Engineering Mathematics Change i to −i, we have 1 sec3x = e3ix (x  i og cos3x ) , then 3 D  3i Chapter 2- Second-Order Linear D.E. P. I. = 1 [ e3ix (x  i og cos3x) − e3ix (x  i og cos3x ) ] = 6i 3ix 3 3ix 3  e3ix 1 e  e3ix  og 3x = x sin 3x + 1 cos 3x .ℓog cos 3x 3 18 3 9 2i 2 Hence complete solution is y = (A cos 3x + B sin 3x) + x sin 3x + 1 cos 3x .ℓog cos 3x. e = x Example 2.33: Solve x 2 Solution: 3 2 d y dx 2 9 x dy  y  x. dx 2 dy 2d y x x  y  x. (1) 2 dx dx t Put x = e , t = ℓog x, denote to d D dt 2 dy dy dy dy dt dy 2d y 1   Dy  x  D(D  1)y .  . = or x x dt dx dt dx dt dx dx 2 3 3d y  D(D  1)(D 2)y . Similarly x 3 dx The equation (l) becomes t 2 t 2 t D(D−1)y + Dy + y = e  (D − D + D + 1) y = e  (D + 1) y = e 2 Auxiliary equation is m + 1 = 0  m = i or C.F. = (A cos t + B sin t) . 1 1 t 1 t P.I. = et  e  e . 2 11 2 D 1 1 The complete solution is y = (A cos t + B sin t) + e t = 2 x y = A cos (ℓog x) + B sin (ℓog x) + = 2 3 2 dy 3d y 2d y Example 2.34: Solve x  3x x  x 3 og x. dx dx 3 dx 2 d t Put x = e , t = ℓog x, denote to D dt 2 dy dy dy dy dt dy 2d y 1   Dy  x  D(D  1)y .  . = or x x dt dx dt dx dt dx dx 2 44 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 3 3d y  D(D  1)(D 2)y Similarly x 3 dx D(D−1)(D−2)y +3D(D−1)y + Dy = t e3t 3 2 2 (D − 3D + 2D + 3D − 3D + D) y = t e3t D y = t e3t 3 Auxiliary equation is m = 0  m = 0 , three times (trio) 2 or C.F. = A + B t + C t . 3 P.I. = 1 D3 te 3t e3t D 3 (1  D)t = (1  ) t = e t =e 27 27 3 (D  3)3 3t 3t 1 1 e3t x3 (t  1)  (og x  1) . = 27 27 Complete solution is 3 e3t 2 x y=A+Bt+Ct + (t  1) = A+ B ℓog x+ C (ℓog x) + (og x  1) . 27 27 2 2 dy 2d y Example 2.35: Solve (1  x)  (1  x)  y  sin 2[ og(1  x)] 2 dx dx t Put 1 + x = e , t = ℓog (x + 1), denote to (1  x) (1  x) dy d2y  Dy  (1  x) 2  D(D  1)y 2 dx dx 3 3d y  D(D  1)(D 2)y Similarly (1  x) 3 d D , dt dx Putting these values in the given differential equation we get 2 D(D − 1)y + Dy + y = sin 2t,  (D −D + D + 1)y = sin 2t, 2 (D + 1)y = sin 2t 2 Auxiliary equation is m + 1 = 0  m =  i C.F. = (A cos t + B sin t) 1 1 1 P.I. = sin 2t  sin 2t   sin 2t 4  1 3 D2  1 1 The complete solution is y = (A cos t + B sin t)  sin 2t 3 1 = A cos ℓog (x + 1) + B sin ℓog (x + 1)  sin 2og  x  1 . 3 45 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. § 2.14 The Method of Undetermined Coefficients A particular solution Y of the nonhomogeneous nth order linear equation with constant coefficients n n−1 L[y] = (ao D + a1 D + ... + an−l D + an ) y = f (x). .................(1) can be obtained by the method of undetermined coefficients, provided that f(t) is of an appropriate form. While the method of undetermined coefficients is not as general as the method of variation of parameters, it is usually much easier to use when applicable. Just as for the second order linear equation, when the constant coefficient m m−1 ++ am, an linear differential operator L is applied to a polynomial ao t + a1t αt exponential function e , a sine function sin β t, or a cosine function cos β t, the result is a polynomial, an exponential function, or a linear combination of sine and cosine functions, respectively. Hence, if f(t) is a sum of polynomials, exponentials, sines, and cosines, or products of such functions, we can expect that it is possible to find g (t) by choosing a suitable combination of polynomials, exponentials, and so forth, multiplied by a number of undetermined constants. The constants are then determined so that Eq. (1) is satisfied. The main difference in using this method for higher order equations stems from the fact that roots of the characteristic polynomial equation may have multiplicity greater than 2. Consequently, terms proposed for the nonhomogeneous part of the solution may need to be multiplied by higher powers of t to make them different from terms in the solution of the corresponding homogeneous equation. The Method of Undetermined Coefficients Simple Form of the Method The method of undetermined coefficients is applicable only if f(x) and all of its derivatives can be written in terms of the same finite set of linearly independent functions, which we denote by {y1(x), y2(x),..., yn(x)}. The method is initiated by assuming a particular solution of the form yp(x) = A1 y1(x) + A2 y2 (x) + + An yn (x), where A1, A2,, An denote arbitrary multiplicative constants. These arbitrary constants are then evaluated by substituting the proposed solution into the given differential equation and equating the coefficients of like terms. th Case 1: f(x) = pn(x), an n degree polynomial in x. Assume a solution of the form n n−1 yp = (An x + An−1 x ++ A1 x + Ao) ...............................................(2) where Ai ( i = 0,1,2,, n) is a constant to be determined. ax Case 2: f(x) = k e where k and a are known constants. Assume a solution of the form ax yp = A e ................................ (3) where A is a constant to be determined. Case 3: f(x) = k1 sin x + k2 cos x where k1, k2, and  are known constants. Assume a solution of the form yp = A sin  x + B cos  x... (4) where A and B are constants to be determined. 46 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. Don’t Forget yp = A sin  x + B cos  x in its entirety is assumed for f(x) = k1 sin  x + k2 cos  x even when k1 or k2 is zero, because the derivatives of sines or cosines involve both sines and cosines. Generalizations If f(x) is the product of terms considered in Cases 1 through 3, take yp to be the product of the corresponding assumed solutions and algebraically combine arbitrary constants where possible. In particular, if ax f(x) = e pn(x) is the product of a polynomial with an exponential, assume x n n−1 yp = e (An x + An−1 x ++ A1 x + Ao)........................................ (5) x where Ai is as in Case 1. If, instead, f(x) = e pn(x) sin x is the product x of a polynomial, exponential, and sine term, or if f(x) = e pn(x) cos x is the product of a polynomial, exponential, and cosine term, then assume x n n−1 yp = e sin x (An x + An−1 x ++ A1 x + Ao) + x n n−1 + e cos x (Bn x + Bn−1 x ++ B1 x + Bo)..................(6) where Ai and Bi ( i = 0,1,2,, n) are constants which still must be determined. If f(x) is the sum (or difference) of terms already considered, then we take yp to be the sum (or difference) of the corresponding assumed solutions and algebraically combine arbitrary constants where possible. Modifications If any term of the assumed solution, disregarding multiplicative constants, is also a term of yh (the homogeneous solution), then the assumed solution must be modified m by multiplying it by x , where m is the smallest positive integer such that the product m of x with the assumed solution has no terms in common with yh. Limitations of the Method In general, if f(x) is not one of the types of functions considered above, or if the differential equation does not have constant coefficients, then the following method is preferred. Variation of parameters is another method for finding a particular solution th of the n order linear differential equation Example 2.36: Solve d2y dx 2 6 dy  9y  5e3x (1) dx Solution: 2 −3x Auxiliary equation is m + 6 m + 9 = 0 or m = −3 twice C.F. = (A + B x) e 3x To find a particular solution Y (t) of Eq. (1), we start by assuming that Y (x) = C e . 3x Thus our final assumption is that Y (t) = C e , where C is an undetermined coefficient. To find the correct value for C, we differentiate Y (x) two times, substitute for y and its derivatives in Eq. (1), and collect terms in the resulting 3x 3x equation. In this way we obtain [9C +(6).(3)C + (9)C] e  5 e 5 e3x . i.e. 36 C = 5  C = . P.I. = 5 36 36 47 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. −3x The complete solution is y = (A + B x) e +5 Example 2.37: d2y e3x . 36 dy  12y  x .................................(1) dx dx 2 Solution: (D − 7 D + l2) y = x, 2 Auxiliary equation is m − 7 m + 12 = 0 or m = 3, 4 3x 4x C.F. = A e + B e x, To find a particular solution Y (t) of Eq. (1), we start by assuming that Y (x) =  + x, where  and  are undetermined coefficients. To find the correct value for  and , we differentiate Y (x) two times, substitute for y and its derivatives in Eq. (1), and collect terms in the resulting equation. In this way we obtain [−7  + 12  + 12  x]  x must be true for all x, the constant terms must sum to zero or 12 − 7  = 0 , 12  = 1. i.e.  = 1/12 and 12 = 7  =7/12   = 7/144 Similarly, all of the terms involving the polynomial x must balance, yielding  = 7 , Solve 2 7 144 and  = 1 . 12 1 (x  7 ) . 12 12 1 3x 4x (x  7 ) . Complete solution is y = A e + B e x + 12 12 d3y d2y dy Example 2.38: Find the general solution of 3  3  y  4e x ......(1) dx dx 3 dx 2 3 2 3 Auxiliary equation is m − 3 m + 3m − 1= 0 or (m − 1) = 0 , m = 1 trio (three times) x 2 x x C.F. = A e + B x e + C x e , ...(2) x To find a particular solution Y (x) of Eq. (1), we start by assuming that Y (x) = K e . x x 2 x However, since e , x e , and x e are all solutions of the homogeneous equation, we 3 must multiply this initial choice by x . Thus our final assumption is that 3 x Y (x) = Kx e , where K is an undetermined coefficient. To find the correct value for K, we differentiate Y (t) three times, substitute for y and its derivatives in Eq. (1), and x x collect terms in the resulting equation. In this way we obtain 6 K e  4 e . Thus 3 x K = 2 and the particular solution is Y (x) = 2 x e . (3) Therefore, the particular solution is P.I. = 3 3 The general solution of Eq. (1) is the sum of C.F. from Eq. (2) and Y (t) from Eq. 3 x x x 2 x (3). y = A e + B x e + C x e + 2 x e , 3 d2y dy  y  x  sin x (1) dx dx 2 2 2 Auxiliary equation is m − 2 m + 1= 0 or (m − 1) = 0 , m = 1 twice x x C.F. = A e + B x e ...(2) Example 2.39: Find the general solution of 48 2 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. From the form of the right side of (1), we guess the particular solution yp(x) =  x +  +  sin x +  cos x, (3) Therefore, y'p(x) =  +  cos x −  sin x, ...(4) and y"p(x) = −  sin x −  cos x, ...(5) y"p(x) − 2 y'p(x) + yp(x) = −  sin x −  cos x −2( +  cos x −  sin x) + + ( x +  +  sin x +  cos x)  x + sin x... (6) −  sin x −  cos x −2( +  cos x −  sin x) + ( x +  +  sin x +  cos x)  x + sin x i.e.  − 2 − 2 cos x +2  sin x +  x  x + sin x coefficient of x :  = 1, coefficient of sin x: 2 = 1  = ½ coefficient of cos x: −2 = 0   = 0 , free term: −2 +  = 0   = 2 = 2, then Since (6) must be true for all x, the constant terms must sum to zero or −2 = 0. Similarly, all of the terms involving the polynomial x must balance, yielding  =1 and  = 2 = 2. Turning to the trigonometric terms, the coefficients of sin x and cos x give 2 = 1 and −2 = 0, respectively. Therefore, the particular solution is yp(x) = x + 2 + + ½ cos x ....(7) x x and the general solution is y = A e + B x e + x + 2 + + ½ cos x Why does this technique work? The reason lies in the set of functions that we have allowed to be included in f(x). They enjoy the remarkable property that derivatives of their sums and products yield sums and products that are also constants, polynomials, exponentials, sines, and cosines. Because a linear combination of derivatives such as a y"p  b y p'  c y p must equal f(x), it seems reasonable to assume that y p has the same form as f(x). The following examples show that our conjecture is correct. Example 2.40: Find the general solution of d2y dy  4y  2e2x  4x  12. ............................................................................ (1) dx dx 2 2 2 Auxiliary equation is m − 4 m + 4 = 0 or (m − 2) = 0 , m = 2 twice 2x 2x C.F. = A e + B x e , ......(2) To find a particular solution Y (x) of Eq. (1), we start by assuming that 2 2x Y (x) =  x e +  x + , where  ,  and  are undetermined coefficients. To find the correct values for ,  and  we differentiate Y (t) two times, substitute for y and its derivatives in Eq. (1), and collect terms in the resulting equation. In this way we obtain 2 2x 2x 2 2x 2 Y (x) =  x e +  x + , Y' (x) = 2e [x + x] + , Y" (x) = 2e [ 2x + 4 x +1 ] 2x 2 2x 2 2 2x i.e. 2e [ 2x + 4 x +1 ]−4[2e [x + x] + ] +4 [ x e +  x + ]  2x  2 e + 4 x −12, 2x 2 2 2 2x e [4 x −8x +4 x +8 x−4x + 2] + 4x + 4  2 e + 4 x −12 2x 2 2 2 2x e [4 x −8x +4 x +8 x−4x + 2] + 4x + 4  2 e + 4 x −12 2x 2x e [4x + 2] + 4x + 4  2 e + 4 x −12 Then, if we take  = 1,  = 1,  = −3 4 49 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. 2x 2 2x The particular solution is Y (x) = x e + x e + x − 3  (3) 2 2x i.e. Y (x) = x e + x − 3  (4) 2x 2x The term x e has been omitted from the P.I. , since B x e is present in the C.F. Example 2.41: Let us find the particular solution to d2y dy  2y  xe x . .....(1) 2 dx dx by the method of undetermined coefficients. From the form of the right side of (1), we guess the particular solution x x Y(x)  Yp(x) =  e +  x e ............................................(2) x x x x x x x x Y(x) =  e +  x e , Y'(x) =  e +  x e +  e , and Y"(x)=  e +  x e + 2 e . x x Substituting into (1), we find that 3 e = x e .....................................................(3) Clearly we cannot choose a constant  such that (2) is satisfied. What went wrong? To understand why, let us find the homogeneous or complementary solution to (1); −2x x 2 it is yH(x) = A e + B e , Since its auxiliary equation is m + m − 2 = 0  (m − 1)(m + 2) = 0 x Therefore, one of the assumed particular solutions,  e , is also a homogeneous solution and cannot possibly give a nonzero left side when substituted into the differential equation. Consequently, it would appear that the method of undetermined coefficients does not work when one of the terms on the right side is also a homogeneous solution. Before we give up, let us recall that we had a similar situation in the case of linear homogeneous second-order ordinary differential equations when the roots from the auxiliary equation were equal. There we found one of the homogeneous solution was We eventually found that the second solution was Could such a solution work here? Let us try. We begin by modifying (3.4.16) by multiplying it by x. Thus, our new guess for the x 2 x particular solution reads yp(x) =  x e +  x e . x x 2 x x Then y'p(x) =  x e +  e +  x e + 2 xe , x x 2 x x x and y"p(x) =  x e + 2 e +  x e + 4 xe + 2 e Substituting into (1) gives  y"p  y'p  2y p =  x e + 2 e +  x e + 4 x e + 2 e + x e +  e +  x e + x x x x 2 x 2 x x x x x x 2 x + 2 x e − 2 x e − 2 x e  x e . 2 2 2 i.e.  x + 2 +  x + 4 x + 2 + x +  +  x + 2 x − 2 x − 2 x  x. 2 2 2 i.e.  x + 2 +  x + 4 x + 2 + x +  +  x + 2 x − 2 x − 2 x  x 2 x [ +  − 2 ] + x[ + 4 +  + 2 − 2  ]+[2 + 2 + ] x, 6  x + 3  + 2  x i.e.  = 1/6 , 3  + 1/3 = 0   = − 1/9 and  = 1/6 x Grouping together terms that vary as x e , we find that 6  =1. Similarly, terms that x vary as e yield 2 + 3 = 0. Therefore  = − 1/9 and  = 1/6 x 2 x i.e. yp(x) =  1 x e + 1 x e , 9 6 50 Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. −2x so that the general solution is y = yH(x) + yp(x) = A e x x 2 x + B e 1 x e + 1 x e . 9 6 In summary, the method of finding particular solutions to higher-order ordinary differential equations by the method of undetermined coefficients is as follows:  Step 1: Find the homogeneous solution to the differential equation.  Step 2: Make an initial guess at the particular solution. The form of yp(x) is a linear combination of all linearly independent functions that are generated by repeated differentiations of f(x).  Step 3: If any of the terms in yp(x) given in Step 2 duplicate any of the homogeneous solutions, then that particular term in yp(x) must be multiplied n by x , where n is the smallest positive integer that eliminates the duplication. Example 2.42: d2y Let us apply the method of undetermined coefficients to solve dy  sin x  e3x cos5x. .(1) dx 2 dx We begin by first finding the solution to the homogeneous version of (1) 2 2 Auxiliary equation is m + 1 = 0 or (m + 1) = 0 , m =  i. C.F. = A cos x + B sin x , ............(2) To find the particular solution we examine the right side of (1) or 3x f(x) = sin x − e cos 5x. ................(3) Taking a few derivatives of f(x), we find that 3x 3x f '(x) = cos x − 3e cos 5x + 5e sin 5x, ..........(4) 3x 3x 3x f"(x)= −sin x − 9 e cos 5x + 30 e sin 5x + 25e cos 5x, .........(5) and so forth. Therefore, our guess at the particular solution is 3x 3x yp(x) = C x sin x +  x cos x + E e cos 5x + F e sin 5x,...........(6)  Why have we chosen x sin x and x cos x rather than sin x and cos x? Because sin x and cos x are homogeneous solutions to (1), we must multiply them by a power of x. Since 3x y"p(x) = 2C cos x − C x sin x − 2 sin x −  x cos x + (30 F − 16E)e cos 5x − 3x −(30 E + 16 F) e sin 5x , and 3x 3x y"p + yp = 2C cos x − 2 sin x + (30 F − 16E) e cos 5x − (30 E + 16 F) e sin 5x 3x  sin x − e cos 5x ,.......................................................................(7) Therefore, 2C = 0, −2 = 1, 30F−15E= −1, and 30E + 15F = 0. Solving this system of equations yields C = 0,  = − 1 , E = 1 and F = 2 . Thus, the general solution is 2 75 3x y(x) = A cos x + B sin x − 1 x cos x + 1 e 2 75 75 3x cos 5x + 2 e sin 5x 75 Example 2.43: Find a particular solution of the equation 51 Advanced Engineering Mathematics d4y 4 2 d2y 2 Chapter 2- Second-Order Linear D.E.  y  3sin x  5cos x. .........................................................................(1) dx dx 4 2 2 2 Auxiliary equation is m + 2m + 1 = 0 or (m + 1) = 0 , m =  i twice. C.F. = A cos x + B sin x + C x cos x +  x sin x , ............(2) The general solution of the homogeneous equation is given in Relation (2). yc(x) = A cos x + B sin x + C x cos x +  x sin x, corresponding to the roots r = i , i , −i, and −i of the characteristic equation. Our initial assumption for a particular solution is Y (t) =  sin x +  cos x, but we must 2 multiply this choice by x to make it different from all solutions of the homogeneous 2 2 equation. Thus our final assumption is Yp (t) =  x sin x +  x cos x. Next, we differentiate Yp (x) four times, substitute into the differential equation (1), and collect terms, obtaining finally −8 sin x − 8 cos x  3 sin x − 5 cos x. Thus  = −3/8 ,  = 5/8 , and the particular solution of Eq. (1) is 2 2 Yp (x) =  3 x sin x + 5 x cos x. The complete solution is 8 8 2 2 y(x) = A cos x + B sin x + C x cos x +  x sin x  3 x sin x + 5 x cos x. 8 8 If f(t) is a sum of several terms, it is often easier in practice to compute separately the particular solution corresponding to each term in g(t). As for the second order equation, the particular solution of the complete problem is the sum of the particular solutions of the individual component problems. This is illustrated in the following example. d3 y dy  x  3cos x  e2x . ...............(1) dx dx 3 3 First we solve the homogeneous equation. The characteristic equation is m − 4m = 0, 2x −2x and the roots are 0, 2; hence C.F. = A + B e + C e ....................................(2) Example 2.44: Solve 4 We can write a particular solution of Eq. (1) as the sum of particular solutions of the differential equations d3y d3 y dy dy d3 y dy  4  e2x . 4 x  4  3cos x , dx dx dx dx 3 dx 3 dx 3 Our initial choice for a particular solution Y1(x) of the first equation is Ao x + A1, but since a constant is a solution of the homogeneous equation, we multiply by x. Thus Y1(x) = x (Ao x + A1). For the second equation we choose Y2(x) = B cos x + C sin x, and there is no need to modify this initial choice since cos x and sin x are not −2x solutions of the homogeneous equation. Finally, for the third equation, since e −2x solution of the homogeneous equation, we assume that Y3(x) = E xe 52 . is a Advanced Engineering Mathematics Chapter 2- Second-Order Linear D.E. The constants are determined by substituting into the individual differential equations; they are Ao=  1 , A1 = 0 , B = 0 , C =  3 , and E = 1 .Hence a particular 8 8 5 2 3 −2x 1 1 solution of Eq. (1) is Yp(x) =  x  sin x + x e . 8 8 5 2x −2x 1 2 3 −2x The complete solution is y = A + B e + C e  x  sin x + 1 x e . 8 8 5 The amount of algebra required to calculate the coefficients may be quite substantial for higher order equations, especially if the nonhomogeneous term is even moderately complicated. A computer algebra system can be extremely helpful in executing these algebraic calculations. The method of undetermined coefficients can be used whenever it is possible to guess the correct form for Y (t). However, this is usually impossible for differential equations not having constant coefficients, or for nonhomogeneous terms other than the type described previously. d2y dy  2y  x 2  1. ....(1) dx dx 2 2 Auxiliary equation is m − 3m + 2 = 0 or (m − 1)(m− 2) = 0 , m = 1 ,2. 2x x C.F. = A e + B e , .................................................(2) The general solution of the homogeneous equation is given in Relation (2). Since the right side of the equation is of order two, we propose for y the quadratic 2 polynomial a x + b x + c, and proceed via substitution in the given equation to determine the values of the constants a, b and c. 2 yp = a x + b x + c , yp' = 2 a x + b , yp" = 2a . 2 2 Substitution yields 2a − 3 (2 a x + b) + 2 (a x + b x + c)  x +1. 2 2 x [2 a] + x[− 6a + 2b] + [2a − 3b + 2c]  x +1 2 2 2a − 3 (2 a x + b) + 2 (a x + b x + c)  x +1. 2 Equating the coefficients of x yields 2a = 1, 2b− 6a = 0 and 2a − 3b + 2c = 1. a = 1 , b = 2 and c = 9 Example 2.45: Solve 2 3 3 4 x 2x 2x 2 2 y = yh + yp = A e + B e + 1 x + 2 x + 9 = A e + B e + 1 [ 2 x + 6 x + 9]. x 2 3 4 53 4 Advanced Engineering Mathematics Chapter 3- Fourier Series CHAPTER 3 FOURIER SERIES §3.1 Periodic Functions: Let f(x) is defined on the real x-axis. If f(x + T) = f(x), for all x(, ), T is a real constant, f(x) is said to be periodic with period T. It is clear that: f(x + n T) = f (x + [n1]T) =  = f(x + T) = f(x), where n is a positive integer. The least value of T > 0 is called the least period or simply the period of f(x). For example: cos x = cos(x + 2) =  = cos(x + 2n), thus the period of cos x is 2. Also the period of sin x is 2, the period of tan x is  and the period of a constant function is any positive number. If f(x) has the period T then the function: g(x) = f(a x) has the period T/a. It is easy to see that g(x + T/a) = f(a{x + T/a}) = f(a x + T) = f(a x) = g(x) . It is clear that cos 2x has the period , the period of sin 4x is /2. If f(x) has the period T, then: CT T C 0  f (x)dx =  f (x)dx (Prove it?) It is apparent that a sum, a difference, a product and a ratio of periodic functions with the same period T are again periodic functions with period T, (not necessary the least period). It is clear that tan x = sin x/ cos x , the period of sin x, cos x is 2, but the period of tan x is . §3.2 Odd and Even Functions: A function f(x) is said to be odd if: f(x) =  f(x) for all x. 5 3 x x It is easy to see that: x , x  2x, sin x, tan x , (e  e ) are odd functions. A function f(x) is said to be even if : f(x) = f(x) for all x . 2 4 2 x x Thus x , x + 3 x + 7, cos x , (e + e ) are even functions. These definitions imply that the graph of an even function is symmetric with respect to the axis of ordinates and the graph of an odd function is symmetric with 54 Advanced Engineering Mathematics Chapter 3- Fourier Series respect to the origin. It is clear that a linear combination of an even (odd) functions is also even (odd) function. If f(x) is integrable on the interval [a , a], then a 0 a a a 0  f (x)dx =  f (x)dx +  f (x)dx 0 a a  a 0 Replacing x by x in the first integral, we get:  f (x)dx =  f ( x)dx . Then a a 2  f (x)dx, if f (x) is an even function, f (x)dx =   [f (x)  f ( x)]dx =  0  a 0 if f (x) is an odd function.  0 In general, any function f(x) can be represented as the sum of even and odd functions. It is obvious that: f(x) = ½[f(x) + f(x)] + ½[f(x)  f(x)] = F(x) + G(x), where F(x) = ½[f(x) + f(x)], G(x) = ½[f(x)  f(x)], F(x)  even function and G(x)  odd function. Every function f(x), f(x) defined on the interval of the form [a, a] can be represented as the sum of the corresponding even and odd functions. Example 3.1: Graph each of the following functions: a) f(x) = |x| , x  [ , ] and has a period 2. Since the period is 2, that portion of the graph in   x   is extended periodically outside this range. It is clear that |x| = |x|, then f(x) is an even function.   x  (  ,0)  4 , period 2. b) f(x) =   , x  (0 , )  4 55 Advanced Engineering Mathematics Chapter 3- Fourier Series It is clear that, f(x) is not defined at x = 0, ± , ±2 ,  etc. These values are the discontinuities of f(x). It is obvious that f(x) is an odd function. c) f(x) = x ,  < x  , period = 2. Exercises 3.1 Graph each of the following functions and classify each of them according as they are even, odd, or neither even nor odd.  2 1) f (x)   2 0 < x 3 , period = 6. 3 x 0 cos x 0  x   ,period =2 2) f (x)   0 x 2       sin x 0  x   , period = 2 4) f (x)    sin x 0  x   , period =2 3) f (x)       cos x x 2    x  2 0  0,  5) f (x)  2,  0,  0  x<1  5, 1  x < 2 , period = 3. 6) f (x)   5, 2  x<3 7) f(x) = x(5x) , 0 < x < 5, period = 5. 56 0<x<3 3 < x < 0 , period = 6. Advanced Engineering Mathematics Chapter 3- Fourier Series §3.3 Fourier Series: Let f(x) be defined in the interval (T, T) and outside of this interval by f(x + 2T) = f(x), i.e. assume that f(x) has the period 2T. The Fourier expansion or Fourier series corresponding to f(x) is given by: S(x) = 1 ao + 2   [ an cos n 1  n T x  + b n   sin n  x ], T (1) where ao , an , bn , n = l, 2,  are called Fourier coefficients and given by : an = CT 1 T n  x dx = 1 f (x) cos f (x) cos n  x dx,   T T T T T CT (2) bn = CT 1 T n  x dx = 1 f (x) sin f (x) sin n  x dx,   T T T T T CT (3)         where C is any real number, n = 1, 2, 3, . 1 T To determine ao, using (2) with n = 0, hence ao =  f (x) dx T T 1 T 1 The constant term in (1) is equal to: ao =  f (x) dx, 2 2T T this is the mean of f(x) over a period. It is clear that bo = 0, since sin 0 = 0. Example 3.2: Find the Fourier coefficients corresponding to the function: 0 f (x)   2 3 < x < 0 0x3 , period = 6 Solution: The graph of f(x) is shown below 57 Advanced Engineering Mathematics Chapter 3- Fourier Series Period = 2T = 6, and T = 3. Then 13 10 13 n  x n  x an =  f (x) cos dx =  (0) cos dx +  (2) cos n  x dx = 3 3 3 3 3 3 3 30       2 3 23 =  cos n  x dx = [ sin n  x ] 30 = 0 for all n = 1, 2, 3, . 3 3 3 n 30  ao =    23  dx = 2. 30 Similarly, bn = 23  sin 30  n 3 x  dx = 2[ 1  cos n ]/n = =2[ 1  (1) n ]/n = if n is even 0 =  4 / n if n is odd i.e. b1 = 4/ , b2 = 0, b3 = 4/3 , b4 = 0, b5 = 4/5, b6 = 0,  etc. Therefore b2 = b4 = b6 = = b2m = 0, b2m+1 = 4/(2m+1) , for all m, ao = 2, a1 = a2 = a3 =  = am = 0. 58 Advanced Engineering Mathematics Chapter 3- Fourier Series Exercises 3.2 Prove that T 1)  T T      sin n  x dx = T T  cos T  n T x  dx = 0 , n = 1, 2, 3, .  2)  sin n  x cos n  x dx = 0, n = 1, 2, 3, . T T T T 3)  T     cos n  x cos m  x dx = T T T  n T x  sin  m T x  dx =  sin T 0 = T if m  n if m = n Find the Fourier coefficients corresponding to the functions: 0 4) f (x)   5 2 <x< 0 0x2 , period = 4. [Answers: ao = 5, an = 0, b2m = 0, b2m+1 = 10/(2m+1) ] 5) f (x) = (  x)/ 2 , x  (0, 2), period =2. [Answers: ao = an = 0, bn = 1/n ] 6) f (x) = | x |, 1 < x < 1, period =2. 2 2 [Answers: ao = 1, a2n = 0, a2n+1 =  4/ (2n+1) , bn = 0.] §3.4 Convergence of Fourier series: Definition: Piecewise continuity: A function f(x) is said to be piecewise continuous in an interval if: 1) the interval can be divided into a finite number of subintervals in each of which f(x) is continuous and, 2) the limits of f(x) as x approaches the endpoints of each subinterval are finite. i.e. the piecewise continuous function is one that has only a finite number of finite discontinuities. 59 Advanced Engineering Mathematics Chapter 3- Fourier Series Now, we shall state a theorem that will yield sufficient conditions for representing a function f(x) by a Fourier series. Theorem 3.1 (Dirichlet conditions): Suppose that: 1) f(x) is defined and single-valued except possibly at a finite number of points in (T, T), 2) f(x) is periodic outside (T, T) with period 2T, 3) f(x) and f'(x) are piecewise continuous in (T, T). Then the series (1) with coefficients (2) converges to: a) f(x) if x is a point of continuity. b) ½[f(x + 0) + f(x0)] if x is a point of discontinuity, where f(x+0) = lim f(x+), f(x0) = lim f(x)  0   0  The conditions (1), (2) and (3) imposed on f(x) are sufficient but not necessary, and are generally satisfied in practice. There are at present no known necessary and sufficient conditions for convergence of Fourier series. In Example (1.2); the Fourier series corresponding to f(x) is: 1+ (4/){ sin(x/3) + (1/3) sin(3x/3) + (l/5) sin(5x/3) + (l/7) sin(7x/3) +.}. Since f(x) satisfies the Dirichlet conditions, we can say that the series converges to f(x) at all points of continuity and to ½[f(x + 0) + f(x0)] at points of 60 Advanced Engineering Mathematics Chapter 3- Fourier Series discontinuity. At x = 3, 0 and 3, which are points of discontinuity, the series converges to ½[2 + 0] = 1 as seen from the graph. If we redefine f(x) as follows: 1 2  f (x)   1 2   1 x =3 3 x  0 x  0 , period = 6. 0x3 x 3 Then the series will converge to f(x) for  3  x  3. Example 3.3: 2 Expand f(x) = x , 0 < x< 2 with period 2, in Fourier series. Solution: The graph of f(x) with period 2 is shown below: Period = 2T = 2 and then T = . 3 2 1  x  2  8 1 2 2 ao =  x dx =   = .  3  0 3  0   2 1 1 2 2 1 x  2 2 2 2 an =  x cos nx dx=  x d(sin nx) = n [ x sin nx 0   2x sin nx dx] =  0 n x  0 0 61 Advanced Engineering Mathematics    = 2 n 1 n Chapter 3- Fourier Series 2π  2  2  x d(cos n x) =  2  [x cos n x 0   cos nx dx ] = n  0 x 0 x  2 sin nx 2    4   4  2n =  22  [ 2 cos 2n  0 ] =  2  (1) =  2  , n  0 n n  n  n  bn = 1 2 2 1 [ x2   cos nx  2x   sin nx  +2  cos nx  ] 2  =  4 . sin n x dx = x      2   3  0 n n  0    n   n  2  4 4 Then f(x) = x = +  [ 42 cos n x  sin n x]. 3 n n n 1 2 This is valid for 0 < x < 2. At x = 0 and x = 2 the series converges to 2 2 ½[0 + 4  ] = 2  . 2  4 +  [ 42 ]. At x = 0 the Fourier series reduces to 3 n 1 n 2 By Dirichlet condition, the series converges at x = 0 to 2 . Then 2  4 2 +  [ 42 ] = 2 , i.e. 3 n 1 n  4 ] = (2  4/3)2 = 2/32. 2 n 1 n [ Example 3.4: Let us take the periodic function f(x) = x, regarded as being defined on the interval (T, T) with period 2T, and expand it into a Fourier series Solution:  It is clear that, this function is an odd function, then f(x) cos n  x 62 T  is an odd Advanced Engineering Mathematics  Chapter 3- Fourier Series  function and f(x) sin n  x is an even function. Hence T 1 T 1 T n  x an = dx = f (x) cos x cos n  x dx = 0, n = 1, 2, 3,    T T T T T T     ao = 1 T  x dx = 0, T T bn = T T 1 T n  x dx = 1 n  x dx = 2 x sin n  x dx = sin sin f (x) x    T T T T T T T T0       2T 2 xT T  T n+1 n  x = [ cos + cos n  x dx] = (1) .  0 T T n T n n 0     Consequently, according to the convergence theorem, we have  n 1 2T  (1) x for  T< x < T ] sin n  x = S(x) = , [ T 0 for x=T  n 1 n   Since S(T) = ½[ f(T + 0) + f(T0)] = ½[ T + T] = 0 , for x = T. It is clear that the function S(x) is periodic with period 2T, and S(x) = x , only for T < x < T. At the points x = (2m+l)T, m = 0, ±1, ±2,  etc., the function x has discontinuities, its left-hand and right-hand limits being, respectively, equal to T and +T, but S(x) assumes the value which is the halfsum of the left-hand and right-hand limits i.e. equal to ½(T+T) = 0 . In the following graph we see the graph of the partial sum S5 (x):  2x  1  3x  1  2T   x  1  4x  S5(x) =   + sin    sin   [ sin    sin   +  T  2  T  3  T  4  T     + 1  5x  sin   ], 5  T  constructed on the interval [T, T]. 63 Advanced Engineering Mathematics Chapter 3- Fourier Series APPENDIX 3.I In the following, we give some important integrals: 1 [cos a x + a x sin a x] + C, 2 a 1)  x cos a x d x = 2) 2  x cos a x d x = 4) 1 [2 a x cos a x + (a2 x2  2) sin a x] + C, 3 a 3 2 2 3 3 1  x cos a x d x = 4 [(3 a x  6) cos a x + (a x  6a x) sin a x] + C, a 4 3 3 4 4 2 2 x cos a x d x = 1 [(4 a x 24ax) cos a x+(a x 12a x +24) sin a x]+ C, 5) 5  x cos a x d x = 6) +(a x 20a x +120) sin a x] + C, n1 n n a  x cos a x d x = x cos a x  n  x cos a x dx, 7)  x sin a x d x = 8) 10) 1 [2 a x sin a x  (a2 x2  2) cos a x] + C, 3 a 3 2 2 3 3 1  x sin a x d x = 4 [(3 a x  6) sin a x  (a x  6a x) cos a x] + C, a 3 3 4 4 2 2 4 1 x sin a x d x = [(4 a x 24ax) sin a x(a x 12a x +24) cos a x] + C, 11) 5  x sin a x d x = 3)  a 5 1 [(5 a4 x460a2 x2+120) cos a x+ 6 a 5 9) 5 3 3 1 [sin a x  a x cos a x] + C, 2 a 2  x sin a x d x =  a 5 1 [(5 a4 x460a2 x2+120) sin a x 6 a 5 5 3 3 (a x 20a x +120) cos a x] + C, 12) n n a  x sin a x d x = x cos a x + n  x 64 n1 cos a x dx, Advanced Engineering Mathematics e ax  e sin nx dx = 13) 2 a n ax 14) ax  e cos nx dx = e 2 ax 2 a n 2 Chapter 3- Fourier Series [ a sin n x  n cos n x] + C, [ a cos n x + n sin n x] + C. Example 3.5: 2 Let us take the periodic function f(x) = x , on the interval [T, T] with period 2T. This function being even, we have: 2 3 T 2 2 2 T 2 ao = x dx = [x ] 0 = T .  3T 3 T 0 an = 2 T 2 n  x dx = 2 T 3  2n  x cos n  x + x cos  T T T 0 T n  T         2 2 T   2 2 +  n  x  2  sin n  x  = 2T [ 2n cos n + (n   2) sin n ] = 3 3 T 0 n   T   = bn =  2 4T ( 1) 2 2 n  n   , 1 T 2 x sin n  x dx = 0.  T T T   2  ( 1) n nx 2 2 1 4T Then S(x) = T + 2  =x , cos 3 2 T  n 1 n 65 T  x  T. Advanced Engineering Mathematics Chapter 3- Fourier Series 2 The function x and the sum S(x) of Fourier's series are continuous everywhere including the points x = (2m+l)T , m = 0, 1, 2, , since for the function 2 2 f(x) = x we have, f(T + 0) = f(T 0) = T . If T=  we get: 2 2 x = 1 + 4 3  ( 1) n  2 cos n x, n 1 n   x  . Substituting with x = 0, we obtain the useful relation: m 1 2  = 1  1 + 1  1 + + ( 1) 2 4 9 16 12 m + §3.5 Fourier series for even and odd functions: From the definition of an even function and odd function it follows that:  T T T 2  f (x)dx, if f (x) is an even function,  f (x)dx =  [f (x)  f ( x)]dx =  0  T 0 if f (x) is an odd function.  0 Now, if an odd function f(x) is expanded in a Fourier series, then the     product f(x) cos n  x is also an odd function, while f(x) sin n  x is an even T T function; hence, 1 T 1 T ao = f (x) dx = 0, an = f (x) cos n  x dx = 0, n = 1, 2, 3, .   T T T T T   1 T 2T n  x dx =  f (x) sin n  x dx. bn = f (x) sin  T T T T T0     Thus the Fourier series of an odd function contains only sines (sine terms); (see Example 3.4). If an even function f(x) is expanded in a Fourier series, then the product     f(x) cos n  x is also an even function, while f(x) sin n  x is an odd function; T hence, 66 T Advanced Engineering Mathematics Chapter 3- Fourier Series 2T 1 T ao =  f (x) dx = T  f (x) dx, T T 0 2T 1 T n  x an = dx =  f (x) cos n  x dx f (x) cos  T T T0 T T     1 T bn = f (x) sin n  x dx = 0, n = 1, 2, 3, .  T T T   Thus, the Fourier series of even functions contains only cosines (see Example 3.5). The formulas obtained permit simplifying computations when seeking Fourier coefficients in cases when the given function is even or odd. It is obvious that not every periodic function is even or odd, for example f(x) = sin x + cos x is neither even nor odd. §3.6 Half range Fourier sine or cosine series: A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are present respectively. When a half range series corresponding to a given function is desired, the function is generally defined in the interval (0, T) and then the function is specified as odd or even, so that it is clearly defined in the other half of the interval, namely (T, 0) . In such case, we have: 2T an = 0 , bn =  f (x) sin n  x dx, for half range sine series, T T0  bn = 0 , an =  2T f (x) cos n  x dx, for half cosine series.  T T0   Example 3.6: Expand f(x) = sin x, 0 < x <  in a Fourier cosine series. Solution: 67 Advanced Engineering Mathematics Chapter 3- Fourier Series A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence we extend the definition of f(x) so that it becomes even. With this extension, f(x) is then defined in an interval of length 2. Taking the period as 2, we have 2T = 2 so that T = . 2 2  4 ao =  sin xdx = (cos x) = , 0   0 an = = 2 1 = sin x cos nx dx   sin(x  nx)  sin(x  nx)  dx = 0 0 1  cos(n  1)x cos(n  1)x   1 1  cos(n  1) cos(n  1)  1     =  =   n 1 n 1 n 1 n  1  0   2 1  cos n  1 1  cos n cos n  1  1  cos n  1 1   =   , n 1.  = =  n  1 n  1  2   n 1 n  1   (n  1) 1 2  a1 =  sin 2x dx =  [cos 2x] = 0. 0  0 Then s(x) = 2 4 =    2 2  1  cos n  cos nx =     n  2 (n 2  1)  cos 2x cos 4x cos6x      .  2  2  1 42  1 62  1  Example 3.7: Expand f(x) = x, 0  x  T in a half range a) sine series , b) cosine series. Solution: 68 Advanced Engineering Mathematics Chapter 3- Fourier Series a) Extend the definition of the given function to that of the odd function of period 2T (See Example 3.4 ) i.e. f(x) = x in the interval [T , T] . n 1 2T  (1) ] sin n  x . Then S(x) = [ T n  n 1   b) Extend the definition of the given function f(x) to that of the even function of period 2T as shown below: Thus bn = 0, n = 1, 2, 3, , ao = 1 2 T 2T xdx = (x ) = T,  0 T T0 2 2 T  2T nx an =  x cos n  x dx=   [cos T T Tn  T0  = 2T 2 2 n    [ ( cos n + n sin n )  ( 1 + 0)] = Then S(x) = =  +  n T x  sin  n T x  ]= 2T 2 2 n  n [(1) 1 ].  2T T n +  [(1) 1 ] cos n  x = T 2 n 1 n 2 2     T 4T  1 (2m 1)   x= cos  T 2 2 m 1 (2m  1) 2 x 3x cos 5x cos 7 x   T 4T  cos T cos T T T      =  . 2 2 2 2 2  12  3 5 7   It should be noted that the given function f(x) =x, 0 < x < T, is represented equally well by two different series in (a) and (b). 69 Advanced Engineering Mathematics Chapter 3- Fourier Series Example 3.8: 2 Expand f(x) = x , 0  x < T in a half range a) cosine series, b) sine series. Solution: a) Extend the definition of the given function to that of the even function of period 2 2T (See Example 3.5 ) i.e. f(x) = x , T < x < T. 2  cos x cos 2x cos 3x cos 4x  4T T T T T 1       . Then S(x) = T  2 2 2 2  2 3  2 3 4   1  2 b) Extend the definition of the given function f(x) to that of the odd function of  x 2 period 2T: i.e. f (x)   2  x 0xT Tx0 . Then ao = an = 0, n = 1, 2, 3, , 2 T 2 bn = x sin n  x dx =  T T 0  = 2T 2 3 3 n    2n  x n  x   n  x 2  2  cos n  x  T = sin  T  T  T T  0      2 2 [ (n   2) cos n  2] =  70   2T 2 3 3 n   2 2  n [ (n   2)(1) + 2]=  Advanced Engineering Mathematics = 2T 2 3 3 n   n 2 2 Chapter 3- Fourier Series n [(1) n  + 2{1(1) }] = (1) Then S(x) =  [ (1) n+1 2T 2 n  4T 2 3 3 {1+(1) n+1 n+1 2T 2 n   4T 2 3 3 n  {1+(1) n+1 }.  }] sin n  x = T n 1 n  x 2x sin 3x sin 4x sin 5x 2   2T  sin T sin T T T T        =   1 2 3 4 5    x 3x sin 5x sin 7 x sin 9x 2   8T  sin T sin T T T T       .  3  3 3 3 3 3  3 5 7 9   1  §3.7 Harmonic analysis: This is the determination of the Fourier coefficients ao, a1, a2 ,; b1,b2,. In technology it is frequently used to analyse periodic phenomena. An oscillation is split up by harmonic analysis into a sum of pure sine oscillations (harmonic oscillations) and a constant part. Apart from the fundamental oscillations there occur the so-called "harmonics" whose frequency is twice, three times etc, the fundamental frequency. Much labour can be saved in harmonic analysis of one observes certain symmetry properties of the function f(x) to be analysed. i) For the function with the property f(x + T) =  f(x) (odd harmonic) the absolute term is ao = 0 and only coefficients with an odd index occur i.e. a2 = a4 =  = b2 = b4 =  = 0. It is clear that: 71 Advanced Engineering Mathematics Chapter 3- Fourier Series T  10 1 T f (x) f (x)dx f (x)dx dx =      T  T T T 0  ao = in the first integral, put x = u  T and dx = du 0 T T T T 0 0 0   f (x)dx =  f (u  T)du =  f (u  T)du =   f (u)du . T  1 T  ao =    f (x)dx   f (x)dx  = 0, T  0  0 an = T 0  1 T n  x dx = 1 n  x dx  f (x)cos n  x dx = f (x) f (x)cos cos     T T T T  T T T 0        Put x = u  T and dx = du, in the first integral 0   T  f (x) cos n  x T  T     T n(u  T) n(u  T) dx =  f (u  T)cos du =   f (u)cos du = T T 0 0 T T 0 0 n =   f (u) cos nu cos n  sin nu sin n  du =  (1) f (u)cos nu du = T T T   = (1) n+1 T  f (u)cos 0 nu du . T 1T Then an =  1  (1)n 1 f (x)cos n x dx . T T0         2T (2m 1)  x dx . Therefore a2m = 0 for all m = 1, 2, 3, , and a2m1 =  f (x)cos T T0 2T (2m 1)  x dx . Similarly b2m = 0 for all m = 1, 2, 3, , and b2m1 =  f (x)sin T T0 ii) For the function with f(x + T) = f(x) (even harmonic) the absolute term ao and only coefficients with an even index occur. 72 Advanced Engineering Mathematics Chapter 3- Fourier Series T  10 1 T f (x)  ao = dx = f (x)dx f (x)dx      T  T T T  0 Put x = u  T and dx = du, in the first integral, we get 0 T T T T 0 0 0  f (x)dx =  f (u  T)du =  f (u  T)du =  f (u)du .  ao = T  2T 1 T f (x)dx  f (x)dx   =  f (x)dx .  T  0 0  T 0 Also T  10 1 T n x n  x dx =  n  x f (x)  an = cos dx = f (x)cos dx f (x)cos      T T T T  T T T  0       Put x = u  T and dx = du, in the first integral 0   T   f (x) cos n x T T    T T 0 0 n =  f (u) cos nu cos n  sin nu sin n  du =  (1) f (u)cos nu du = T T T   n T = (1)  f (u)cos nu du . T 0 1T Then an =  1  (1) n 1 f (x)cos n x dx . T T0   T n(u  T) n(u  T) dx =  f (u  T)cos du =  f (u)cos du = T T 0 0    73 Advanced Engineering Mathematics Chapter 3- Fourier Series 2T Therefore a2m =  f (x)cos 2m x dx , and a2m1 = 0 for all m = 1, 2, 3, . T T0   2T Similarly b2m =  f (x)sin 2m x dx ,and b2m1 = 0 for all m = 1, 2, 3, . T T0   iii) For the function with the properties f(x + T) =  f(x), f(x) = f(x) (odd cosine harmonic functions) we have bn = 0 for all n = 1, 2, 3, , ao = a2n = 0  and f(x) =  a 2n 1 cos n 1  (2n 1)  x T  , where a 2n1 = T 2   4 (2n 1)  x f (x)cos dx .  T T0 iv) For the function with the properties f(x + T) =  f(x) , f(x) =  f(x) (odd sine harmonic functions) We have ao = an = 0, b2n = 0 for all n = 1, 2, 3,  b2n1 = T 2   4 (2n 1)  x f (x)sin dx .  T T0 v) For the function with the properties f(x + T) = f(x) , f(x) = f(x) 74 Advanced Engineering Mathematics Chapter 3- Fourier Series (even cosine harmonic functions) We have bn = 0, a2n1 = 0, for all n = 1, 2, 3,  a2n = T 2   4 f (x)cos 2n x dx .  T T0 vi) For the function with the properties f(x + T) = f(x) , f(x) = f(x) (even sine harmonic functions). We have an = 0 for all n = 1, 2, 3, , b2n1 = 0, b2n = T 2   4 f (x)sin 2n x dx .  T T0 Example 3.9: Expand f (x)   x xT 0 x<T , T  x  2T 75 Period = 2T in Fourier series. Advanced Engineering Mathematics Chapter 3- Fourier Series Solution: It is easy to see that f(x + T) = f(x), then f(x) - even harmonic; i.e. a1 = a3 =  = b1 = b3 = = 0, and 1 2 T 2T 2T f (x)dx =  xdx = ao = x = T.  0 T T0 T0 2T 2T 2n x  dx = x cos 2n x dx = a2n =  f (x)cos  T T T0 T0     2 2 T   2n x 2n x 2n x  T =   cos T  T sin T  = 0 T  2n       T  = cos 2n  2n sin 2n  1 = 0,  2 2   2n     2 T 2 T   2T 2n x  sin 2n x  2n x cos 2n x  = b2n =  f (x)sin dx =   T T T T 0 T  2n   T0        T  T  sin 2n  2n cos 2n =  . =    n  2n 2 2  Therefore  2nx  = T  T  sin 2Tx  sin 4Tx  sin 6Tx  sin 8Tx   . T T  sin T f(x) =   n 2  n 1 2    1 2 3 4 Example 3.10:  x Expand f (x)     x  T  0 x<T , T  x  2T 76 Period = 2T in Fourier series.   Advanced Engineering Mathematics Chapter 3- Fourier Series Solution: It is obvious that f(x + T) = f(x), then f(x) is odd harmonic; i.e. ao = a2 = a4 =  = b2 = b4 = = 0, and a2n1 =     2T 2T (2n 1)  x (2n 1)  x f (x)cos dx x cos dx = =   T T T0 T0 2       2 T (2n 1)  x (2n 1)  x (2n 1)  x  T  =  = cos sin   0 T T T T  (2n  1)       2T 4T  cos(2n  1)  (2n  1) sin(2n  1)  1 =   = ,    (2n  1)2 2   (2n  1) 2 2           2T (2n 1)  x dx = b2n1 =  f (x)sin T T0 2       2 T (2n 1)  x (2n 1)  x (2n 1)  x  T  =  = sin cos   0 T T T T  (2n  1)     2T 2T  sin(2n  1)  (2n  1) cos(2n  1)  = = .     (2n  1)2 2  (2n  1)    Therefore      4T 2T (2n 1) x (2n 1) x cos sin f(x) =    2 2 T T (2n  1) n 1   (2n  1) 77    =  Advanced Engineering Mathematics  Chapter 3- Fourier Series   2T   1 2 (2n 1) x (2n 1) x sin cos =   T 2 T  n 1  (2n  1) (2n  1)     .  Example 3.11: Expand f (x) = x , 0  x < T/2 , period = 2T and satisfies the following properties f(x + T) =  f(x) , f(x) = f(x) , in Fourier series. Solution: Extend the definition of the given function to that of the odd cosine harmonic function of period 2T, and then we get:  x x T  f (x)   T  x 2T  x if 0  x  T / 2 if T / 2  x  T if T  x  3T / 2 if 3T / 2  x  2T and ao = a2 = a4 = = a2n = b1 = b2= b3 = = bn = = 0  n = 1, 2, 3, . T   T   4 2 4 2 (2n 1)  x (2n 1)  x a2n1 =  f (x)cos dx = x cos dx =  T T T 0 T 0 2       4 T (2n 1)  x (2n 1)  x (2n 1)  x  T 2 cos sin =  =   0 T T T T  (2n  1)       4T (2n 1)  (2n 1)  cos  (2n  1)  sin  1 = =   2 2 2  (2n  1) 2 2     78 Advanced Engineering Mathematics Chapter 3- Fourier Series n 1    4T 1 (1)n 1 (2n  1)   1 =  4T   ( 1)   = ,     2  (2n  1) 2 2     2   2(2n 1) (2n 1)2       4T    ( 1)n 1   (2n 1)  x 1 cos Therefore f(x) =  =      2 T  2  n 1  2(2n 1) (2n 1)  x 3x cos 5x cos 7 x   2T   cos T cos T T T =       2 1 3 5 7     x 3x cos 5x cos 7 x    4T   cos T cos T T T       .  2 2 2 2      1 3 5 7   Example 3.12: Expand f(x) = 1 , 0 < x < /2 period = 2 with a) odd cosine harmonic, c) odd sine harmonic, b) even cosine harmonic, d) even sine harmonic. Solution: a) Extend the definition of the given function to that of odd cosine harmonic function of period 2 as shown in the figure. f(x) satisfies the properties: f(x + ) = f(x) , f(x) = f(x). Therefore ao = a2 = a4 = = a2n = 0,and b1 = b2 = b3 = = bn = = 0  n = 1, 2, 3, , 79 Advanced Engineering Mathematics Chapter 3- Fourier Series  4 2 4 4  sin(2n  1)x   2 and a2n1 =  cos(2n  1)xdx =  sin(2n  1)  = =  2 0   (2n  1)  0 (2n  1) = 4 n 1 . (1) (2n  1) n 1 4  (1) Thus f(x) =  cos(2n  1)x =  n 1 (2n  1) = 4   cos x cos3x cos5x cos7x cos9x   1  3  5  7  9   . b) The case of an even cosine harmonic function i.e. f(x) satisfies the following properties f(x + ) = f(x) , f(x) = f(x). Therefore bn = a2nl = 0,  n = 1, 2, 3, ,  4  sin 2nx   2 2 4 2 a2n =  cos 2nx dx =  = sin n = 0,  n = 1, 2, 3, ,   2n  0 n 0  4 4 2 = 2. ao =  dx = 2 0 Thus f(x) = 1 , and it is obvious since f(x) is constant function from the beginning. c) The case of an odd sine harmonic, i.e. f(x) satisfies the following properties , f(x + ) = f(x) , f(x) = f(x). Therefore ao = an = b2n = 0,  n = 1, 2, 3, , 80 Advanced Engineering Mathematics Chapter 3- Fourier Series  4   cos(2n  1)x   2 4 2 b2n1 =  sin(2n  1)xdx =  =   (2n  1)  0 0 = 4  4 . cos(2n  1)   1 = 2  (2n  1) (2n  1)  Thus f(x) = = 4  sin(2n  1)x =   n 1 (2n  1) 4  sin x sin 3x sin 5x sin 7x sin 9x        .   1 3 5 7 9  d) The case of an even sine harmonic i.e. f(x) satisfies the following properties f(x + ) = f(x) , f(x) = f(x). Therefore ao = an = b2n1 = 0,  n = 1, 2, 3, ,  4 2 2  4   cos 2nx   2 2 n   1 . b2n =  sin 2nx dx =  =   = cos n 1 ( 1)      0 n    2n  0 n 81 Advanced Engineering Mathematics Chapter 3- Fourier Series n 2  [1  (1) ] Then f(x) =  sin 2nx =  n 1 n = 4  sin 2x sin 6x sin10x sin14x sin18x         .   1 3 5 7 9 Example 3.13: Find the Fourier series for the function given by : f(x) = cos x , 0 < x < 2. Solution: It is clear that f(x + ) = f(x) , f(x) = f(x). Therefore the function is even cosine harmonic. Thus bn = a2n1= 0,  n = 1, 2, 3, ,     4 4 4 2 4 2 4 2 ao =  f (x)dx =  cos x dx =  cos x dx = [sin x] 2 = .  0  0 0 0   4 2 4 2 a2n =  f (x) cos 2nx dx =  cos x cos 2nx dx = 0 0  2 2 =   cos(2n  1)x  cos(2n  1)x  dx = 0    2  sin(2n  1)x sin(2n  1)x   2 2  sin(2n  1) 2 sin(2n  1) 2   = =   (2n  1)    (2n  1) (2n  1)  0   (2n  1)   n n 1  n 2(1) 2  (1) (1) = =      (2n  1) (2n  1)    n 1  1 1  4(1)  .  (2n  1) (2n  1)  = 2   (4n  1)  n = 1, 2, 3,. 82 Advanced Engineering Mathematics Chapter 3- Fourier Series n 1 2 4  (1) Therefore f(x) = +  cos 2nx =   n 1 (4n 2  1)  2 4  cos 2x cos 4x cos6x cos8x       . = +    4(1)2  1 4(2) 2  1 4(3)2  1 4(4)2  1   Example 3.14: Find the Fourier series for the function given by: f(x) = sin x , 0 < x < 2. Solution: It is clear that f(x + ) = f(x) , f(x) = f(x). Therefore the function is even cosine harmonic. Thus bn = a2n1 = 0,  n = 1, 2, 3, ,     4 4 4 2 4 2 4 2 ao =  f (x)dx =  sin x dx =  sin x dx = [ cos x] 2 = . 0   0 0 0   4 2 4 2 a2n =  f (x) cos 2nx dx =  sin x cos 2nx dx = 0 0  2 2 =  sin(2n  1)x  sin(2n  1)x  dx = 0 4 1 1  2  cos(2n  1)x cos(2n  1)x   2 2  1 =    =  = .    4n 2  1   (2n  1) (2n  1)  0   (2n  1) (2n  1)  Therefore f(x) = 2 4  1   cos 2nx =   n 1 (4n 2  1) 83 Advanced Engineering Mathematics Chapter 3- Fourier Series  2 4  cos 2x cos 4x cos6x cos8x      . =      4(1) 2  1 4(2) 2  1 4(3)2  1 4(4)2  1   Compare this result with Example 3.6 Example 3.15: Find the Fourier series for the function given by: f(x) = x for 0  x < T/2, with period = 2T, a) Odd sine harmonic, b) Odd cosine harmonic, c) Even sine harmonic, d) Even cosine harmonic. Solution: a) The case of odd sine harmonic: It is clear that f(x + T) =  f(x) , f(x) = f(x), i.e. ao = an = b2n = 0,  n = 1, 2, 3, , and T 2  n 1  4T(1) 4 (2n 1)  x dx = b2n1 =  x sin . 2 2 T T0 (2n  1)  Therefore f(x) = n 1 4T  (1) 2   n 1 (2n  1) 2 sin (2n  1)x = T 3x sin 5x sin 7 x sin 9 x  x  4T  sin T sin T T T T       = 2 2 2 2 2 2  3 5 7 9   1  b) The case of odd cosine harmonic. See Example 3.11. 84 Advanced Engineering Mathematics Chapter 3- Fourier Series c) The case of even sine harmonic. It is clear that f(x + T) = f(x) , f(x) =  f(x) . and i.e. ao = an = b2n-1 = 0,  n = 1, 2, 3, , and T 2  n 1  T(1) 4 b2n =  x sin 2n x dx = T n T0 n 1 T  (1) Therefore f(x) =   n 1 n sin . 2nx = T  2x sin 4x sin 6x sin 8x sin 10x  T  sin T T T T T       . = 2 3 4 5  1    d) The case of even cosine harmonic. It is clear that f(x + T) = f(x) , f(x) = f(x) . and i.e. bn = a2n-1 = 0,  n = 1, 2, 3, , and 85 Advanced Engineering Mathematics T Chapter 3- Fourier Series T 2  2 T2 T 4 2 4 2 x ao =  f (x)dx =  x dx = = T  0 T 0 T 0 2 T 4 2 T n [(1)  1]. a2n =  f (x) cos 2nx dx = 2 2 T T 0 n  Therefore   n (1)  1  2(2m  1)x 2 2nx T T  T T cos f(x) = + = + = cos   2 2 2 T 4 2 n 1 4 T n  m 1 (2m  1) 2x cos 6x cos 10x cos 14 x cos 18x   T 2T  cos T T T T T       . = + 2 2 2 2 4 2  12  3 5 7 9   §3.8 Parseval's Identity: We can say that the series u1(x) + u2(x) + converges to the sum S(x) in a region R , if given  > 0 there exists a number N, which in general depends on both  and x, such that S(x)  Sn (x)|<  whenever n > N , where Sn (x) = u1(x)+ u2(x)+  + un(x). If we can find N depending only on  and not on x, we say that the series converges uniformly to S(x) in R. Uniformly convergent series have many important advantages as indicated in the following theorems. Theorem 3.8.1  If un(x), n = 1, 2, 3,  are continuous in the interval [a, b] and  u n (x) is n 1 86 Advanced Engineering Mathematics Chapter 3- Fourier Series uniformly convergent to S(x) in the interval [a, b], then S(x) is continuous in the interval [a, b]. Theorem 3.8.2  If  u n (x) converges uniformly to S(x) in the interval [a, b] and un(x) , n 1 n = 1, 2, 3,  are integrable in [a, b], then: b b a a  S(x)dx =   u1 (x)  u 2 (x)  u 3 (x)   dx = b b b a a a =  u1 (x)dx +  u 2 (x)dx +  u 3 (x)dx + Theorem 3.8.3 If un(x) , n = 1, 2, 3,  are continuous and have continuous derivatives in   n 1 n 1 ' the interval [a, b] and if  u n (x) converges to S(x) while  u n (x) is uniformly convergent in the interval (a, b), then S'(x) = d ' ' ' u1 (x)  u 2 (x)  u 3 (x)   = u1 (x)  u 2 (x)  u 3 (x)   .  dx Theorem 3.8.4: Parseval's identity If the Fourier series corresponding to f(x) converges uniformly to f(x) in the interval 2  ao 1 T 2 2 2 f (x) dx = [T, T], then     (a n  b n ) .  2 n 1 T T Where the integral is assumed to exist. Proof: Let us consider 87 Advanced Engineering Mathematics f(x) = 1 ao + 2   [ an cos n 1 Chapter 3- Fourier Series  n T x  + b sin  n T x  ], n Multiplying by f(x) and integrating term by term from T to T (which is justified since the series is uniformly convergent) We obtain T  T  T T n 1 T 2 T f (x) dx = 12 ao  f (x) dx +  [ an  f (x) cos  n T x  dx + 2  ao 2 2 n  x + bn  f (x) sin dx] = T  T  (a n  b n ) , T 2 n 1 T T   where we have used the results T  f (x) dx = T ao, T T  T  T  f (x) cos n  x dx=T an, T  f (x) sin T  n T x  dx= T b n Obtained from the Fourier coefficients. Therefore 2  ao 1 T 2 2 2  = f (x) dx    (a n  b n ) . ■  2 n 1 T T Note: Parseval's identity is valid under less restrictive conditions than the imposes here. Example 3.16: Write Parseval's identity corresponding to the Fourier series of the function defined in Example 3.7 ; and then determine the sum of the series: 1 4 1  1 2 4  1 4 3  1 4 4  1 5 4  1 6 Solution: From the Example 3.7 (b) :- f (x)  and S(x) = 4   x x 1 n 4  0xT . Tx0   1 T 4T  (2m 1)  x,  cos  T 2 2 m 1 (2m  1) 2 88 Advanced Engineering Mathematics Chapter 3- Fourier Series 1 T 4T 1 ao = , a2m-1 = 2 2 2 2  (2m  1) 3 2 1  x  T 2T 1 T 2 1 T 2 Therefore   f (x) dx = T  x dx = T  3  T = 3 . T T   T Hence 2 2 2 2 2  16T 2T 1 16T  1 T T +  + = = .  4 4 3 2 m 1 4 (2m  1)4 2  m 1 (2m  1) 1 2 1 1 16  ,   = =  3 2 6 4 m 1 (2m  1)4 4 4  1   1 1 1 1 i.e. =      = . = 4 6  16 m 1 (2m  1)4 14 34 54 96 (2m  1) 1 1 1 4 1      = Hence . 4 4 4 4 96 1 3 5 (2m  1)  1 1 1 1 1 1 1 1 =        = 4 4 4 4 4 4 4 4 n 1 n 1 2 3 4 5 6 n Thus S =  odd part even part      1  1  1 1 1 1 1 1    =          +      = 4 4 4 4 4 4 4 4  1   2 3 5 (2m  1) 4 6 (2m) 1  1 1 1    + =  14 34 54  24 1  1  1 1 1 1 1 1      =      4   4  + 4 S. 4 4 4 4 4 1    2 2 3 4 3 5  1 1  4 1 1    = i.e. S[1 ] =   4 4 4 4 3 5 2 1  96 1 4 4  16  = S= . 96 15 90 89 Advanced Engineering Mathematics Chapter 3- Fourier Series Example 3.17: From the Fourier's expansion of the function f(x) = sin x,    x   (See Example 3.6) find the sum of the following series S= 1 2 2 1 .3  1 2 2 3 .5  1 2 2 5 .7  1 2 2 7 .9  Solution: Using the result of Example 3.6 we get: 2 4 f(x) =     cos 2x cos 4x cos6x       2 .  2  1 42  1 62  1   4 4 1  . i.e. ao = , a2n =    (2n)2  1     Therefore = 1  1  1  x sin x cos x   2 2 sin x dx =   f (x) dx =       =     2 2 1     ( )  = 1  2   2 From Parseval's identity we get: 1= 16 22 +  16  1 1 1 1        2  (22  1)2 (42  1)2 (62  1)2 (82  1)2 2 1 1 1 1 1   = +     16 2 12.32 32.52 52.7 2 7 2.92 2 2  1  8 Therefore S = . − = 16 2 16 §3.9 Differentiation and integration of Fourier series: Differentiation and integration of Fourier series can be justified by using the theorems 1.8.2, 1.8.3. It must be emphasized however, that those theorems provide sufficient conditions and are not necessary. 90 Advanced Engineering Mathematics Chapter 3- Fourier Series The following theorem for integration is especially useful. Theorem 3.9: The Fourier series corresponding to f(x) may be integrated term by term from a x to x, and the resulting series will converge uniformly to  f (u)du provided that a f(u) is piecewise continuous in −T  u  T and x [−T , T]. Example 3.18: 2 Find a Fourier series for f(x) = x , −T < x < T, by integrating the series of  ( 1) n 1 . Example 3.4, and use it to evaluate the sum of the series  2 n 1 n Solution: From Example 3.4, n 1 2T  (1) x= sin n  x , −T < x < T  T  n 1 n   Integrating both sides from 0 to x ( applying Theorem 1.9.1) and multiplying by 2, we find : n 4T  (1) x =C+   n 1 n 2  nT  cos  n T x  , 2  n (1) =C+ cos n  x . 2  T 2  n 1 n  4T  2  n (1) Put x = 0 , we get C = − . 2  2  n 1 n 4T 2  n 1 2 n 1 (1) 4T  (1) − cos n  x i.e. x = 2 2  2  T  n 1 n  n 1 n 2  4T 2 , and this represents the Fourier cosine series for x in [−T, T]. Therefore: 91 Advanced Engineering Mathematics Chapter 3- Fourier Series 1 1 2 1 T 2 1 T 2 2C = ao = f (x) dx =  x dx = x3 TT = [ T3− (−T3)] = T2.  3T 3 3T T T T T 2 n 1 2 4T  (1) i.e. C = T and since C = − . 2 2  3  n 1 n 2 2  ( 1) n 1 2   = T . =   2 2 3 12 n 1 n 4T §3.10 Complex notation for Fourier series: Using Euler's identities: ix = cos x + i sin x = cis x, i = ix = cos x  i sin x = cis (x) exp (i x) = e exp (i x) = e and cos x = sin x = exp  ix   exp(ix) 2 exp  ix   exp(ix) 2 = = 1 cis  ix   cis(ix) 2 cis  ix   cis(ix) 2 = e e = ix ix e 2 e 2 ix ix , . Supposing that the Dirichlet conditions are satisfied for the function f(x) and further that f(x) is continuous at x. Then      f(x) = 1 ao +  [ an cos n  x + bn sin n  x ], 2 T T n 1 Putting    , we get T f(x) = 1 ao + 2   [ an n 1 e inx e 2 inx + bn e inx e 2 inx  inx a n  ib n inx a n  ib n = 1 ao +  [ e ]. +e 2 2 2 n 1 a  bn a  ib n = Cn, n = Cn, 1 ao = Co , we get Put n 2 2 2 92 ]= Advanced Engineering Mathematics  inx  Cn e f(x) = n  =   Cn n  Chapter 3- Fourier Series e inTx  , and a n  ib n 1 T f (x) [sin n  x  i sin n  x ]dx Cn = =  T T 2 2T T  =       inx T dx,  f (x) e 1 T 2T T a n  ib n 1 T Cn = = f (x) [sin n  x + i sin n  x ]dx =  T T 2 2T T      inTx  dx. 1 T =  f (x) e 2T T 1 T Co =  f (x) dx 2T T Therefore the Fourier series for f(x) can be written as: f(x) =   Cn n  e inTx  , where  1 T Cn =  f (x) e 2T T (4)  inTx  dx, (5) If f (x) is discontinuous at x, the left hand side of (4) shall be replaced by [f(x+0) + f(x0)] / 2 . Example 3.19: ax Expand the function f(x) = e , x  [ , ], period = 2 in complex Fourier series. Solution: 93 Advanced Engineering Mathematics Chapter 3- Fourier Series It is clear that: 1  ax inx 1  (a  in)x 1 1  (a  in)x   dx = e e e e dx = = Cn =     2 a  in  2  2  = 1 e(a  in)   e(a  in)   ,  2(a  in)  in and since e Cn = n = cos n  i sin n = cos n = (1) , then n  a a  n 1 ea .ein  ea .ein  = (1)  e  e  = (1) sinh a ,  (a  in)  2 2(a  in)   (a  in)  x e e where sinh x = 2 ax f(x) = e = x . Therefore sinh a  (1)n inx e .   n  (a  in) Exercises 3.3 1. Expand the following function in a Fourier series in the interval ( , ), period 2 f (x)   2x x 0x . x 0 [Answer:   2  cos x cos3x cos5x cos 7x      +   2 2 2 4   12  3 5 7  sin x sin 2x sin 3x sin 5x       ] . +3  2 3 5  1  2. Expand f(x) = 1, 0 < x < , period = 2 in a Fourier sine series and using it to calculate the sum of the series 1  1 1 1 +  + . 3 5 7 94 [Answer:  ]. 4 Advanced Engineering Mathematics 3. Expand f (x)   2x x6 Chapter 3- Fourier Series 0<x<4 4x < 8 in a Fourier series of period 8. [Answer: 16  x  1 cos 3x  1 cos 5x  1 cos 7 x   ] cos  2 2 2 2 4 4 4 4 3 5 7    4. Expand f(x) = cos x , 0 < x < , in a Fourier sine series. How should f (x) be defined at x = 0 and x =  so that the series will converges to f(x) for [Answer: f(x) = 0x ? 8  n sin 2nx , f(0) = f() = 0].   n 1 4n 2  1 5. Expand in a Fourier series f(x) = cos x , 0 < x <  if the period is  and compare with the result of problem 4, explaining the similarities and difference if any. [Answer: there is no difference between them.]. 6. Expand f (x)   x 8x 0<x<4 4x<8 in a series of (a) sines (b) cosines. [Answer: (a) 32  1 2   n 1 n 2 sin n sin nx , (b) 2 + 2 8 n 16  2cos 2  cos n  1 cos nx ] +  2  n 1 7. Prove that for 0x 2    cos 2x cos 4x cos 6x cos8x       (a) x(  x) = 2 2 2 6  12  2 3 4 95 n 2 8 Advanced Engineering Mathematics (b) x(  x) = 8  Chapter 3- Fourier Series  sin x sin 3x sin 5x sin 7x        3  3 3 3 3 5 7  1  and using it to show that: 2  1  i)  ,  2 6 n 1 n 8. i) Prove that for 2  1   , ii)  2 6 n 1 n 2  (1) n 1  . iii)   3 32 n 1 (2n  1)   x  ,  sin x sin 2x sin 3x sin 5x       , x = 2 2 3 5  1  ii) By integrating the result of (i), show that for  < x < , 2  cos x cos 2x cos3x cos 4x    4 x =      , 2 2 2 3  12 2 3 4  2 iii) By integrating the result of (ii), show that for  < x < ,  sin x sin 2x sin 3x sin 4x  x(  x) ( + x) = 12       . 3 3 3 3 2 3 4  1  9. (i) Show that for  < x < , x cos x =  1 3 4 5  2  sin 3x  sin 4x  sin 2x   . sin x + 2  sin 2x  2 2.4 3.5 4.6 1.3  10. By differentiating the result of problem 7 (b) prove that for 0  x  , 96 Advanced Course in Mathematics x= Chapter 1- Fourier Series   4  cos x cos3x cos5x cos7x        . 2 2 2 2   12 3 5 7  11. By using problem 7 and Parseval's identity, show that : 4  1  i)  ,  4 90 n 1 n 6  1  ii)   6 945 n 1 n 12. Using the expansion of f(x) = sin x , 0 < x <  in a Fourier cosine series to  1  2  8 1 1 1 show that       = . 2 2 2 2 2 2 2 2 16 3 .5 5 .7 7 .9 1 .3  13. Show that  1 4  1 6  ii)   6 960 n 1 (2n  1)   i)  4 96 n 1 (2n  1) 2 4  39 14. Show that .     = 2 2 2 2 2 2 2 2 2 2 2 2 16 1 .2 .3 2 .3 .4 3 .4 .5 4 .5 .6 1 1 1 2 1 2  x 15. Show that, the expansion of f (x) =  in a Fourier series in the interval 12 4  cos x cos 2x cos3x cos 4x       . ( , ), period = 2 is  2 2 2 2 2 3 4  1  16. Expand the function f(x) = cos2x in a series of sines in the interval (0, ) . [Answer:  4  x 17*. Expand the function f (x) = e  sin x  3sin 3x 5sin 5x 7sin 7x     ].  2 2 2 2 2 2 2 2  2  1  2 3 2 5 2 7 in a Fourier series in the interval (T, T). 97 Advanced Course in Mathematics [Answer: Chapter 1- Fourier Series T e e 2T T T T T T  (1) + T(e  e )  n 2 2 2 n 1 T  n   +T(e  e )  n (1) n 2 2 2 n 1 T  n  cos( nx )+ T sin( nx ) ]. T   18*. Prove that if the function f(x) is even and we have f(x+ ) = f(  x) then its 2 2 Fourier series in the interval (, ) represents as odd cosines harmonic, and if the   function f(x) is odd and we have f(x + ) = f(  x), then its Fourier series in the 2 2 interval (, ) represents as odd sines harmonic. 19. Expand f(x) = | x |, 1< x < l in Fourier series period = 2. [Answer: 4  cos(2n+1)x 1 ].   2 2 n  0 (2n+1)2 2 20. Using the expansions of the functions x and x in the interval (0, ) in cosines  cos nx 3x 2  6x  22 , (0  x  ). = series, prove the equality:  2 12 n 1 n 4 21. Expand the function f (x) = sin x in a Fourier series. [Answer: 3 1 1  cos 2x + cos4x]. 8 2 8 22. For the periodic functions with the property f( 2  x) = f(x), prove that: bn = 0 for all n = 1, 2, . 23. For the periodic functions with the property f(2  x) = f (x) prove that: 98 Advanced Course in Mathematics Chapter 1 ao = an = 0 for all n = 1, 2, 3, . 3 24. Expand the function f (x) = cos x, in a Fourier series. [Answer: 3 1 cos x + cos3x] 4 4 25. Expand the function f (x) = sin3x in a Fourier series x  [ , ], period = 2. [Answer: sin 3x] 99 Advanced Engineering Mathematics Chapter 4- Fourier Integrals CHAPTER 4 FOURIER INTEGRALS § 4.1 Theorem 4.1: Let f(x) satisfies the Dirichlet conditions in every finite interval (T, T) and   f (x) dx converges (i.e. f(x) is absolutely integrable in (, ), then:   f(x) =   A(u)cos ux  B(u)sin ux  du , 0 (1) 1   f (x) cos ux dx ,   (2) 1  B(u) =  f (x) sin ux dx .   (3) where A(u) = The result (1) holds if x is a point of continuity of f(x). If x is a point of discontinuity, we must replace f(x) by [f(x+0) +f(x0)]/2 as in the case of Fourier series. Note that the above conditions are sufficient but not necessary. The right hand side of (1) is sometimes called Fourier integral expansions of f(x). Example 4.1: Find the Fourier integral of 1  f (x)   1 2 0  x a x =a x a Solution:  f(x) =   A(u)cos ux  B(u)sin ux du , 0 1 1  1 a f (x) cos ux dx =  cos ux dx = where A(u) = sin ux  aa =  u    a = 1 2 sin u a, [sin u a  sin (u a)] = u u 100 Advanced Engineering Mathematics Chapter 4- Fourier Integrals 1 a 1  B(u) =  f (x) sin ux dx =   sin ux dx = 0.   a 1  Therefore f(x) =  sin ua cos ux du =  1 2 0u 0  21 x a x =a x a §4.2 Equivalent forms of Fourier's integral theorem: Fourier's integral theorem can also be written in the forms:  1  f(x) =   f(t) cos u(t  x) dt du ,  u  0 t  f(x) = 1  iux  1   iut iu(t  x) = e du f (t) e dt f (t) e dtdu .     2  2     (4) (5) Where it is understood that if f(x) is not continuous at x the left side must be replaced by [f(x+0) +f(x0)]/2. The case of even functions:  1  f(x) =  cos ux  f(t) cos ut dt du , if f(x) is even.  u 0 t 0 (6) The case of odd functions: f(x) =  1  sin ux   f(t) sin ut dt du , if f(x) is odd.  u 0 t 0 If we put F (u) = 1  iut  f (t) e dt , 2  in (5) , we get : f(x) = (7) (8) 1  iux  F(u) e du 2  (9) The function F (u) is called the Fourier transform of f (x) and is sometimes written F(u) = F{f (x)}. The function f (x) is the inverse Fourier transform of F (u) 1 and is written f(x) = F {F(u)}. The constants preceding the integral signs in (8) and (9) were here taken as equal to 101 Advanced Engineering Mathematics Chapter 4- Fourier Integrals 1 . However, they can be any constants different from zero so long as their 2 product is 1 . The above is called the symmetric form. 2 Equation (8) can be written as follows: F(u) = = 1  iut  f (t) e dt = 2  1   f (t)  cos ut  isin ut  dt = 2  1   f (t) cos ut dt + 2  i   f (t)sin utdt . 2  (10) Equation (9) can be written as follows: f(x) = 1  iux  F(u) e du = 2  1   F(u)  cos ux  isin ux du = 2  1  i  =  F(u) sin ux du .  F(u) cos ux du + 2  2  (11) If f(x) is an even function, then f(t) sin u t is odd and f(t) cos u t is even. Then the second integral on the right of (10) is zero and the result can be written Fc(u) = 2   f (t) cos ut dt = 2 0 2  f (t) cos ut dt . 0 (12) From (12) it is clear that Fc (u) = Fc (u) = so that Fc (u) is an even function and Fc(u) cos u x is even and Fc(u) sin u x is odd. Then the second integral on the right of (11) is zero and the result can be written f(x) = 2  2  Fc (u) cos ux du =   Fc (u) cos ux du . 2 0 0 (13) and we call Fc(u) and f(x) Fourier cosine transforms of each other. A similar result holds for odd functions Fs(u) = 2  f (t) sin ut dt . 0 (14) 102 Advanced Engineering Mathematics f(x) = Chapter 4- Fourier Integrals 2  Fs (u) sin ux du . 0 (15) Example 4.2: Find the Fourier integral of the function 1  f (x)   1 2 0  x a x =a x a Solution It is clear that f(x) - an even function, then the Fourier transform of f (x) is Fc(u) = 2  f (t) cos ut dt = 0 For u = 0, we obtain Fc(0) = 2a  cos ut dt = 0 2  sin ux  a 2  sin ua  = , u  0.   u  0   u  2a 2  f (t) dt =   dt = 0 0 2 a.  The graphs of f(x) and Fc(u) for a = 3 are shown in the figure below 103 Advanced Engineering Mathematics Chapter 4- Fourier Integrals Since we have Fc (u) = 1 2  sin ua cos ux  du =  1 i.e.  2 u 0 0   sin ua cos ux 2 Hence u 0 2  2  sin ua  cos ux du   0   u  2  sin ua  , then f(x) =   u  x a x = a. x a x a   du =   2 0  x =a x a  sin u Putting x = 0 and a = 1, we get 2  0 u  sin u du =   u du = . Example 4.3: Solve the integral equation   1 u  f (x) cos ux dx = 0 0 0  u 1 u >1 and then prove that  sin 2 u  0 u 2  du = . 2 Solution: 2   f (x) cos ux dx = Fc(u), and then Fc(u) =  0 Then f(x) = = Hence x 2 0  u 1 u >1 2 1 2 21 (1  u)cos uxdu =   (1  u)cos uxdu = 0  0 2  Fc (u) cos ux du = 0 2(1  cos x)  2  1  u     0 .  2 2  2(1  cos x)  cos ux dx =   1  u   2 0  0 x  (1  cos x) Taking the limit as u 0+ , we find  0 x 2 But this integral can be written as : 104 0  u  1. u >1 dx =  . 2 Advanced Engineering Mathematics  (1  cos x)  0 x 2 Chapter 4- Fourier Integrals 2  2sin x  sin 2 u  2 du = du =  dx =  2 2 2 0 u 0 x Example 4.4: ax Find the Fourier cosine (sine) transform of f(x) = e  cos ux  x du = e , x 0, that  2 2 0 u 1 (a > 0, x > 0), and then prove  u sin ux  x du = e , x 0.  2 2 0 u 1 Solution: From (12) we determine the Fourier cosine transform Fc(u) = 2  f (t) cos ut dt = 0 2  at  e cos ut dt = 0 at 2 e  = [a cos ut + u sin ut] 0 =  a2  u2 2 a ,  a2  u2 and from (13) , we get f(x) = 2 2 a 2a  cos ux cos ux du du , =    0  a2  u2  0 a2  u2 2a  cos ux ax i.e. du = e   0 a2  u2 (i) From (14) we determine the Fourier sine transform: 2  f (t) sin ut dt = 0 Fs(u) = = at 2 e  [a sin ut  u cos ut] = 0  a2  u2 and therefore f(x) = i.e. 2  at  e sin ut dt = 0 2 u ,  a2  u2 2 2 u sin ux du ,   0  a2  u2 2  u sin ux ax du = e  2 2  0 u a (ii) 105 Advanced Engineering Mathematics Chapter 4- Fourier Integrals In (i), (ii) put a = 1, we get  cos ux  2 0 u 1 du =  x e , 2 x 0  u sin ux  x du = e ,  2 2 0 u 1 x 0. §4.3 Parseval's identities for Fourier integrals: If Fc(u) and Gc(u) are Fourier cosine transforms of f(x) and g(x) respectively, then:   0 0  Fc (u) G c (u)du =  f (x) g(x) dx (16) Similarly if Fs(u) and Gs(u) are Fourier cosine transforms of f(x) and g(x), then   0 0  Fs (u) Gs (u)du =  f (x) g(x) dx (17) In the special case where f(x) = g(x), (16) and (17) becomes respectively :   2 2 (18) 2 (19)   Fc (u) du =   f (x) dx , 0 0   2   Fs (u) du =   f (x) dx , 0 0 Relations (16)  (19) are known as Parseval's identities for integrals. Similar relations hold for general Fourier transforms. Thus if F(u) and G(u) are Fourier transforms of f(x) and g(x) respectively, we can prove that      F(u) G(u) du =  f (x) g(x) dx , (20) where the bar signifies the complex conjugate. Example 4.5: Verify Parseval's identity for Fourier integral for the Fourier transforms of Example 2.2. Solution: 106 Advanced Engineering Mathematics  Chapter 4- Fourier Integrals  2   f (x) dx =   F(u) du , We must show that 2  1  where f (x) =  1 2 0    x a x = a and F (u) = x  a a 2 2 sin ua  u 2 L.H.S. =   f (x)  dx =  1 dx = 2a a   2 2 2a 2  sin ua 2a  sin v . = 2a. du dv = = R.H.S. =   F(u) du =   2 2      u  v 2 (Using the result of Example 2.3) then L.H.S = R.H.S.  sin 2 u du , directly. The method can also be used to find  2  u § 4.4 The convolution theorem: If F(u) and G(u) are the Fourier transforms of f(x) and g(x) respectively, then   F(u)G(u)e iux du =    f (t) g(x  t) dt (21)  If we define the convolution, denoted by f* g , of the functions f and g to be f*g = 1   f (u) g(x  u) du . 2  (22) Then (21) can be written in the form F { f*g } = F{f} F{g} i.e. the Fourier transform of the convolution of two functions is equal to the product of their Fourier transforms. This is called the convolution theorem for Fourier transforms. Example 4.6: Solve for g(x) the integral equation   g(u) 2  (x  u)  a 2 du = 2 1 x b 2 ; a, b are constants and 0 < a < b. 107 Advanced Engineering Mathematics Chapter 4- Fourier Integrals Solution: Taking the Fourier transform of both sides of the given integral equation, we have the convolution theorem 2 F {g} F { we have F { = (Since 2 2 1 x a 1 x b 2 2 }=F{ 2 1 x b }, 2 (i) iut 1  e dt =  2  t 2  b 2 }= 2  cos ut 1 2  cos buv = dt dv =    0 t 2  b2 b  0 v2  1 1 2   ub 1   ub = .e . .e b 2 b 2 cos ut 2 t b 2 is even and Similarly: F { 2 1 x b 2 sin ut 2 t b }= 2 iut is odd and e = cos ut + i sin ut ). 1   ub .e b 2 Substituting in (i) we get 2 F {g} Then G(u) = 1   ua 1   bu .e e = . b 2 a 2 a (a  b)u . e b 2 Thus g(x) = 1  iux a  (a  b)u iux e du =  e G(u) du = 2b  e 2   (a  b)u a e a  (a  b)u  = e cos ux du = [(b  a)cos ux  u sin ux] =  2 2 0 b (b  a)  x b 0 = a (b  a) . b (b  a) 2  x 2 i.e. g(x) = a(b  a) 2 2 b  x  (b  a)    . 108 Advanced Engineering Mathematics Chapter 4- Fourier Integrals Exercises 4.1 1. Prove that the Fourier transforms of 1  x 2 f(x) =   0 x <1 , is 2 2  u cos u  sin u  and then prove that   3   x >1 u    sin x  x cos x    0 2. If f(x) = x  3 x .  cos dx = 16 2  3 1 0 0  x <1 , find x 1 (a) Fourier sine transform, (b) Fourier cosine transform of f(x). 2  1  cos u    , b)  u  [Answers: a) 1  3. Solve for g(x) the integral equation  g(x) sin ux dx = 2  0 0  2 sin u ]  u 0  u 1 1 u  2 . u2 [Answers: g(x) = (2 + 2 cos x  4 cos2x)/( x).] x 4. Using the Fourier sine and cosine transforms of e  , x > 0 , and Parseval's identity  dx b)   1  cos x 2   5. Use problem (2) to show that a)   0   x cos x  sin x 2 6. Show that  0 x 6 2 x dx  = . 2 2 4 0 (x  1)  to prove that: a)  = , 2 2 4 0 (x  1) dx = x   dx = , 2   sin 2 x b)  0 x 2 dx =  . 2  . 15 7. Find the Fourier transform of the following functions: sgn x a) f(x) =  0 x 1 , x 1 x b) f(x) =  0 109 x 1 x 1 x  c) f (x)  0 1 ,  2 x 1 x >1 , x 1 Advanced Engineering Mathematics  cos x d) f(x) =   0 x 2  x  2 Chapter 4- Fourier Integrals sin x  e) f(x) =   0 x  2  x  2 [Answers: a) f(x) = 2   sin u cos u   b) f(x) =  sin ux du , x  1,   0  u 2 u  2   sin u cos(u  1)   c) f(x) =  cos ux du ,  2  0  u u  u 2  cos 2 d) f(x) = cos ux du ,   0 1  u2 2  sin u e) f(x) = sin ux du .]   0 1  u2 110 2  1  cos u sin ux du , x  1,  u  0 Advanced Engineering Mathematics Chapter 5 - Laplace transforms CHAPTER 5 LAPLACE TRANSFORMS § 5.1 The Laplace Integral: Let f (t) be a function of t specified for all t, 0 < t < , and consider T lim lim  est f (t) dt , (5.1)  0 T   for some complex s. If the double limit exists, we shall write this as the improper integral  e 0 st (5.2) f (t) dt , and denote it by L{f (t)} = F(s) . If the function of t is indicated in terms of a small (lower case) letter, such as f(t), g(t), y(t) , etc., the Laplace integral or the Laplace transform of the function is denoted by the corresponding capital letter, i.e. F(s), G(s), Y(s), etc. In other cases, a tilde (~) can be used to denote the Laplace integral (or the Laplace transform). Thus, for example, the Laplace transform of f(t) is f (s) or L{f(t)} or F(s). We shall write f(t) .  . F(s) to denote that L{f(t)} = F(s). Definition: Exponential order A function f(t) is said to be of exponential order for t > T if we can find t constants M and  such that |f(t)|  M e for t > T . Theorem 5.1 If f(t) is piecewise continuous in every finite interval 0  t  T and is of exponential order for t > T, then the double limit (7.1) exists. It must be emphasized that the stated conditions are sufficient to guarantee the existence of the Laplace transform. If the conditions are not satisfied, however, the Laplace transform may or may not exist. Thus the conditions are not necessary for the existence of the Laplace transform. 111 Advanced Engineering Mathematics Chapter 5 - Laplace transforms § 5.2 Some important properties of Laplace transforms: In the following list of theorems we assume that all functions f(t) possesses a Laplace transform i.e. the double limit (7.1) exists. 1. Linearity property: Theorem 5.2: If L{f(t)} = F(s) and L{g(t)} = G(s), then for any constants a and b (7.3) L{a f(t) + b g(t)} = a F(s) + b G(s). Proof:  L{a f(t) + b g(t)} =  est [ a f (t)  b g(t)] dt = 0  = a e st  f (t) dt + b  est g(t) dt .■ 0 0 The result is easily extended to more than two functions. This implies that the taking of a Laplace transform is a linear operation. 2. Translation property or Shifting property: Theorem 5.3: First translation theorem at If L{f(t)} = F(s), then L{e f(t) } = F(s  a). Proof:  at L{e f(t) } =  e 0 st e at  f (t) dt =  es(t  a) f (t) dt = F(s  a). ■ 0 Theorem 5.4 Second translation theorem f (t  a) If L{f(t)} = F(s), and g(t) =  0 Proof:  L{g (t) } =  e a 0 st a g(t)dt =  e  0 st t>a t<a  g(t)dt +  est g(t)dt = a  =  est (0)dt +  est f (t  a)dt =  es(u  a) f (u)du = 0 a 0 112 as , then L{g(t) } = e F(s). Advanced Engineering Mathematics =e sa  e su 0 as f (u)du = e Chapter 5 - Laplace transforms F(s).■ Theorem 5.5 Change of scale property If L{f (t)} = F(s), then L{f (at)} = Proof:  L{f (at)} =  e st 0 1 s F( ). a a  s ( u ) f (at)dt =  e a 0   su 1 s u 1 f (u) d( ) = ( )  e a f (u) du = F( ).■ a a a a 0 Theorem 5.6 Laplace transform of derivatives: If f (t) is continuous for 0  t  N and of exponential order for t > N while f '(t) is piecewise continuous for 0  t  N and L{f(t)} = F(s), then L{f '(t)} = s F(s)  f(0). Proof: Using integration by parts, we have: T  st  T L{f '(t)} =  e f '(t)dt = lim  e f '(t)dt = lim e f (t)  s  est f (t)dt  = 0 T  0 T    0 0  T  sT   st st = lim e f (T)  f (0)  s  e f (t)dt  = s  e f (t)dt  f (0)  lim esT f (T) = T  T    0 0   st T st    = s  est f (t)dt  f (0) = s F(s)  f (0). 0   Using the fact that f(t) is of exponential order  as t  , so that lim esT f (T) = 0 T  for s > . ■ Theorem 5.7: If in Theorem 5.6, f (t) fails to be continuous at t = 0, but lim f (t) =f(0+) exists (but t 0 is not equal to f(0), which may or may not exist) , then L{f '(t)} = s F(s)  f(0+). Theorem 5.8: If in Theorem 5.6, f (t) fails to Continuous at t = a, then: as L{f '(t)} = s F(s)  f(0)  e [f(a+)  f(a)], 113 Advanced Engineering Mathematics Chapter 5 - Laplace transforms where [f (a+)  f (a)] is called the jump at the discontinuity t = a. Theorem 5.9: If f (t) and f '(t) are continuous for 0  t  N and of exponential order for t > N while f "(t) is piecewise continuous for 0  t  N, and L{f (t)} = F(s), then 2 L{f "(t)} = s F(s)  s f (0) – f '(0). Proof: By Theorem 5.6: L{g '(t)} = s G(s)  g (0). Let g (t) = f '(t). Then L{f "(t)} = s G(s)  g(0) = s L{g(t)}  f '(0) = s L{f '(t)}  f '(0) = 2 = s{s F(s)  g (0)}  f '(0) = s F(s)  s f(0)  f '(0). ■ The generalization to higher order derivatives can be proved by using Mathematical induction. If f (t) and f '(t) have discontinuities, appropriate modification can be made as in Theorems 7.7 and 7.8. Theorem 5.10: If f (t), f '(t),, f t > N while f L{f (n) (n) (n1) (t) are continuous for 0  t  N and of exponential order for (t) is piecewise continuous for 0  t  N, and L{f (t)} = F(s), then n (t)}= s F(s)  s n1 f (0) s n2 f '(0)    s f (n2) (0)  f (n1) (0). Theorem 5.11: Multiplication by powers of t: If L{f (t)} = F(s), then n L{t f (t)} = (1) Proof: n dn ds n n (n) F(s) = (1) F (s), where n = 1, 2, 3, .  We have F(s) =  est f (t)dt . Then by Leibnitz's rule for differentiating under the 0 integral sign,    dF d  st st = est {t f (t)}dt = L{t f (t)}. e f (t)dt =  = F '(s) = e f (t)dt    ds ds 0 0 0 s 114 Advanced Engineering Mathematics Thus L{t f (t)} =  Chapter 5 - Laplace transforms dF =  F '(s), which proves the theorem for n = 1.We shall using ds Mathematical induction. Assume the theorem true for n = k, i.e. assume  e st k 0 k (k) [t f (t)]dt = (1) F (s) Then (1)  d  st k e [t f (t)]dt =   est {t k 1 f (t)}dt ,  ds 0 0  i.e.  est {t k 1 f (t)}dt = (1) k+1 0 F (k+1) (s) (2) It follows that if (1) is true, i.e. if the theorem holds for n = k, then (2) is true, i.e. the theorem holds for n = k + 1. But the theorem is true for n = 1. Hence it is true for n = 2 and n = 3, etc., and thus for all positive integer value of n. ■ Theorem 5.12 Laplace transform of integrals: t If L{f (t)} = F(s) , then L{  f (u)du } = 0 Proof: F(s) . s t Let g(t) =  f (u)du . Then g '(t) = f (t) and g (0) = 0. Taking the Laplace transform of 0 both sides, we have L{f (t)} = L{g '(t)} = s L{g (t)}  g(0) = s L{g(t)}. t 1 F(s) F(s) Thus L{g (t)} = L{f (t)} = , or L{  f (u)du } = .■ s s s 0 Theorem 5.13 Division by t:  f (t) f (t) If lim exists and L{f (t)} = F(s), then L{ } =  F(u)du . t t 0 t s Proof: Let g (t) = f (t) / t or f (t) = t g (t) , d d F(s) = L{f (t)} = L{t g (t)} =  L{g (t)} =  G(s). ds ds s c c s Thus G(s) =   F(u)du =  F(u)du . 115 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Now since g (t) satisfies the conditions of Theorem 5.1, it follows that lim G(s) = 0. s   f (t) } =  F(u)du . ■ Then c must be infinite and so G(s) = L{ t s Theorem 5.14 Convolution theorem: t If L{f (t)} = F(s), L{g (t)} = G(s), then L{  f (u)g(t  u)du } = F(s) G(s). 0 We call the above integral the convolution of f and g; and write t f * g =  f (u)g(t  u)du . 0 § 5.3 Laplace Transforms of elementary functions: Example 5.1: 2 Prove that (a) L{1} = 1/s, s > 0, (b) L{t} = 1/s , s > 0, at (c) L{e } = 1/(s  a), s > a Solution:  (a) L{1} =  e 0  (b) L{t} =  e st (1)dt = lim  e st T  0 st 0 T T (t)dt = lim  t e dt = st T  0 st lim e T  ( s) T 0 (1 e sT ) 1 = , s > 0. s s T  = lim T st st dt = lim  t e  (1) e 2  = s 0 T   ( s) st  sT  1 1 = lim   e 2  T e  = , if s > 0. s s T   s 2  s2 at  (c) L{e } =  e 0 st at e T dt = lim  e T  0 (s  a)t T  (s  a)t  dt = lim  e = T   (s  a)  0  (s  a)T  1 = , if s > a. = lim 1 e  T   (s  a)  s  a Another method: It is clear that L{f ' (t)} = s L{f (t)]  f (0) ( see Theorem 5.6 ) (a) Let f(t) = 1 , then f '(t) = 0 , f (0) = 1 then L{0} = 0 = s L{1}  1 or L{1} = 1/s. 116 Advanced Engineering Mathematics Chapter 5 - Laplace transforms (b) Let f(t) = t , then f '(t) = 1 , f (0) = 0, using part (a) we get L{1} = 1/s = s L{t}  0 2 or L{ t } = 1/ s . By using mathematical induction we can similarly n show that L{ t } = n!/s (n+1) for any positive integer n. at at at at (c) Let f(t) = e . Then f '(t) = a e , f (0) = 1, then L{a e } = s L{e }  1. at at at i.e. a L{e } = s L{ e }  1 or L{ e } = 1/(s  a). Example 5.2: Prove that (a) L{ sin at } = Solution: a a 2  s2 , (b) L{ cos at } =  T 0 T  0 s a 2  s2 , s > 0, a) L{ sin at } =  est sin at dt = lim  est sin at dt =  esT [ssin aT  a cos aT] a a   = = lim   T   s 2  a 2  a 2  s 2 s2  a 2  Another method: 2 i) From Theorem 5.9 we have L{f "(t)} = s L{ f (t)}  s f(0) – f '(0). 2 (a) Let f (t) = sin at, f "(t) =  a sin at, f (0) = 0, f '(0) = a, then 2 2 2 L{ a sin at} = = s L{sin at}  a. 2 i.e. (s + a ) L{sin at} = a or L{sin at} = a a 2  s2 . 2 (b) Let f (t) = cos at, f "(t) = a cos at, f (0)= 1, f '(0) = 0 , then 2 2 L{ a cos at} = s L{cos at}  s. 2 2 i.e. (s + a ) L{cos at} = s or L{sin at} = a 2  s2 ii) From Example 5.1 part (c) we have 1 s  ia iat = . L{e } = s  ia s 2  a 2 iat But e s . (*) = cos at + i sin at. Hence 117 Advanced Engineering Mathematics  iat L{e } =  e st Chapter 5 - Laplace transforms  [cosat  isin at]dt =  e  st cos atdt  i  est sin atdt = 0 0 = L{cos at} + i L{sin at} = s  ia s2  a 2 0 . On equating real and imaginary parts, we get: s a L{cos at} = , L{sin at} = . s2  a 2 s2  a 2 Example 5.3: Using the linearity property to prove that: a s (a) L{sinh at} = , (b) L{cosh at} = if s > |a|. 2 2 2 2 s a s a Solution: at at at at a) L{sinh at} = L{½(e  e )} = ½L{e }  ½L{e } = = a 1 1 1  =  for s > |a|. 2  s  a s  a  s 2  a 2 at at at at b) L{cosh at} = L{½(e + e )} = ½L{e } + ½L{e } = = s 1 1 1  =  for s > |a|. 2  s  a s  a  s 2  a 2 Short Table Of Special Laplace Transforms o N f(t) L{f(t)} = F(s) conditions 1 1 1/s s>0 2 t 1/s 2 s>0 3 tn n!/sn+1 s > 0, n=0,1,2, 4 eat 1/(sa) s>a 5 sin at a/(s + a ) 6 cos at s/(s + a ) 7 sinh at a/(s  a ) 8 cosh at s/(s  a ) 9 t p 2 2 s>0 2 2 s>0 2 2 s > |a| 2 2 s > |a| (p+1)/s (p+1) 118 s > 0 , p > 1 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Example 5.4: p Prove that: L{ t } = Solution: (p  1) s p 1 , if s > 0 and p > 1.  L{ t } =  est t p dt . Let s t = u and note that in order for the integral to converge we p 0 p must have s > 0. Then L{ t } = (p  1) 1  p u u e du =  s p 1 s p 1 0 the restriction p > 1 occurs because the integral defining the gamma function converges if and only if p > 1. Example 5.5: Find the Laplace transforms of each of the following: 3t a) 4 e , 2 b) 3 t , c) 5 cos 4t , d) sin t , e) L{ 5 }. t Solution: 4 4  , s > 3. s  (3) s  3 2! 6 2 2 , s > 0. b) L{3 t } = 3 L{ t } = 3 = 3 3 s s s 5s c) L{5 cos 4t} = 5 L{cos 4t} = 5  . 2 2 2 s 4 s  16  d) L{ sin t} = , s > 0. s 2  2 ( 1 )  5 ½ e) L{ , s > 0. } = 5 L{ t } =  5 5 2  5 s s t 3t a) L{4 e } = 4 L{e 3t }= Example 5.6: Find the Laplace transforms of each of the following: t a) e cos 2t, 2 2 3t t b) 4t  3cos2t + 5 e , 2t e) e (3 cos 6t  5 sin 6t), Solution: a) Using Theorem 5.3 we get 119 4t c) t e , d) e cosh 5t, f) t sin at , g) t cos at . 2 Advanced Engineering Mathematics s 1 t L{e cos 2t} = F(s+1) = 2 Chapter 5 - Laplace transforms (s  1) 2  4  s 1 s 2  2s  1 , 2 t t b) L{4t  3cos2t + 5 e } = 4L{ t } 3 L{ cos2t } + 5L{e } = 1 8 3s 5 ,    s  1 s3 s 2  4 s  1 s3 s2  4 2! 2 2 3t 2 c) L{t e } = F(s3) = , where F(s) = = L{t }, s3 (s  3)3 s s4 s4 4t , d) L{cosh 5t} = . Then L{e cosh 5t} =  s 2  25 (s  4)2  25 s 2  8s  9 =4 2! 3 s 5 e) L{3 cos 6t  5 sin 6t} = 3L{cos 6t}  5L{sin 6t} = =3 2 s s  36 5 2 6  3s  30 2 . s  36 s  36 3s  24 3(s  2)  30 2t Then L{e (3 cos 6t  5 sin 6t)} = F(s+2) = = , (s  2)2  36 s 2  4s  40 a , we have f) Since L{sin at} = 2 2 s a d  a  2as  , L{t sin at} =    ds  s 2  a 2  2 2 2 s a  g) Since L{cos at} = 2 L{t cos at} = d2 ds 2 2 s s a 2  , we have '   "  s   s   a 2  s 2  2s(s 2  3a 2 ) . =  2  =  = 2 2 3  s  a 2   s2  a 2   s2  a 2 2  s a       Example 5.7:  f (t) a) Prove that  0 t  sin t b) Show that  0 Solution: t  dt   F(s) ds provided that the integrals converge. 0 dt   . 2  a) From Theorem 5.13,  e 0 st f (t) t  dt =  F(u)du . s 120 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Then taking the limit as s 0+, assuming the integrals converge and the required  f (t)  result is obtained. i.e.  dt   F(s) ds . t 0 0 2 b) Let f (t) = sin t . Then F(s) = 1/(s + 1).  sin t  ds   = tan 1 s  . dt   Hence  2 2 0 0 t 0 s 1 Example 5.8: Find L{f (t)} if 3 a) f(t) =  0 cos(t   )3  3 b) f(t) =   0 0<t<5 t 5 t>  3 t  3 Solution: 5 est = a) By definition, L{f (t)} =  est f (t)dt =  est (3)dt   est (0)dt = 3 s  0 5 0 5   0 3(1  e5s ) , = s 3 b) L{f (t)} =  e st  (0) dt   e  0 =e   s  su 3 e 0 st 3 cos udu = cos(t   )dt = 3 2 s s 1 e  s 3  s(u   ) e 3 0 cos udu = . Theorem 5.15: Periodic functions: Let f (t) have period T > 0 then L{f (t)}  1 T e (1  esT ) 0 st f (t)dt . Proof: f (t) periodic function with period T, then f(u) = f(u + T) = f(u + 2T) == f(u + n T). We have  L{f (t)} =  e 0 st T f (t)dt =  e 0 st 2T f (t)dt   e st T 3T f (t)dt   est f (t)dt   . 2T In the second integral let t = u + T, in the third integral let t = u + 2T, etc. Then 121 Advanced Engineering Mathematics T L{f (t)} =  e su 0 T T f (u)du   e s(u  T) 0 T Chapter 5 - Laplace transforms T f (u  T)du   es(u  2T) f (u  2T)du   T 0 =  esu f (u)du  esT  esu f (u)du  e2sT  esu f (u)du   = 0 0 sT = [1 + e 2sT +e 3sT +e 0 T +]  esu f (u)du = 0 1 T 1 e 0 sT e su f (u)du .■ Example 5.9: Find the Laplace transform of the function of period 2 which in the interval 0  t<  sin t . 0  t < 2 is given by f (t)     t < 2 0 Solution: The graph of this function, often called a rectified sine wave or a half wave rectified sine curve. 2s Since T = 2, let R = 1/(1  e ) , we have:   est (ssin t  cos t)  st su sin t dt = R  L{f(t)} = R  e f (u)du = R  e  = 2 s 1 0 0   0 2  1  es  1 . = =R   2 s 2  s  1  (1  e )(s  1) Theorem 5.16: Behaviour of F(s) as s   : If L{f(t)} = F(s) , then lim F(s) = 0. s  Theorem 5.17: Initial-value and Final-value theorem: If the indicated limits exists, then 122 Advanced Engineering Mathematics Chapter 5 - Laplace transforms a) lim f(t) = lim s F(s) , t 0 b) lim f(t) = lim sF(s). s  t  s 0 Example 5.10: 3t Illustrate Theorems 7.16 and 7.17 for the function f(t) = 2 e . Solution: 3t 3t We have f(t) = 2 e , F(s) = L{2 e } = 2/(s+3). It is clear that: 2 = 0, s  s  s  3 2s = 2, lim sF(s)  lim s  s  s  3 2s = 0, lim sF(s)  lim s 0 s 0 s  3 (1) lim F(s)  lim (2) (3) lim f (t)  lim 2e3t = 2, (4) lim f (t)  lim 2e3t = 0, (5) t 0 t 0 t  t  i.e. lim F(s)  0 , lim f (t) = lim sF(s) ; lim f (t) = lim sF(s) s  t 0 s  t  s 0 § 5.4 Some Special Functions: I. Bessel Functions: We define a Bessel function of order n by   tn t2 t4 Jn (t) = 1      . n    2.(2n 2) 2.4.(2n 2)(2n 4) 2 (n  1)   Some important properties are n 1. Jn (t) = (1) Jn(t) if n is a positive integer. 2n 2. Jn+1(t) = Jn (t)  Jn1(t), t d n J '0 (t) = J1(t). 3. t J n (t)  t n J n 1 (t) , dt   t(u  1)  =  J n (t)u n . 4. exp    2u  n    This is called the generating function for the Bessel functions. 5. Jn (t) satisfies Bessel's differential equation: 123 Advanced Engineering Mathematics 2 2 Chapter 5 - Laplace transforms 2 t y"(t) + t y'(t) + (t  n ) y(t) = 0. It is convenient to define Jn (it) = i n In(t), where In(t) is called the modified Bessel function of order n. Example 5.11: a) Find L{Jo (t)} and then use it to find a) L{Jo(at)}, b) Find L{J1(t)}. Solution: It is clear that Jo(t) = 1  t2 22  t4 22.42  t6 22.42.62 . Then by using the binomial theorem, we get 1 1 2! 1 4! t6 6!    = L{ Jo (t)} =  s 22 s3 22.42 s5 22.42.62 s7  1 1 1  1 1 1.3 1 1.3.5 1 = 1      = 1   s  2 s 2 2.4 s 4 2.4.6 s6  s  s2  1 2 1 = 2 , s 1 1 s To find L{Jo (at)} we use Theorem 5.5, i.e. L{f (at)} = F( ) . a a 1 1 . = L{Jo (at)} = 2 2 2 s a s a   1 a b) Since J '0 (t) = J1(t), L{f ' (t) } = s F(s)  f(0), then L{J1 (t)} = L{ J '0 (t) } =  [ s L{ Jo(t)}  1] = 1  s s 1 Another method: Using differential equations, a) The function Jo(t) satisfies the differential equation, Taking the Laplace transform of both sides and using Jo (0) = 1, J 'o (t) = 0, y = L{ Jo(t)}, we have  d  2 dy 0, s y  s.1  0   [sy  1]   ds  ds 124 2 = s2  1  s 2 s 1 . t J"o (t) + J 'o (t) + t J o (t) = 0 . Advanced Engineering Mathematics from which y= c 2 s 1 Chapter 5 - Laplace transforms dy sy dy sds . Thus and by integrating we get   ds y s2  1 s2  1 . Now lim sy(s)  lim cs s  s 2  1 s   c , and lim J o (t)  1 . Thus by the initial-value t 0 Theorem 5.17(a), we have c = 1 and so L{Jo (t)}= 1 2 s 1 . II The Unit Step Function: The unit step function, also called Heaviside's unit step function, is defined as 0 U(t  a) =  1 t<a t>a It is possible to express various discontinuous functions in terms of the unit step function. It is clear that: est st st st L{U(ta)} =  U(t  a)e dt   e (0)dt   e (1)dt = 0+ s 0 0 a a    a e as = , if s > 0. s Example 5.12: 4 Express the function f (t) =  2 t<3 , in terms of the unit step function and thus t>3 obtain L{f (t)}. Solution: 0 We have f (t) = 4 +  2 t<3 =42 t>3 0  1 t<3 = 4  2 U(t3). t>3 4 e3s 4  2e3s . L{f(t)} = L{4  2 U (t3)} = L{4}  2 L{U (t3)}=  2 = s s s It is clear that, the second translation Theorem 5.4 can be written as follows: 125 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Theorem 5.4': If L{f (t)} = F(s) and g (t) = f (t  a) U (t  a), then as L{g (t)} = L{f (t  a) U (t  a)} = e F(s). Example 5.13: 0  Find L{f (t)} where f(t) =  t a  t<0 0 < t < a , a > 0. t>0 Solution: L{f(t)}=   f (t)e st 0 a dt =  t e 0 st  dt +  a est dt = a a   test est  aest = =     2 s s a   a 0 aeas aeas 1 aeas 1  eas    = = . s s s2 s2 s2 Using the unit function, we can express f(t) as follows : f(t) = t [ U(t)  U(t  a)] + a U(t  a) = t U(t)  (t  a)U(t  a). Hence L { f (t) } = 1 s2  eas s2 . Example 5.14: 0 Find L { f(t) } where f(t) =  n t<0 , n = 0, 1, 2, 3, na < t < (n+1)a Solution: 126 Advanced Engineering Mathematics Chapter 5 - Laplace transforms  L{f(t)} =  f (t)est dt = 0  (n 1)a =   n 0 na  ne st nest dt   n  0 s  n =  e nas  e(n 1)as  =   n 0 s 1   nas = = (n 1)a  na e s n 0 eas =  1  s(1  eas ) s eas  1 . We can express f(t) as follows: f(t) = [U(t  a)  U(t  2a)] + 2[U(t  2a)  U(t  3a)] + + + n [U(t  n a)  U(t  {n+1}a)] + ... =  = U(t  a) + U(t  2a) + U(t  3a) ++ U(t  n a) =  U(t  na) .   e  nas and L{f(t)} =  L{U(t  na)} =  n 1 n 1 s = e n 1 as = 1 s(1  eas ) s(eas  1) . III The Unit Impulse Function: The unit impulse function, also called Dirac delta function. Consider the function 0  t  1 /  F(t) =  t> 0 where  > 0 It is geometrically evident that as   0 the height of the rectangular region increases indefinitely and the width decreases in such a way that the area is always equal to 1,  i.e.  F (t)dt  1 . 0 127 Advanced Engineering Mathematics Chapter 5 - Laplace transforms This idea has led some engineers and physicists to think of a limiting function denoted by approached by F (t) as   0. This limiting function they have called the impulse function of Dirac delta function. Some of its properties are   1)  (t)dt = 1,  2)  (t)g(t)dt = g(0), 0 3)  (t  a)g(t)dt = g(a), 0 0 for any continuous function g (t). It is clear that  L{F(t)} =  e st 0  F (t)dt =  e st 1 0  ( )dt +  e   st  st 1  st (0) dt = ( )  e dt = 1 e , s  0 s 1  (1  s  s 2  2 / 2! ) = lim (1   ) = 1. 2! s  0  0  0 Although mathematically speaking such a limit: lim F(t) does not exist, so and lim L{F (t)}  lim  0 that L{ lim F(t) } is not defined, manipulations or operations using it can be made  0 rigorous. It proves useful to consider (t) = lim F(t) to be such that L{(t) } = 1.  0 Example 5.15: Show that L{(ta)} = e Solution: st Let g(t) = e and from  as .   (t  a)g(t)dt = g(a), we get 0 L{(ta)} =  (t  a)est dt = e . st 0 IV Null Functions: t If N (t) is a function of t such that for all t > 0,  N(u)du = 0 0  1 t=1 4  We call N(t) a null function. The function f (t) =  1 t=2  0 otherwise  128 Advanced Engineering Mathematics Chapter 5 - Laplace transforms is a null function. In general, any function which is zero at all but a countable set of points is a null function. Example 5.16: Indicate which of the following null functions are:  12 t=0  2 t = 3  a) f(t) =  , b) f(t) =  1 1  t  0 , c) f(t) = (t). 2 otherwise  1 otherwise  2 Solution: t a) Since  f (u)du = 0, for all t > 0, f(t) is a null function. 0 t b)  f (u)du = 0, if t < 1, t 0 t  f (u)du =  (1)du = t  1, if 1  t  2, 0 t 1 2  f (u)du =  (1)du = 1, if 0 t 1 t > 2. Since  f (u)du  0 for all t > 0, f(t) is not a null function. t 0 c)  (u)du = 1, for all t > 0, (t) is not a null function. 0 Example 5.17: Find L{f(t)} where f(t) is given as shown in the figure. Solution: It is clear that t 1 1  t  2  f(t) = 1 2 t 4, 5  t 4  t  5  129 Advanced Engineering Mathematics  L{f(t)} =  f (t)e = 0 s e s 2 st e  dt =  (t  1)e 2s s 2 2  1 5s e s 2  e st 4s s 2 Chapter 5 - Laplace transforms 4 dt   e st 2 5 dt   (5  t)est dt = 4 . We can express f(t) by using the unit step function U(t  a) as follows : f(t) =(t  1)[U(t  1)  U(t  2)]+[U(t  2)  U(t  4)] + (5  t)[U(t  1)  U(t  2)]= = (t  1)U(t  1)  (t  2)U(t  2)  (t  4)U(t  4) + (t  5)U(t  5). Then L{f (t)} = e s s 2  e2s s 2  e5s s 2  e4s s 2 . Example 5.18: Find L{f (t)} where f (t) is the square wave function  1  f (t) = f (t + 2a) = 1  1  2na < t < (2n+1)a, (2n+1)a < t < (2n+2)a, , t < 0, n = 0, 1, 2, 3, . Solution: f (t) = [U(t)  U(t  a)]  [U(t  a)  U(t  2a)] + [U(t  2a)  U(t  3a)]  = = U(t)  2U(t  a) + 2 U(t  2a)  2U(t  3a)+ =  = U (t)  2  (1) n 1U(t  na) . n 1 L{f (t)} = L{ U(t)}    2  (1)n 1 L{U(t  na)} = n 1  nas  1 n 1 e = =  2  (1) s s n 1 1 2eas . = 2 as s s(1  e )  L{f (t)} = L{U (t)}  2  (1) n 1 L{U(t  na)} = n 1 130 Advanced Engineering Mathematics Chapter 5 - Laplace transforms  nas  1 1 2eas n 1 e =  2  (1) = 2 . as s s s  s(1 e ) n 1 Example 5.19: Find L{f (t)} where f (t) is saw tooth wave function t   n, f(t) = f(t + a) =  a  0, na < t < (n+1)a , n = 0, 1, 2, 3, . t<0 Solution: f (t) is a periodic function, then a e 0 st a f (t)dt   est t dt = a 0 1 eas eas 1  eas eas 1  1  st a 1 1 1 1 as  .   = t e =   (a  )e = 0 as s as s s as  s  s as 2 as 2 as 2 1  eas eas  1 eas as  L{f (t)} =  .  / (1  e ) = 2  2 as s  as s(1 e )  as  Example 5.20: t sin u Prove that L{ 0 u Solution: du}  1 1 1 tan ( ) s s t sin u du . Then f(0) = 0 and u 0 sin t f '(t) = or t f '(t) = sin t. Taking the Laplace transform, t d 1 . L{ t f '(t)} = L{sin t} or  [sF(s)  f (0)] = ds s2  1 d 1 i.e. [sF(s)] =  . ds 1  s2 Let f (t) =  1 Integrating we get s F(s) =  tan s + C. By the initial value Theorem 5.17(a), 131 Advanced Engineering Mathematics Chapter 5 - Laplace transforms lim s F(s) = lim f(t) = f(0) = 0, so that C = /2. s  t 0 Thus s F(s) = /2  tan 1 1 s = tan 1 (1/s) or F(s) = tan (1/s) / s. § 5.5 The Inverse Laplace Transform: If the Laplace transform of a function f(t) is F(s) , i.e. if L{f(t)} = F(s), then f(t) is called an inverse Laplace transform of F(s) and we write symbolically f(t) = 1 1 is called the inverse Laplace transformation operator. For 2t } = 1/(s+2), we can write L {1/(s+2)} = e . L {F(s)}, where L example, since L{e 1 2t Since the Laplace transform of a null function N(t) is zero, it is clear that if L{f (t)} = F(s), then also L{f(t)+N(t)} = F(s). From this it follows that we can have two different functions with the same Laplace transform. If we disallow null functions (which do not in general arise in cases of physical interest). This result is indicated in Theorem 5.18: Lerch's theorem: If we restrict ourselves to functions f(t) which are piecewise continuous in every finite interval 0  t  N and of exponential order for t > N, then the inverse Laplace 1 transform of F(s) [i.e. L {F(s)} = f(t)] , is unique . We shall always assume such uniqueness unless otherwise stated. o Short Table Of Special Inverse Laplace Transforms N F(s) f(t) = L 1 2 1/s 1 t s>0 s>0 1/sn 1/(sa) tn1/ (n1)! eat s > 0, n = 1,2, 1/(s + a ) (sin at)/a s/(s + a ) cos at s > 0, a  0 s>0 1/(s  a ) (sinh at)/a s > |a| s/(s  a ) cosh at s > |a| 3 4 5 6 7 8 9 1/s 2 2 2 2 2 2 2 2 2 1/s p t p1 1 F(s) /(p) 132 Conditions s>a s>0,p>0 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Some Important Properties Of Inverse Laplace Transforms: In the following list we have indicated various important properties of inverse Laplace transforms. Note the analogy of properties 1-8 with the corresponding properties on Laplace transforms. 1. Linearity Property: 1 1 1 L {a F(s) + b G(s)} = a L {F(s)} + b L {G(s)}. Proof: Since L{a f (t) + b g (t)} = a L{f(t)} + b L{g(t)} = a F(s) + b G(s). 1 1 1 Then L {a F(s) + b G(s)} = a f (t) + b g (t) = a L {F(s)} + b L {G(s)}. 2. First translation or shifting property: 1 at 1 If L {F(s)} = f (t), then L {F(s  a)} = e f (t). 3. Second translation or shifting property: 1 1 as If L {F(s)} = f(t) , then L {e F(s)}= f (t  a)U(t  a). 4. Change of scale property: 1 1 If L {F(s)} = f (t), then L {F(as)} = f (t/a) / a. 5. Inverse Laplace transform of derivatives: 1 1 (n) If L {F(s)} = f (t), then L {F n n (s)} = (1) t f (t). 6. Inverse Laplace transform of integrals: 1 1  If L {F(s)} = f (t), then L {  F(u)du } = f (t) / t. s 7. Multiplication by sn: 1 1 If L {F(s)} = f (t) and f (0) = 0, then L {s F(s)} =f '(t). Thus the multiplication by s has the effect of differentiating f (t). 1 * If f(0)  0, then L { s F(s)  f(0)} = f '(t) . 8. Division by s: 1 1 If L {F(s)} = f (t), then L { F(s) }= s t  f (u)du . 0 133 Advanced Engineering Mathematics Chapter 5 - Laplace transforms 9. The convolution property: 1 1 If L {F(s)} = f (t) and L {G(s)} = g (t), then t 1 L {F(s) G(s)} =  f (u)g(t  u)du = f*g. 0 We call f*g the convolution or faltung of f and g. The proofs are left to the students. Example 5 .21: 1 Find L { 6 5s  4 2s  18 3  4s +  } = A. + 2s  3 s3 s 2  9 9s 2  16 Solution: A=L 1   5 4 2s 18 1 4 1  3          3 2 s3 s 2  9 s 2  9 3(s 2  16 ) 9 (s 2  16 )   s  2 s 9 9  3t/2 + 5t + 4(t /2!)  2cos 3t + 18(sin 3t)/3 + (1/3)(3 sinh 4t/3)/4 + 4(cosh 4t/3)/9 = 3t/2 + 5t + 2t  2cos 3t + 6 sin 3t + (1/4) sinh (4t/3) + (4/9) cosh (4t/3) = =3e =3e 2 2 Example 5.22: Find each of the following:  3s  2   s3  a) L1  , b) L1   , 2 2  s  8s  16   s  4s  20   1   s 1  d) L1  e) L1  , ,  2s  3   s2  s  1  3s  7  c) L1  , 2  s  2s  3    1   f) L1  . 5   (s  4) 2  Solution:  3s  2   3s  2  1  3(s  2)  4  1  L = a) L1   = L  =  2 2 2  s  4s  20   (s  2)  16   (s  2)  16    1   s2 4 = 3 L1  + L   = 2 2   (s 2) 16 (s 2) 16       2t 2t 2t = 3 e cos 4t + e sin 4t = e [3 cos 4t + sin 4t].  (s  4)  1   s  3   s3  = L1  = L1  = b) L1    2 2   s 2  8s  16   (s  4)   (s  4)  134 Advanced Engineering Mathematics Chapter 5 - Laplace transforms  1  4t  1  4t 4t 1  L  = e  t e = e (1  t). = L1     2 s  4   (s  4)   3(s  1)  10   3s  7   3s  7  = L1  = L1  c) L1   =  2 2  s 2  2s  3   (s  1)  4   (s  1)  4    s  1  2  1  = 3 L1  L + 5  =  2 2  (s  1)  4   (s  1)  4  t t t 3t t = 3e cosh 2t + 5 e sinh2t = e [3cosh 2t + 5 sinh2t] = 4 e  e .   1  1 3t / 2 t 1/ 2 1 1 1 1 3t / 2  12 3 e = = . d) L  e L (s  ) = 1 2 2s 3  2 2 t  2 ( )    1  2  (s  1 )  1    1  2 2 = =L    3 (s  1 )2  3    2 4 4  3    s 1  1    1  1  2 2 + =L  L   = 1 )2  3 1 )2  3 3  (s  (s    2 4 2 4   3 1 t/2 3 1 t/2 3 3 t/2 t+ t= = e cos t + sin t ]. e sin e [ 3 cos 2 2 2 2 3 3   1 4 3/ 2 4t  4t 1  1  4t 3/2 1  f) L  = e L  5  = e t / (5/2) = t e . 5  3  2 2  (s  4)   s   s 1  s 1 1  e) L  =L   1 2  s2  s  1  (s  2 )  1  Example 5.23: Find each of the following: s 1 a) L1{ }, b) L1{n(1  )} , (s 2  a 2 ) 2 s2 1 3s  71 d) L1{ }, }, e) L1{ 2 3 2 s  2s  3 s (s  1) c) L1{ 2 s (s  1) }, 2s 2  4 f) L { } (s  1)(s  2)(s  3) 1 Solution: 2s d 1  s 1 d 1   , i.e. a)      ds  s 2  a 2  (s 2  a 2 )2 2 ds  s 2  a 2  (s 2  a 2 )2  1  sin at , we have by property (5) Then since L1   a  s2  a 2  135 2 1 Advanced Engineering Mathematics Chapter 5 - Laplace transforms    d  1   t sin at s 1 L1   =  L1   .  2 2 2 2 2  2 ds 2a (s a )  s a         1 1  s  2 = 2   b) Let F(s) = n 1   = L{f (t)}. Then F'(s) = . 2 2 2 s s(s  1) s  1  s   2(1  cos t) 1 . Since L {F'(s)} = 2(1  cos t) = t f (t), then f (t) = t  1  c) Since L1    sin t , we have by repeated application of property 8, 2  s  1  1  t L1     sin udu = 1  cos t, 2  s(s  1)  0   t 1 L1     (1  cos u)du = t  sin t. 2 2  s (s  1)  0  t 1 1  1    (u  sin u)du = t 2 + cos t  1. d) L   3 2 2  s (s  1)  0 3s  7 4 1 e) By using the method of partial fractions we find that   . 2 s 3 s 1   s  2s  3  3s  7   1   1  3t t Then L1  =4e e .  4 L1   L1     s  3  s  1  s 2  2s  3  f) By partial fractions method we get 2s 2  4 1 4 7 .    (s  1)(s  2)(s  3) 6(s  1) 3(s  2) 2(s  3)   2s 2  4 1  1  t 4 2t 7 3t Thus L   = 6 e  e  e . 3 2  (s  1)(s  2)(s  3)  Example 5.24: Find each of the following:  2  1  5s  15s  11  , a) L  3  (s  1)(s  2)    3s  1 b) L1  , 2   (s 1)(s 1)     s 2  2s  3 c) L1  , 2 2  (s  2s  2)(s  2s  5)    s e) L1  , 2  (s  1)(s  1)   2s3  10s 2  8s  40  d) L1  , 2 2 s (s  9)     s  2  f) L1 es   . 2 s  4   136 Advanced Engineering Mathematics Solution: a) It is clear that 5s 2  15s  11  Chapter 5 - Laplace transforms 4 1 1 7    . 3(s  1) (s  2)3 (s  2)2 3(s  2) (s  1)(s  2)3  2  1 7 1 1  5s  15s  11  Thus L  =  e t  t 2 e2t  4te2t  e2t .  3 3 2 3  (s  1)(s  2)  2s  1 3s  1 2 , therefore b) Since   2 2 s 1  (s  1)(s  1) s 1    s   1  3s  1  1   2 L1   L1  L1  = 2L1    =  2 2 2  s 1   (s 1)(s 1)     s 1 s 1       t = 2 e  2 cos t + sin t. c) It is clear that s 2  2s  3 = 1 2 = (s 2  2s  2)(s 2  2s  5) 3(s 2  2s  2) 3(s 2  2s  5) 1 2 = .  3[(s  1)2  1] 3[(s  1) 2  4]  Then   1  t 2 1 1 t s 2  2s  3 1  L  = e sin t  . e t sin 2t = e [sin t + sin 2t].  2 2 3 3 2  (s  2s  2)(s  2s  5)  3 1 1 1 1  d) Since     , we have 2 2 2 2 9 s (s  9) s  9 s 2s3  10s 2  8s  40 10s  50  1 8 40  (2s  10   )  (2s  10  ) = 9 s s2 s 2 (s 2  9) s2  9  1  8 40 10s 50  =     . 9  s s 2 s 2  9 s 2  9  and so 2  3  50  1  2s  10s  8s  40  1  L  = 8  40t  10cos3t  sin 3t  =  3  s 2 (s 2  9)   9  1 [24 + 120 t + 30 cos 3t + 50 sin 3t]. = 27 We can also use the method of partial fractions.  1  t 1  s  e) We can apply property 8. Since L1    cos t ,   e and L  2  s  1  s  1 137 Advanced Engineering Mathematics Chapter 5 - Laplace transforms t   t (t  u) t u cos udu = e  e cosdu = then L   2    e  s  1   s  1   0 0 1 t = [ cos t + sin t  e ]. 2  s2  s 2 f) It is clear that F(s) = es  = e s  e s  s2  4 s2  4  s2  4   s   2  Now L1  = cos 2t, L1    = sin 2t. 2 2 s  4  s  4  1  1   s Application of property 3 yields   s  2  L1 es . = cos 2(t  1) U(t  1), L1 es .   = sin 2(t  1) U(t  1), s2  4  s2  4    Then property 1 (linearity) gives: L1 F(s) = [cos 2(t  1) + sin 2(t  1)] U(t  1). § 5.6 Solution of differential equations: The method of Laplace transforms is particularly useful for solving linear differential equations with constant coefficients and associated initial conditions. To accomplish this we take the Laplace transform of the given differential equation (or equations in the case of a system), making use of the initial conditions. This leads to an algebraic equations (or system of algebraic equations) in the Laplace transform and then taking the inverse, the required solution is obtained. Example 5.25: Solve y"(t) + y(t) = t, given y'(0) = 2, y(0) = 1. Solution: Take the Laplace transform of both sides of the given differential equation and let 2 2 Y = Y(s) = L{y(t)}. Then L{y"(t) + y(t)} = L{t} or s Y  s y(0)  y'(0) + Y=1/s . Since y(0) = 1, y'( 0 ) = 2 this becomes 1 1 s2 2 s Ys+2+Y= . ,Y=  s2 s 2 (s 2  1) s 2  1 1 1 s 2 1 s 3 Then Y =    =   , 2 2 2 2 2 2 2 s s 1 s 1 s 1 s s 1 s 1 1 and y = L {Y} = t + cos t  3 sin t. 138 Advanced Engineering Mathematics Chapter 5 - Laplace transforms Example 5.26: 2t Solve y"  3y' + 2y = 4 e , y(0) = 3, y'(0) = 5. Solution: 2t We have L{y"}  3 L{y'} + 2L{y} = 4 L{e } 2 [s Y  s y (0)  y'(0)] 3[ s Y  y(0)] + 2Y = 2 [s Y + 3s  5]  3[s Y + 3] + 2Y = 2 (s  3s + 2)Y + 3s  14 = Y= 4 (s 2  3s  2)(s  2)  4 . s2 14  3s 4 s2 s 2  3s  2 t 1 = 4 s2 3s 2  20s  24 2t (s  1)(s  2)2 = 4 4 7   s  1 s  2 (s  2)2 2t Thus y (t) = L {Y} = 7 e + 4 e + 4t e . Example 5.27: Solve y" + y = cos t given y(0) = 0, y'(0) = 0. Solution: L{y"} + L{y} = L{cos t} Hence Y = i.e. 2 s Y+Y= s s2  1  s  1  =   (s 2  1)2  s 2  1  s 2  1  s 1 t Then y (t) = L {Y} =  cos u sin(t  u)du = 0 1 1 1  t 1 = sin t( u  sin u cos u)  cos t( sin 2 u)  = t sin t. 2 2 2  0 2 Example 5.28: Solve y' + 4y + 4z = 0, z' + 2y + 6z = 0 given y(0) = 3, z(0) =15, Solution: [s Y  y(0)] + 4Y +4Z = 0 , [s Z  z(0)] + 2Y + 6Z = 0 (s + 4)Y + 4 Z = 3, (s + 6) Z + 2Y = 15 139 Advanced Engineering Mathematics 8 3s  42 11   , (s  2)(s  8) s  2 s  8 s 2  10s  16 3(2)  15(s  4) 15s  54 4 11    Z= . 2 (s 2)(s 8) s 2 s 8     s  10s  16 Then Y = 3(s  6)  15(4) Chapter 5 - Laplace transforms 1  2t Hence y (t) = L {Y} = 8 e + 11 e 8t 1 , z(t) = L {Z} = 4 e 2t 8t + 11 e . Example 5.29: 2 t Find the general solution of y'"  3y" + 3y'  y = t e . Solution: In this case, we assume y(0) = A , y'(0) = B, y"(0) = C, and we have: 3 2 2 [s Y  s y(0)  s y'(0)  y"(0)] 3[s Y  s y(0)  y'(0)] +3[sY  y(0)]  Y = 3 2 2 i.e. (s Y  As  B s  C)  3(s Y  As  B) + 3(sY  A) Y = Y= As 2  (B  3A)s  3A  3B  C (s  1)3 + 2 (s  1)6 2 (s  1) 3 2 (s  1)3 , , . Since A, B and C are arbitrary, so also is the polynomial in the numerator of the first 2    term on the right. We can thus write Y = , and    (s  1)3 (s  1)2 s  1 (s  1)6 invert to find the required general solution 1 1 1 2 t 5 t 2 t t t 5 t t t y =  t e +  t e +  e + 60 t e =  t e +  t e +  e + 60 t e , 2 where , , ,  are arbitrary constants. It should be noted that finding the general solution is easier than finding the particular solution since we avoid the necessity of determining the constants in the partial fraction expansion. 140 Advanced Engineering Mathematics Chapter 5 - Laplace transforms EXERCISES 5 1. Find the Laplace transforms of each of the following functions. In each case specify the values of s for which the Laplace transform exists. 4t 2t b) 3 e a) 2 e , 2 , c) 5t  3 , d) 2t  e , t e) 3 cos 5t, g) 6 sin2t  5 cos2t, h) (t + 1) , 2 f) 10 sin 6t, 2 i) sin t  cos t, j) 3 cosh 5t  4 sinh 5t. [Answers: a) 2/(s  4), s > 4, b) 3/(s+2) , s > 2, 3 3 2 d) (4 + 4s  s )/s (s+1), 2 2 2 e) 3s/(s + 25), 4 g) (12  5s)/(s + 4), 2 c) (5  3s)/s , s > 2, f) 60/(s + 36), 2 3 h) (s + 4s + 24)/s , 2 2 i) (s  2s + 4)/s(s + 4), s > 0, j) (3s  20)/(s  25), s > 5. ] 2. Find the Laplace transforms of each of the following functions: 4 3 3t a) 4t  2t + 4e 3t 4t b) t e , 2 t d) 2e sin4t, g) e 3 3t  2sin 5t + 3cos 2t, t c) e cos2t, 2t e) (t + 2) e , f) e (3 sin 4t  4 cos4t), t cosh 2t, h) e (3 sinh2t 5cosh2t). 5 4 2 2 [Answers: a) 72/s  12/s  4/(s + 3) + 3s/(s + 4)  10/(s + 25), 4 2 2 2 3 b) 6/(s+3) , c) (s+1)/(s + 2s + 5), d) 8/(s  6s + 25), e) (4s  4s + 2)/(s  1) , 2 2 2 f) (20  4s)/(s  4s + 20), g) (s + 4)/(s + 8s + 12), h) (1  5s)/(s + 2s  3).] t 3. Find a) L{e sin 2t}, t 3 b) L{(1 + t e ) }. 2 2 3 4 [Answers: a) 2/[(s + 1)(s + 2s + 5)], b) 1/s + 3/(s + 1) + 6/(s + 6) +6/(s + 3) .] 4. Show that: s 3 a) L{f (t)} = 2e /s (t  1)2 if f (t) =   0 2 t > 1, 0 < t < 1. 2 2 2 b) L{f (2t)} = (s  2s + 4)/4(s + l) (s  2) , if L{f (t)} = (s  s + 1)/(2s + 1) (s  1), t 3/(s+1) c) L{e f (3t)} = e 2 1/s /(s+1), if L{f (t)} = e s s 2, d) L{f (t)} = 2/s  e /s e /s 2t if f (t) =  t 5. Verify that 141 /s, 0  t  1, t > 1. Advanced Engineering Mathematics s a) L{f "(t)} = 2(1  e )/s Chapter 5 - Laplace transforms  t 2 f(t) =   0 if 0 < t  1, t > 1.  t 2  2 t b) L   (u  u  e u )du  = L{t  t + e }/s. 0   t 1  e u  1 1 c) L   du   n(1  ) . s 0 u  s d) L{t cos at} = s2  a 2 e) L{t sin at} = , (s 2  a 2 )2 2 2 2as (s 2  a 2 )2 2 f) L{t(3 sin2t  2cos2t)}= (8 + 12s  2s )/(s + 4) , 2 2 2 3 2 2 2 2 2 g) L{t sin t} = (6s  2)/(s + 1) , h) L{t sinh 2t} = 4s/(s  4) , 2 3 2 3 i) L{t cosh3t} = (s + 9)/(s  9) , j) L{t cos t} =(2s  6s)/(s + 1) , 2 4 3 2 2 3 k) L{(t  3t + 2) sin3t} = (6 s  18s +126 s  162 s + 432)/(s + 9) . at 6. Show that: a) L{e bt  e )/t} = ℓn[(s + b)/(s + a)], 2 2 2 2 b) L{(cos at  cos bt)/t}= ½ ℓn [(s + a )/(s + b )], c) L{(sinh t)/ t] = ½ ℓn[(s + 1)/(s − 1)],  sin 2 t 1 dt =  . 2 2 0 t d)  7. Find L{f (t)} if 3t a) f (t) =  6 2 b) f (t) = t , t c) f (t) =  0 0<t<2 2 t < 4 , where f (t) has period = 4. 0 < t < 2 and f (t + 2) = f (t). 0<t<1 , where f (t + 2) = f (t) for t > 0. 1 t < 2 2s [Answers: a) (3  3e 2s 2s 2 2s 3 2s b) (2  2e  4se  4s e )/s (l  e  t sin u  tan 1 (s  2) du  = 8. Show that a) L e2t  , u  s 2  0  142 s 2 4s  6s e 4s )/s (1 e 2 ), 2s ), c) [l  e (s + l)]/s (1  e ) .] Advanced Engineering Mathematics Chapter 5 - Laplace transforms  tan 1 s 1 .  u du  = 2  2 s s(s 1)  0  0<t<a g(t) 9. Show that f (t) =  t>a  h(t)  b) L  t  t sin u can be written as f(t) = g(t) + [h(t)  g(t)] U(t  a). e t 0 < t < 3 3(s + l) 10. Find L{f (t)} if f (t) =  [Answer: L{f (t)} = (1  e /(s+1)] t>3  0 11. Express in terms of Heaviside's unit step functions 0<t<  sin t  t 2  0<t<2 b) f (t) = sin 2t  < t < 2 a) f (t) =  4t t > 2   sin 3t t > 2  2 2 [Answers: a) f (t) = t + (4t  t ) U (t  2), b) f (t) = sin t + (sin 2t  sin t) U (t  ) + (sin 3t  sin 2t) U(t  2).]  2 t  e U(t  2)dt = e , 12. Show that a)  2 3 2 2s b) L{t U(t  2)} = (2/s )  2e 3 (1 + 2s + 2s )/s , s>0 13. Find L{f(t)} if sin t a) f (t) =  0 0<t< t> s [Answers: a) (l + e  cos t b) f (t) =  sin t , 2 0<t< t> s )/(s + 1), b) [s + (s 1) e 3 c) f(t) = sin t. , 2 2 2 ]/(s + 1), c) 6/(s + 1)(s + 9).] 14. Find the inverse Laplace transform of the following functions: a) 3/(s + 4), 2 2 2 e) (3s  12)/(s + 8), f) (2s  5)/(s  9), i) 12/(4  3s), j) (s+1)/s , 4t 4 2/3 + 3t , 2 sin 2 2 t, h) 8 t /15, 1/3 )/(1/3), 7/2 h) 1/s , ℓ) (2s + 1)/s(s + 1). c) 8 cos 4t, 5/2 g) t /24, 5 g) 1/s , k) ( s  1) /s , e) 3 cos2 2 t  3 d) 3 sin 2t, j) (t 5t/2 b) ½e d) 6/(s + 4), 2 2 3/4 [Answers: a) 3e , 2 c) 8s/(s + 16), b) 1/(2s  5), ½ k) 1+t4t /  , f) 2 cosh 3t  (5 sinh 3t)/2, 4t/3 i)  4e , t ℓ) 1 + e .] 15. Find the inverse Laplace transforms of the following functions: 2 a) (3s  8)/(4s + 25), 2 2 2 b) (5s + 10)/(9s  16), c) (3s8)/(s + 4)(4s  24)/(s  16), 143 Advanced Engineering Mathematics d) (3s  2)/s 2 5/2 Chapter 5 - Laplace transforms 5/2 7/(3s + 3) , 2 e) s/(s + 1) , 5 2 ½ 5/2 f) 3(s  1) /2s + (4s  18)/(9  s ) + (s+1)(2  s )/s , 5 2 g) s/(s + 1) , h) (3s  14)/(s  4s + 8), 2 i) 1/(8s  27) 1/3 2 2 j) (8s + 20)/(s  12s + 32), k) (3s + 2)/(4s + 12s + 9), ℓ) (5s  2)/(3s + 4s + 8). [Answers: a) (3/4) cos (5t/2)  (4/5) sin (5t/2) , b) (5/9) cosh (4t/3) + (5/6) sin (4t/3), c) 3cos 2t  4sin2t  4cosh 4t + 6 sinh 4t, ½ 3/2 2t/3 d) 6t /   8 t /3   7e 2 4 ½ f) ½  t  3t /2 + t /16 + 4 t /  + 8t 3 t 4 3/2 ½ /3 , e) 2t (3  2t)/3  , /3   4cosh 3t + 6 sinh 3t, 2t g) e (4t  t )/24, h) e (3 cos 2t  4 sin 2t), i) t 2/3 27t/8 e / [2 (1/3)], 6t 8t 4t j) 2e (4 cosh 2t + 7 sinh 2t) = 11 e  3 e , 3t/2 k) (3/4) e 2t/3 ℓ) e 3t/2  (5/8) t e , [25 cos (2 5 t/3)  24 5 sin (2 5 t/3)]/15.] 16. Find the inverse Laplace transforms of the following functions: 2s 2 a) e 2 3s /s , b) 8e ½ s 2 2 3s /(s + 4), c) e / (s + 1) , d) e 2 2 2s /(s  2s + 5) , e) se 3 /(s  3s + 2), 2 f) (s + 1)/(s + 2s+2) , g) ℓn[(s + 2)/(s + 1)], h) 1/s (s + 1), i) (s + 2)/s (s + 3) , 3 5 5 j) 1/s(s+1) , k) 1/(s  1) (s + 2), ℓ) s/(s  2) (s+1). [Answers: a) (t  2)U(t  2), (t  3) d) ½U(t  3) e t 2t g) (e  e 2 (t  2) e t j) 1 e (1 + t + ½ t ), 3 ½ U(t  2)/  , t ] U(t  2), f) ½t e sin t, 3t i) 2t/3 + (l  e )/9, et 4 3 2 2t [ t  (4t /3) + (4t /3)  (8t/9) + (8/27)  e /243, k) 72 2 4 e) [2e h) l  t + ½ t e , t 2t 2(t  2) sin 2(t  3), )/t, c) (t  1) b) 4U(t  3) sin 2(t  3) , 2 t ℓ) e (t /36) + (t /54) + (t /54) + (t/81)  (1/243) + e / 243.] 17. Find the inverse Laplace transforms of the following functions: 1 2 a) tan (2/s ), 2 2 d) 1/(s + 1)(s + 1), 2 2 c) 1/(s + 2) (s 2), b) 1/(s + 3)(s  1), 2 2 2 e) s /(s + 4) , 3 2 g) s/(s + 4) , 3 f) 1/(s + 1) , h) (3s + 16)/(s  s  6), 3 i) (2s 1)/(s  s), 2 j) (s + 1)/(6s + 7s + 2). t [Answers: a) (2 sin t sinh t)/t, b) ¼(e  e t d) ½(sin t  cos t + e ), 2 3t 2t ), c) ¼ t e e) ½ t cos 2t + ¼ sin 2t , g) (t sin 2t)/64  (t cos 2t)/32, 3t 2t h) 5e  2e , 144 2t + (e  e 2t )/16, 2 f) [(3  t ) sin t  3t cos t]/8 , t i) 1  3e , t/2 j) ½e 2t/3 e /3.] Advanced Engineering Mathematics Chapter 5 - Laplace transforms 18. Use partial fractions to show that: 2   2t t/2 t a) L1  11s  2s  5  = 5 e  3 e /2 + 2 e ,  (s  2)(2s 1)(s 1)    4t b) L1  27 12s  = 3 e  3 cos 3t, 2  (s  4)(s  9)   3  c) L1  s 16s  24  = ½ sin 4t + cos 2t  sin2t,  s4  20s2  64  d)  2  t t L1  s  2s  3  = ½(2t  1) e + 3 e /2, 2  (s 1) (s 1)   3  2 2t 2t e) L1  3s 3s  40s  36  = (5t + 3) e  2t e , (s2  4)2     3t 2t t t s2  3 f) L1  = 3e /50  e /25  (e cos 2t)/50+ (9e sin 2t)/25.   (s  2)(s 3)(s2  2s  5)    s g) L1   = ½ sin t sinh t, 2 2  (s  2s  2)(s  2s  2)    3 2 t h) L1  2s s 1  = ½ sin t + ½ t cos t  t e . 2 2 2  (s 1) (s 1)  2   t 2t 3t i) L1  2s  6s  5  = ½ e  e + 5 e /2,  s3  6s2 11s  6    t s 5 j) L1  = 2 e + 3 sin t  2 cos t.  2  (s 1)(s 1)  t 19. Solve y" + 2y' + 5y = e sin t, y(0) = 0, y'(0) = 1. 2 t 2 2 [Answer: y= e (sin t + sin 2t), Y = (s + 2s + 3)/(s + 2s + 2)(s + 2s + 5).] 20. Solve y" + 9y = cos 2t, if y(0) =1, y(½) = 1. 2 2 2 [Answer: Y = (4s)/5(s + 9) + 12/5(s + 9) + s/5(s + 4), y = (4/5) cos 3t + (4/5) sin3t + (1/5) cos2t.] 2 ½ 21. Solve ty" + y' + 4ty = 0, y(0) = 3, y'(0) = 0. [Answer: Y = 3/(s + 4) , y =3Jo(2t).] 22. Solve y"  t y' + y = 1, y(0) =1 , y'(0) = 2. 2 [Answer: Y = (l/s) + (2/s ) + (c 145 e  12 s2 s 2 ), y = 1 + (c + 2) t, c = 0.] Advanced Engineering Mathematics Chapter 5 - Laplace transforms 23. Solve each of the following by using L.T. a) y"(t) + 4y (t) = 9t, y (0) = 0, y'(0) = 7, t b) y"(t)  3y'(t) + 2y = 4t + 12 e , y (0) = 6, y'(0) = 1, 2 c) y"  4y' + 5y = 125t , y(0) = y'(0) = 0, d) y" + y = 8 cos t, y(0) = 1, y'(0) = 1, t e) y"'(t)  y = e , y(0) = y'(0) = y"(0) = 0, f) y (4) + 2y" + y = sin t , y(0) = y' (0) = y"(0) = y'"(0) = 0. t 2t t [Answers: a) 3t + 2sin 2t, b) 3e  2e + 2t + 3 + 2e , 2 2t c) 25t + 40 t + 22 + 2e (2 sin t  11cos t), d) cos t  4 sin t + 4t cos t, t ½t e) t e /3 + e t (½ cos ( 3 t/2) + 5 3 sin ( 3 t/2)/36 ½e , 2 f) ((3  t ) sin t  3t cos t)/8.] t 24. Solve y' + 2z' = t, y"  z = e , subject to conditions y(0) = 3 , y'(0) = 2 , z(0) = 0. 2 t t [Answer: y = 2 + ½t + ½e  (3 sin t)/2 + ½cos t, z = 1½e + (3 sin t)/2 ½cos t.] 25. Solve y' = 2y  3z, z' = z  2y subject to y(0) = 8, z(0) = 3. 4t t 4t t [Answer: y = 5e + 3e , z = 5e  2e .] t 26. Solve y" + z' + 3y = 15 e , z"  4y' + 3z = 15sin 2t subject to y(0) = 35, y'(0) =  48, z(0) = 27, z'(0) = 55. t [Answer: y = 30 cos t  15 sin 3t + 3e + 2 cos 2t, t z = 30cos 3t  60 sin t 3e + sin 2t.] 27. Solve y'  z'  2y + 2z = sin t, y" + 2z' + y = 0 if y(0) = 0, y'(0) = z(0) = 0. 2t t t/3 [Answer: y = e /9 + 4e /45  (1/5) cos t  (2/5) sin t + t e 2t t , t z = e /9  e /9 + t e /3.] t 28. Solve y' + 2z" = e , y' + 2y  z = 1, if y (0) = z(0) = z'(0) = 0. t at [Answer: y = 1 + e  e bt e t at , z = l + e  be where a = ½ (2  bt ae 2 ), b= ½(2 + , 2 )] t 29. Solve t y"  z' = sin t, 3y" + 3z" = te 3cos t, given that y(0) = 1, y'(0) = 2, z(0) = 4, z"(0) = 0. 2 t [Answer: y = (2t /3) + (5t/3)  (2/3)  (e /3), 146 Advanced Engineering Mathematics Chapter 5 - Laplace transforms 2 t t z = (2t /3) + (8/3) + (e /3) + (t e /3) + cos t.] 30. Solve 2y"  y' + 9y  z"  z' 3z = 0, 2y" + y' + 7y  z" + z'  5z = 0, given that y(0) = y'(0) = 1, z(0) = z'(0) = 0. t t [Answer: y = (e + 2cos 2t + sin 2t)/3, z = (2e 2cos 2t  sin2t)/3.] 147 Advanced Engineering Mathematics Chapter 5 - Laplace transforms REFERENCES (1) Advanced Mathematics for Engineers and Scientists By Murray R. Spiegel. (2) Advanced Calculus, By Murray R. Spiegel. (3) Laplace Transforms, By Murray R. Spiegel. (4) Vector Analysis, By Murray R. Spiegel. (5) Differential Equations, By Jr. Frank Ayres. (Schaum's outline series in Mathematics-McGraw-Hill book company) (6) A course in Mathematics Parts I, II, III. By Dr. Naguib G. Bakhoom. (7) Mathematics for Engineers, By Dr. Abdel-Latif El-Sedeek. (8) Mathematics for Engineering Students (1984) By Dr. A. A. Azim, Dr. M. Shams, and Dr. Labib Iskandar (9) Solved Examples for Engineering Students (1983) By Dr. A. A. Azim, Dr. M. Shams, and Dr. Labib Iskandar. (10) Mathematics for Engineering Students Part II By Dr. Abdel-Lateef El Sedeek and Dr. S. Elnahwy (Faculty of Engineering  Cairo University) (11) Special functions By Earl D. Rainville. (12) Special functions of Mathematical Physics and Chemistry By Ian N. Sneddon (Oliver & Boyd LTD.) (13) The theory of ordinary differential equations By J. C. Burkill (Oliver and Boyd LTD.) (14) Integration of ordinary differential equations By E. L. Ince (Oliver and Boyd LTD.) (15) An elementary treatise on differential equations and their applications. By H.T.H. Piaggio. (16) Volume and integral By W. W. Rogosinski (0 & B) (17) Vector Methods By D.E. Rutherford (0 & B) (18) Multiple Integrals, Field theory and Series By B. M. Budak and S. V. Fomin. (19) University Mathematics By J. Blakey (A textbook for students of Science and Engineering) (20) Laplace Transform theory By M. G. Smith. (21) Problems in Mathematical Analysis - Under the editorship of B. Demidovich. (22) Differential and Integral Calculus By N. Piskunov. 148 Advanced Engineering Mathematics Chapter 6 - Special Functions CHAPTER 6 SPECIAL FUNCTIONS § 6.1 The Gamma Function: The gamma function denoted by (s) is defined by the 2  (s) =  e x x s 1dx , nd Eulerian integral (s > 0) 0 (6.1) This is improper integral with respect to the upper limit, s < l improper with respect to the lower limit. This integral convergent for all s > 0. Integrating by parts we get  (s+1) =  e x s 0 x dx =  e x s  x 0  + s  e x x s 1dx = s (s), 0 (6.2) this is recurrence formula for the gamma function. The recurrence relation (6.2) is a difference equation which has (6.1) as a solution. By repeating the application of the recurrence relation (6.2) we get: (s+1) = s (s) = s(s  l) (s1) = s(s  l)(s  2) (s2) = = s(s  l)(s  2) (s  k)(sk) , (k < s) (6.3) It is clear that (n+1) = n(n  l)(n  2) 2.1 (1) , if n is positive integer,  But (1) =  e 0  x 0 x dx =  e 0 x dx =  e x  = 1. 0 Therefore (n + l) = n! , n = 1, 2, 3,  (6.4) For this reason (n) is sometimes called the factorial function. 2 Putting x = t in the integral (6.1) we get   0 0 2 (s) =  e x x s 1dx = 2  e t t 2s 1dt , s>0 149 (6.5) Advanced Engineering Mathematics It is clear that ( 1 ) = 2 2  e 0 t2 Chapter 6 - Special Functions dt . To calculate ( 1 ) , we shall prove that Gauss integral 2      2 t  e dt =  . 0 2 2 Put A =  e x dx =  e y dy .     2 2 2 Then A =  e x dx .  e y dy =   =   e  (x 2  y 2 )  2 2 dxdy = 4   e(x  y ) dxdy 00   Changing to polar coordinates (r, ) where x = r cos , y = r sin , the last integral becomes  2 2 A =4    e r 2  0 r  0 and so A =  0  . It follows that ( 1 ) = 2 It is clear that (n+ 1 ) = (n– 1 )(n– 3 ) 1 ( 1 ) = 2 n 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00 2 r dr dθ =   e r d(r 2 ) = , 2 (n) 1.0000 0.9514 0.9182 0.8975 0.8873 0.8862 0.8935 0.9086 0.9314 0.9618 1.0000 2 2 2 . 1.3.5 (2n  1) 2n = (2n)! 2(2n) n! . Table of values and graph of the gamma function 150 (6.6) Advanced Engineering Mathematics Chapter 6 - Special Functions For any positive p (p+1) = p (p) For a positive integer n (n+1) = n! For p = 1/2 ( 1 )   For p = n + (n+ 1 ) = (n+ 1 ) = 2 2 2 (2n)! 2(2n) n!  Values of Gamma Function  (s)   e x x s 1dx, for 0 (1  x  2) The value of the (x) for x < 1 (x  0, 1, 2, ... ) and x > 2 can be calculated by means of the following formulas: (x+1) = x (x), and (x) = (x1). (x1) How To Use: • You need to find the (l.075) • Find the row with x = 1.07 • Find the column 5, the number Inside the cell is 0.961918 See How To Read Numbers In Tables. • The result is 0.961918 x 0 1 2 3 4 5 6 7 8 9 1.00 1 0.999424 0.99885 0.998277 0.997707 0.997139 0.996572 0.996008 0.995445 0.994885 1.01 0.994326 0.993769 0.993214 0.992661 0.99211 0.991561 0.991014 0.990469 0.989925 0.989384 1.02 0.988844 0.988306 0.987771 0.987236 0.986704 0.986174 0.985645 0.985119 0.984594 0.984071 1.03 0.98355 0.983031 0.982513 0.981997 0.981484 0.980972 0.980461 0.979953 0.979446 0.978941 1.04 0.978438 0.977937 0.977437 0.97694 0.976444 0.975949 0.975457 0.974966 0.974477 0.97399 1.05 0.973504 0.97302 0.972538 0.972058 0.971579 0.971103 0.970627 0.970154 0.969682 0.969212 1.06 0.968744 0.968277 0.967812 0.967349 0.966887 0.966427 0.965969 0.965512 0.965057 0.964604 1.07 0.964152 0.963702 0.963254 0.962807 0.962362 0.961918 0.961476 0.961036 0.960598 0.960161 1.08 0.959725 0.959292 0.958859 0.958429 0.958 0.957573 0.957147 0.956723 0.9563 0.955879 1.09 0.955459 0.955042 0.954625 0.954211 0.953797 0.953386 0.952976 0.952567 0.95216 0.951755 151 Advanced Engineering Mathematics x 0 1 2 Chapter 6 - Special Functions 3 4 5 6 7 8 9 1.10 0.951351 0.950948 0.950548 0.950148 0.94975 0.949354 0.948959 0.948566 0.948174 0.947784 1.11 0.947396 0.947008 0.946623 0.946238 0.945856 0.945474 0.945095 0.944716 0.944339 0.943964 1.12 0.94359 0.943218 0.942847 0.942477 0.942109 0.941743 0.941378 0.941014 0.940652 0.940291 1.13 0.939931 0.939574 0.939217 0.938862 0.938508 0.938156 0.937805 0.937456 0.937108 0.936761 1.14 0.936416 0.936072 0.93573 0.935389 0.935049 0.934711 0.934374 0.934039 0.933705 0.933372 1.15 0.933041 0.932711 0.932382 0.932055 0.931729 0.931405 0.931082 0.93076 0.93044 0.930121 1.16 0.929803 0.929487 0.929172 0.928858 0.928546 0.928235 0.927925 0.927617 0.92731 0.927004 1.17 0.9267 0.926397 0.926095 0.925794 0.925495 0.925197 0.924901 0.924606 0.924312 0.924019 1.18 0.923728 0.923438 0.923149 0.922862 0.922575 0.92229 0.922007 0.921724 0.921443 0.921164 1.19 0.920885 0.920608 0.920332 0.920057 0.919783 0.919511 0.91924 0.91897 0.918702 0.918435 x 0 1 2 3 4 5 6 7 8 9 1.20 0.918169 0.917904 0.91764 0.917378 0.917117 0.916857 0.916599 0.916341 0.916085 0.91583 1.21 0.915576 0.915324 0.915073 0.914823 0.914574 0.914326 0.91408 0.913834 0.91359 0.913348 1.22 0.913106 0.912865 0.912626 0.912388 0.912151 0.911916 0.911681 0.911448 0.911216 0.910985 1.23 0.910755 0.910526 0.910299 0.910072 0.909847 0.909623 0.909401 0.909179 0.908959 0.908739 1.24 0.908521 0.908304 0.908088 0.907873 0.90766 0.907447 0.907236 0.907026 0.906817 0.906609 1.25 0.906402 0.906197 0.905992 0.905789 0.905587 0.905386 0.905186 0.904987 0.904789 0.904593 1.26 0.904397 0.904203 0.904009 0.903817 0.903626 0.903436 0.903247 0.90306 0.902873 0.902688 1.27 0.902503 0.90232 0.902137 0.901956 0.901776 0.901597 0.901419 0.901242 0.901067 0.900892 1.28 0.900718 0.900546 0.900375 0.900204 0.900035 0.899867 0.8997 0.899533 0.899368 0.899204 1.29 0.899042 0.89888 0.898719 0.898559 0.898401 0.898243 0.898086 0.897931 0.897776 0.897623 152 Advanced Engineering Mathematics x 0 1 2 3 Chapter 6 - Special Functions 4 5 6 7 8 9 1.30 0.897471 0.897319 0.897169 0.89702 0.896872 0.896724 0.896578 0.896433 0.896289 0.896146 1.31 0.896004 0.895863 0.895723 0.895584 0.895446 0.89531 0.895174 0.895039 0.894905 0.894772 1.32 0.89464 0.89451 0.89438 0.894251 0.894123 0.893997 0.893871 0.893746 0.893623 0.8935 1.33 0.893378 0.893257 0.893138 0.893019 0.892901 0.892784 0.892669 0.892554 0.89244 0.892327 1.34 0.892216 0.892105 0.891995 0.891886 0.891778 0.891671 0.891565 0.89146 0.891356 0.891253 1.35 0.891151 0.89105 0.89095 0.890851 0.890753 0.890656 0.89056 0.890464 0.89037 0.890277 1.36 0.890185 0.890093 0.890003 0.889913 0.889825 0.889737 0.88965 0.889565 0.88948 0.889396 1.37 0.889314 0.889232 0.889151 0.889071 0.888992 0.888914 0.888836 0.88876 0.888685 0.888611 1.38 0.888537 0.888465 0.888393 0.888323 0.888253 0.888184 0.888116 0.888049 0.887983 0.887918 1.39 0.887854 0.887791 0.887729 0.887667 0.887607 0.887548 0.887489 0.887431 0.887375 0.887319 x 0 1 2 3 4 5 6 7 8 9 1.40 0.887264 0.88721 0.887157 0.887105 0.887053 0.887003 0.886953 0.886905 0.886857 0.88681 1.41 0.886765 0.88672 0.886676 0.886633 0.88659 0.886549 0.886509 0.886469 0.88643 0.886393 1.42 0.886356 0.88632 0.886285 0.886251 0.886217 0.886185 0.886153 0.886123 0.886093 0.886064 1.43 0.886036 0.886009 0.885983 0.885958 0.885933 0.88591 0.885887 0.885865 0.885844 0.885824 1.44 0.885805 0.885787 0.885769 0.885753 0.885737 0.885722 0.885708 0.885695 0.885683 0.885672 1.45 0.885661 0.885652 0.885643 0.885635 0.885628 0.885622 0.885617 0.885612 0.885609 0.885606 1.46 0.885604 0.885603 0.885603 0.885604 0.885606 0.885608 0.885611 0.885616 0.885621 0.885626 1.47 0.885633 0.885641 0.885649 0.885658 0.885668 0.885679 0.885691 0.885704 0.885717 0.885732 1.48 0.885747 0.885763 0.88578 0.885798 0.885816 0.885836 0.885856 0.885877 0.885899 0.885922 1.49 0.885945 0.88597 0.885995 0.886021 0.886048 0.886076 0.886104 0.886134 0.886164 0.886195 153 Advanced Engineering Mathematics x 0 1 2 3 Chapter 6 - Special Functions 4 5 6 7 8 9 1.50 0.886227 0.88626 0.886293 0.886328 0.886363 0.886399 0.886436 0.886474 0.886512 0.886551 1.51 0.886592 0.886633 0.886675 0.886717 0.886761 0.886805 0.88685 0.886896 0.886943 0.88699 1.52 0.887039 0.887088 0.887138 0.887189 0.887241 0.887293 0.887346 0.8874 0.887455 0.887511 1.53 0.887568 0.887625 0.887683 0.887742 0.887802 0.887863 0.887924 0.887986 0.888049 0.888113 1.54 0.888178 0.888243 0.888309 0.888376 0.888444 0.888513 0.888582 0.888653 0.888724 0.888796 1.55 0.888868 0.888942 0.889016 0.889091 0.889167 0.889244 0.889321 0.8894 0.889479 0.889559 1.56 0.889639 0.889721 0.889803 0.889886 0.88997 0.890055 0.89014 0.890226 0.890313 0.890401 1.57 0.89049 0.890579 0.890669 0.89076 0.890852 0.890945 0.891038 0.891132 0.891227 0.891323 1.58 0.89142 0.891517 0.891615 0.891714 0.891814 0.891914 0.892015 0.892117 0.89222 0.892324 1.59 0.892428 0.892533 0.892639 0.892746 0.892854 0.892962 0.893071 0.893181 0.893292 0.893403 x 0 1 2 3 4 5 6 7 8 9 1.60 0.893515 0.893628 0.893742 0.893857 0.893972 0.894088 0.894205 0.894323 0.894441 0.894561 1.61 0.894681 0.894801 0.894923 0.895045 0.895169 0.895292 0.895417 0.895543 0.895669 0.895796 1.62 0.895924 0.896052 0.896182 0.896312 0.896443 0.896574 0.896707 0.89684 0.896974 0.897109 1.63 0.897244 0.897381 0.897518 0.897655 0.897794 0.897933 0.898074 0.898215 0.898356 0.898499 1.64 0.898642 0.898786 0.898931 0.899076 0.899223 0.89937 0.899518 0.899666 0.899816 0.899966 1.65 0.900117 0.900269 0.900421 0.900574 0.900728 0.900883 0.901039 0.901195 0.901352 0.90151 1.66 0.901668 0.901828 0.901988 0.902149 0.90231 0.902473 0.902636 0.9028 0.902965 0.90313 1.67 0.903296 0.903464 0.903631 0.9038 0.903969 0.904139 0.90431 0.904482 0.904654 0.904827 1.68 0.905001 0.905176 0.905351 0.905527 0.905704 0.905882 0.90606 0.90624 0.90642 0.9066 1.69 0.906782 0.906964 0.907147 0.907331 0.907515 0.907701 0.907887 0.908074 0.908261 0.90845 154 Advanced Engineering Mathematics x 0 1 2 3 Chapter 6 - Special Functions 4 5 6 7 8 9 1.70 0.908639 0.908829 0.909019 0.909211 0.909403 0.909596 0.909789 0.909984 0.910179 0.910375 1.71 0.910572 0.910769 0.910967 0.911166 0.911366 0.911567 0.911768 0.91197 0.912173 0.912376 1.72 0.912581 0.912786 0.912991 0.913198 0.913405 0.913613 0.913822 0.914032 0.914242 0.914453 1.73 0.914665 0.914878 0.915091 0.915306 0.915521 0.915736 0.915953 0.91617 0.916388 0.916607 1.74 0.916826 0.917046 0.917267 0.917489 0.917712 0.917935 0.918159 0.918384 0.918609 0.918835 1.75 0.919063 0.91929 0.919519 0.919748 0.919978 0.920209 0.920441 0.920673 0.920906 0.92114 1.76 0.921375 0.92161 0.921846 0.922083 0.922321 0.92256 0.922799 0.923039 0.923279 0.923521 1.77 0.923763 0.924006 0.92425 0.924494 0.92474 0.924986 0.925233 0.92548 0.925728 0.925977 1.78 0.926227 0.926478 0.926729 0.926981 0.927234 0.927488 0.927742 0.927997 0.928253 0.92851 1.79 0.928767 0.929026 0.929285 0.929544 0.929805 0.930066 0.930328 0.930591 0.930854 0.931119 x 0 1 2 3 4 5 6 7 8 9 1.80 0.931384 0.93165 0.931916 0.932184 0.932452 0.93272 0.93299 0.933261 0.933532 0.933804 1.81 0.934076 0.93435 0.934624 0.934899 0.935175 0.935451 0.935728 0.936006 0.936285 0.936565 1.82 0.936845 0.937126 0.937408 0.937691 0.937974 0.938258 0.938543 0.938829 0.939115 0.939402 1.83 0.93969 0.939979 0.940269 0.940559 0.94085 0.941142 0.941434 0.941728 0.942022 0.942317 1.84 0.942612 0.942909 0.943206 0.943504 0.943803 0.944102 0.944402 0.944703 0.945005 0.945308 1.85 0.945611 0.945915 0.94622 0.946526 0.946832 0.947139 0.947447 0.947756 0.948066 0.948376 1.86 0.948687 0.948999 0.949311 0.949625 0.949939 0.950254 0.95057 0.950886 0.951203 0.951521 1.87 0.95184 0.95216 0.95248 0.952801 0.953123 0.953446 0.953769 0.954093 0.954419 0.954744 1.88 0.955071 0.955398 0.955726 0.956055 0.956385 0.956715 0.957047 0.957379 0.957711 0.958045 1.89 0.958379 0.958714 0.95905 0.959387 0.959725 0.960063 0.960402 0.960742 0.961082 0.961424 155 Advanced Engineering Mathematics x 0 1 2 3 Chapter 6 - Special Functions 4 5 6 7 8 9 1.90 0.961766 0.962109 0.962453 0.962797 0.963142 0.963488 0.963835 0.964183 0.964531 0.964881 1.91 0.965231 0.965582 0.965933 0.966286 0.966639 0.966993 0.967347 0.967703 0.968059 0.968416 1.92 0.968774 0.969133 0.969492 0.969853 0.970214 0.970576 0.970938 0.971302 0.971666 0.972031 1.93 0.972397 0.972764 0.973131 0.973499 0.973868 0.974238 0.974609 0.97498 0.975352 0.975725 1.94 0.976099 0.976473 0.976849 0.977225 0.977602 0.97798 0.978358 0.978738 0.979118 0.979499 1.95 0.979881 0.980263 0.980647 0.981031 0.981416 0.981802 0.982188 0.982576 0.982964 0.983353 1.96 0.983743 0.984133 0.984525 0.984917 0.98531 0.985704 0.986098 0.986494 0.98689 0.987287 1.97 0.987685 0.988084 0.988483 0.988883 0.989285 0.989687 0.990089 0.990493 0.990897 0.991302 1.98 0.991708 0.992115 0.992523 0.992931 0.993341 0.993751 0.994162 0.994573 0.994986 0.995399 1.99 0.995813 0.996228 0.996644 0.997061 0.997478 0.997896 0.998315 0.998735 0.999156 0.999578 2.00 1 1.00042 1.00085 1.00127 1.0017 1.00212 1.00255 1.00298 1.00341 1.00384 156 Advanced Engineering Mathematics Chapter 6 - Special Functions This is the graph of the absolute value of the Gamma function: This is the graph of the real pat of the Gamma function 157 Advanced Engineering Mathematics Chapter 6 - Special Functions This is the graph of the imaginary part of Gamma function: Gamma Function Calculator Since (s+ 1) = s (s) ,then (1.5) = (0.5) (1/2) = 1   1  = 0.886226925453 2 2 (2.5) = (1.5)*(0.5) (1/2) = 3   3  = 1.329340388179 4 4 (3.5) = (2.5)*(1.5)*(0.5) (1/2) = 5 3 1 222   15  = 3.323350970448 8 (1.461) = 0.885603 (1.496) = 0.886104 (5.461) = 4.461*3.461*2.461*1.461*(1.461) = = 4.461*3.461*2.461*1.461*0.885603 = 49.162587369672505923 (0.99713) = 1.00166478 (1.496) = 0.8861042 Since (s ) = (s + 1)/s ,then (-1.496) = (-0.496)/( -1.496) = (0.504)/[( -1.496)* ( -0.496)] = 158 Advanced Engineering Mathematics Chapter 6 - Special Functions = (1.504)/[( -1.496)* ( -0.496) )* 0.504]= (0.886363)/1.496*0.496* 0.504 = (0.886363)/ 0.373976064 = 2.3701062322534096727645114741889 (-5.461) = (-4.461) /[-5.461] = (-3.461) /[-5.461*-4.461] = = (-2.461) /[-5.461*-4.461*-3.461] = (-1.461) /[-5.461*-4.461*-3.461*2.461] = = (-0.461) /[-5.461*-4.461*-3.461*-2.461*-1.461] = (0.539) /[-5.461*-4.461*-3.461*-2.461*-1.461*-0.461]= = (1.539) /[-5.461*-4.461*-3.461*-2.461*-1.461*-0.461*0.539] = = 0.888113/ 75.328187751537172939179 = 0.0117899159 = 0.0117899161514461426821 (-5.461) = 0.0117899161514461426821 (-1.496) = 2.37010599 (-5.461) = 0.011789916 Γ(−5 /2) = −8 /15 √π Γ(−3 /2) = 4 /3 √π Γ(−1 /2) = −2 √π Γ(1 /2) = √π Γ(3 /2) = 1 /2 √π Γ(5 /2) = 3 /4 √π Γ(5.25) = 35.2116118528 Γ(-5.461) = 0.011789916151446 Γ(-1.496) = 2.3701059895317 § 6.2: The Beta function: The beta function, denoted by B (p, q) is defined by the first Eulerian integral 1 B (p, q) =  x p 1 (1  x)q 1 dx , (6.7) 0 159 Advanced Engineering Mathematics Chapter 6 - Special Functions this is convergent for p > 0, q > 0. In case p < 1 this integral is improper with respect to the lower limit, in case of q < 1 it is improper with respect to the upper limit of integration. 2 2 If we put x = cos , we get 1  x = sin , dx = 2 cos  sin  d  2 B(p, q) = 2  cos 2p 1 sin 2q 1  d , (p > 0, q > 0) 0 The beta function is connected with the gamma according to the relation (p)(q) , B(p,q)  (p  q) and (6.8) (6.9) To prove this relation, we use equation (6.5)  2 (p) = 2  e x x 2p 1dx , 0  (q) = 2  e  x 2 2q 1 0 x p>0,  2 dx = 2  e y y 2q 1dy ,  0 q>0 2 2 Then (p) (q) = 4   x 2p 1y 2q 1e(x  y ) dxdy 00 Transforming to polar coordinates, x = r cos , y = r sin , it follows that the Jacobean of the transformation J = r and  2 2 2 2 (p) (q) = 4   e(x  y ) x 2p 1y 2q 1dxdy = 4  e(x  y ) x 2p 1y 2q 1dxdy = 00 D 2 = 4  e r r 2p  2q 1 cos 2p 1  sin 2q 1 dr d =  160 Advanced Engineering Mathematics Chapter 6 - Special Functions      r 2 2p  2q 1   2 2p 1 2q 1  = 2 e r  sin  d  = dr  2  cos     0  0  = (p + q) B (p, q). It is clear that the beta function is symmetric with respect to p and q. i.e. B (p, q) = B (q, p). Since (n) = (n  1)! , (m) = (m  1)! (m  1)!(n  1)! Then B (m, n) = , m, n are positive integers. (m  n  1)!  Also B ( 1 , 1 ) = 2 2 2 2 0  0 (6.10) 2  cos  sin  d = 2  d = . 0 0 2 [( 1 )] 2 1 1 = [( 1 )]2 = , then ( 1 ) = But B ( , ) = 2 2 2 2 (1)  (another proof.) If p > 0 any real number, n > 0 positive integer then (p)(n) (p).(n  1)! = = B (p, n) = (p  n) (p  n  1)(p  n  2) p.(p) (n  1)! = . (p  n  1)(p  n  2) p (6.11) 2 If we put x = y/(l + y) then we get dx = dy/(l + y) and  y  B (p, q) =    0 1 y  p 1  1    1 y  q 1  dy (1  y) 2 =  y p 1 pq 0 (1  y) dy , (6.12) this is the third expression for the beta function. Example 6.1: Given  x p 1    1  x dx = sin p , show that (p)(1  p) = sin p , where 0 Solution: Letting q = 1  p in equation (6.12) we get  y p 1 B (p, 1  p) =  0 1 y dy =  . sin p From (6.9) we get B (p, 1  p) = (p)(1  p)  . = sin p (1) 161 0 < p < 1. Advanced Engineering Mathematics Then (p)(1  p) = Chapter 6 - Special Functions  , sin p (6.13) Example 6.2: 2q1 Prove the duplication formula: 2 Solution:   2 (q).(q+ 1 ) = 2  (2q). 2 Let I =  sin 2p xdx , J =  sin 2p 2xdx . From (6.8), (6.9) it is clear that  0 0 2 (p)(q) 1 2p 1 = B (p , q).  sin 2q 1  d =  sin 2 (p q)   2 0 (p  1 )( 1 ) 1 1 1 2 2 , Then I = B (p+ , ) = 2 2 2 2(p  1) (6.14) In the integral J, let 2x = u, we get  2 1  2p J =  sin 2udu =  sin 2p udu = I. 20 0  2 But J =  (2sin x cos x) 0 =2 2p 1 B(p  1 ,p  1 ) = 2 2 2p dx = 2 2p  2 2p 2p  sin x cos x dx = 0 2(2p 1) [(p  1 )]2 (2p  1) 2 . (6.15) Then since I = J, (p  1 )( 1 ) 2(2p 1) [(p  1 )]2 2 2 = 2 2(p  1) (2p  1) Therefore 2p 2 (p + 1 ) (p+1) = 2  (2p + 1). Let p + 1 = q, we get 2 2 2q1 (q).(q+ 1 ) = 2 (6.16)  (2q). 162 Advanced Engineering Mathematics Chapter 6 - Special Functions §6.3.1 Bessel Functions: Bessel functions were first introduced by Bessel, in 1824, in the discussion of a problem in dynamical astronomy. Bessel differential equation appears in the solution of Laplace's equation  2 U = 0 in cylindrical coordinates. Now we shall show how Bessel's equation emerges from physical problems. The wave-equation, with x, y, z as Cartesian coordinates and t as time, is 2V x 2  2V y 2  2V z 2  1 2V (6.17) C2 t 2 In cylindrical coordinates with x y « r sin G the equation is x = r cos  , y = r sin , the equation is 2V r 2  1 V 1  2 V 1  2 V .   r r r 2 2 C2 t 2 (6.18) Seek solutions of the form where V = R(r).().T(t), (6.19) where, by the method of separation of variables, we get after dividing both sides by V: R " 1 R ' 1  " Z" 1 T"     R r R r2  Z C2 T (6.20) The left-hand side is independent of t, being only a function of r,  and z; and the right-hand side is independent of r,  and z, being only a function of t. Since the two sides are identically equal, this implies that each is independent both of r, , z and t, 2 and must therefore be constant. Putting this constant equal to p , we find 2 2 T" =  C p T, (6.21) 2 The constant (p ) which we introduced, is known as the separation constant. There 2 was no reason why the separation constant should have a negative value of (p ) except that this enabled us to obtain harmonic solutions. Equation (6.20) reduces to R " 1 R ' 1 " Z"     p2 (6.22) R r R r2  Z 2 Similarly, put Z" =  q Z R " 1 R ' 1 " We get    q 2  p2  n 2 . 2 R r R r  163 (6.23) (6.24) Advanced Engineering Mathematics Chapter 6 - Special Functions 2 Now let " =  m  (6.25) Equation (6.24) becomes R " 1 R ' m2    n2 R r R r2 2 2 2 or r R" + r R' + (r  n ) R = 0 (6.26) The equation (6.26) for R is Bessel's differential equation. Bessel function arise as solution of the differential equation 2 2 2 x y" + x y' + (x  n ) y = 0, n  0 (6.26') To obtain a solution in series by the Frobenius method, let  y =  a r x (C  r) r 0 Substituting we get   r 0 r 0  [(C  r)2  n 2 ]a r x (C  r) +  a r x (C  r  2) = 0. Equating coefficients of various powers of x to zero, we obtain the indicial equation is 2 2 C  n = 0, i.e. C =  n. 2 2 and [(C+1)  n ] a1 = 0. Hence a1 = 0 and a r  2 , r = 2, 3, 4, . ar = (C  r) 2  n 2 a r  2 Now, let C = n and ao = 1, we get ar = , r = 2, 3, 4, , and 2 2 (n  r)  n  1 1 1 n x2  x4  x 6   y1 = x 1  42.2!(n  1)(n  2) 43.3!(n  1)(n  2)(n  3)  4(n  1)  a r  2 , r = 2, 3, 4, , If C = n, ao = 1, we get ar = 2 2 (r  n)  n  1 1 1 k  y2 = x 1  x2  x4  x 6   42.2!(1  n)(2  n) 43.3!(1  n)(2  n)(3  n)  4(1  n)  The coefficients of y1, are finite except when n is a negative integer. Excluding this n case, we standardise the solution by taking ao = l/2 (n + l) in general and n ao = l/(2 . n!) when n is a positive integer. Since 164 Advanced Engineering Mathematics a 2r a 2r  2 = Chapter 6 - Special Functions 1 0 as r, 4r(n  r) The series converges for any finite value of x, and thus we have the first solution y = Jn(x), where (1)r x Jn(x) =    r  0 r!(n  r  1)  2   n  2r , (6.27) in general. In particular, where n is a positive integer, (1) r  x  Jn(x) =    r  0 r!(n  r)!  2   n  2r . This function Jn(x) is known as the Bessel Function of the first kind of order n. When n is not integer the second solution may be obtained by replacing n by n; it is therefore (1)r x Jn(x) =    r  0 r!(r  n  1)  2   2r  n , and the general solution of the Bessel equation may be expressed as y = A Jn(x) + B Jn(x). But when n is an integer, and since n appears in the differential equation only 2 as n there is no loss of generality in taking it to be a positive integer, Jn (x) is not distinct from Jn(x). For the denominators of the first n terms of the series Jn(x) contains respectively the factors (l  n), (2  n), , (0) which are infinite, and so these terms vanish. Thus (1)r x Jn(x) =    r  n r!(r  n  1)  2   let r = k + n, we get 2r  n (1)k  n x Jn(x) =    k  0 (k  n)!(k  1)  2   , 2k  n n = (1) Jn(x), (6.28) In this case a modification of the method of Frobenius is necessary to give an independent second solution. We shall define the Bessel function of the second kind of order n as 165 Advanced Engineering Mathematics Chapter 6 - Special Functions  J n (x)cos n  J  n (x)  sin n  Yn (x)   J (x)cos p  J  p (x)  lim p p  n sin p n  0,1,2,3, (6.29) n = 0,1,2,3, The Bessel function of the second kind of order n is given also by 1  J  (x)  (1)n  J  (x)   Yn(x) =       n      n (6.30) The general solution of equation (6.26') is given by y = A Jn(x) + B Yn(x). § 6.3.2 Asymptotic formulas: In certain physical problem it is desirable to know the value of a Bessel function for large values of its argument. In the following we shall derive the asymptotic formulas. We put y = z/ x in Bessel's differential equation 2 2 2 3/2 z, x y" + x y' + (x  n )y = 0, we get y' = z' x ½ ½x ½ 3/2 +¾zx y" = z" x  ½ z' x  1  4n 2  z" 1  z  0 2   4x   5/2 , and For large values of x the differential equation is reduced to z" + z = 0 then z = A sin x + B cos x sin x cos x yA B , Therefore x x cos x , where J n (x)  x Yn (x)  sin x x §6.3.3 Recurrence formulas: It is clear that  ( 1) r x 2n  2r 1 d  (1)r x 2n  2r d  n  =  = x J n (x) = n  2r n  2r 1  dx r dx  r!(n  r) r!(n  r  1) r  0 2 0 2 n  (1)r x (n 1)  2r n =x  = x Jn1(x). (n 1)  2r r![(n  1)  r  1)] r 0 2 i.e. d  n x J n (x)  = x n J n 1 (x) .  dx  (6.31) 166 Advanced Engineering Mathematics Chapter 6 - Special Functions Similarly, we can prove that d  n x J n (x)  =  x  n J n 1 (x) .  dx  (6.32) From (6.31) it is clear that x n J 'n (x)  nx n 1J n  x n J n 1 (x) . i.e. xJ 'n (x)  xJ n 1 (x)  nJ n (x) . (6.33) From (6.32) we get x  n J 'n (x)  nx  n 1J n   x n J n 1 (x) . or xJ 'n (x)  nJ n (x)  xJ n 1 (x) . (6.34) Adding (6.33) and (6.34) and dividing by 2x gives 1 J 'n (x)  [J n 1 (x)  J n 1 (x)] . 2 Subtracting (6.34) from (6.33) and dividing by x we get 2n J n 1 (x)  J n 1 (x)  J n (x) . x 2n J n 1 (x)  J n (x)  J n 1 (x) . i.e. x (6.35) (6.36) The functions Yn(x) satisfy exactly the same results as those above (6.31) to (6.36), where Yn(x) replaces Jn(x). Example 6.3: Find J3(x) in terms of J0(x) and J1(x). Solution: In (6.36), put n = 2 we get 4 J3(x) = J2(x)  J1(x). x 2 In (6.36), put n = 1, J2(x) = J1(x)  J0(x). x 4 2 4 8 Therefore J3(x) = [ J1(x)  J0(x)]  J1(x) = ( 1) J1(x)  J0(x). x x x x2 8 4 1) J1(x)  J0(x). J3(x) = ( x x2 Example 6.4: 167 Advanced Engineering Mathematics Prove that a) J 1 (x)  2 Solution: Chapter 6 - Special Functions 2 sin x , x 2 b) J  1 (x)  cosx . 2 x 1 2r  r 2  ( 1) ( x ) 2 a) J 1 (x)   3 2 r  0 r!(r  2 ) Since (r+3/2) = (r+½) (r+½) = (r  1 )(2r)!  2 22r r! 1 , 2r  2r r 22  (1) ( x ) 2x  (1)r x 2r 2 then J 1 (x)   = =  1 2  r  0 (2r  1)! r  0 (r  2 )(2r)!  2  (1)r x 2r 1 2 =  x r  0 (2r  1)! = x sin x . 1 1 r x 2r  2 2r r x 2r  2  ( 1) ( )  (1) ( ) 2 2 2 b) J  1 (x)   =  = 1) 2 (2r)!    r! (r r 0 r 0 2 = 2  (1)r x 2r =  x r  0 (2r)! 2 cos x . x §6.3.4 Integrals involving Bessel functions: From equations (6.31) and (6.32) we have the relations: n n  x J n 1 (x)dx = x Jn(x) + C, (6.37) n n  x J n 1 (x)dx = x Jn(x) + C. (6.38) Example 6.5: Evaluate a)  x 3J 2 (x)dx , J (x) dx , b)  1 x Solutions: a) In (6.37) put n = 3, we get  x 3J 2 (x)dx = x J3(x) + C 2 b) In (6.36) put n = 1, we get J1(x) = J0(x) + J2(x). x J (x) 1 1 dx =  J 0 (x)dx +  J 2 (x)dx , Therefore  1 x 2 2 3 168 c)  x 2 J1 (x)dx Advanced Engineering Mathematics Chapter 6 - Special Functions J (x)  J (x)   J (x)  and  J 2 (x)dx =  x  2  dx =   x d  1  =  J1(x) +  1 dx . x  x   x  J (x) 1 1 J (x) 1 Hence  1 dx =  J 0 (x)dx  J1(x) +  1 dx . 2 x x 2 2 J (x) dx =  J 0 (x)dx  J1(x). i.e.  1 x c) In (6.32) put n = 0 and in (6.37) put n = 1, we get d J0(x) =  J1(x),  xJ 0 (x)dx = x J1(x) + C. dx and  x 2 J1 (x)dx =   x 2 d[J 0 (x)] =  x J0(x) +  2x J 0 (x)dx = =  x J0(x) + 2  x J 0 (x)dx =  x J0(x) + 2 x J1(x) + C. Example 6.6: J (x) b)  2 dx , 2 x Evaluate a)  J3 (x)dx , Solutions: J (x) J (x)  J (x)  a)  J3 (x)dx =  x 2 3 2x dx = dx =   x 2 d  2  =  J2(x) +  2  x  x2 x2 J (x) J (x) + C. =  J2(x) + 2  2 dx =  J2(x)  2 1 x x 4 b) In (6.36) put n = 2 we get J3(x) = J2(x)  J1(x). x J 2 (x) J1 (x) J3 (x) . i.e.   4x 4x x2 J3 (x) J3 (x) J 2 (x) J 2 (x)  J 2 (x)   x dx =  x 2 dx =   x d  2  =  x   2 2x dx . x  x  x From Example (6.5)(b) we have J1 (x)  x dx =  J1(x) +  J 0 (x)dx . Therefore J 2 (x)  and x2 dx =  J (x) 1 J 2 (x) 1 1 +  J1(x) + J 0 (x)dx  0 dx  4x 4 4 4 x2 J (x) 1 3 J 2 (x) 1 +  J 0 (x)dx , dx =  J1(x)  0  4x 4 x2 4 4 169 Advanced Engineering Mathematics i.e.  J 2 (x) x2 dx =  Chapter 6 - Special Functions J (x) 1 1 J1(x)  0 +  J 0 (x)dx . 3 3 3x §6.3.5 Orthogonality of Bessel functions: It is clear that, y = Jn(  x), z = Jn(  x) are solutions of the equations 2 2 2 2 2 2 2 2 x y" + x y' + ( x  n ) y = 0, x z" + x z' + ( x  n ) z = 0, Multiplying the first equation by z/x, the second by y/x and subtracting, we find 2 2 x [z y"  y z"] + [z y'  y z'] = (   ) x y z, this can be written as d 2 2 [ x (z y'  y z')] = (   ) x y z. dx Then by integrating we get 2 2 (   )  x y z dx = x (z y'  y z') + constant. d 2 2 [ x (z y'  y z')] = (   ) x y z. dx d d 2 2 Since y' = Jn( x) =  Jn( x) =  J 'n (x) , and if (   )  0, dx d(x) 1 ' '  xJ n (x)J n (x)dx = 2 2 x[Jn( x) J n ( x)  Jn( x) J n ( x)] + C.   Thus 170 then Advanced Engineering Mathematics 1  xJ n (x)J n (x)dx = 0 1 2   2 Chapter 6 - Special Functions .[Jn( x) J 'n ( x)  Jn( x) J 'n ( x)] (6.39) Let  , and using L'Hospital's rule, (  is considered constant) we find J n ()J 'n ()  J n ()J 'n () 2 =  xJ n (x)dx = lim 2 2    0 J 'n ()J 'n ()  J n ()J 'n ()  J n ()J"n () = lim = 2  [J 'n ( )]2  J n ()J 'n ()  J n ()J"n () . = 1 2 But since  J"n () +  J 'n () + (    ) Jn() = 0, we find 2 2 [J 'n ()]2  J n ()J 'n ()  J n ()J"n () 1 ' 2 (  n ) 2 = [J n ()]  J n ( ) . 2 2 2 2 2 2 2 1  ' n2 2  2 2 i.e.  xJ n (x)dx = [J n ()]  (1  2 )J n ()  . 2  0   1 (6.40) Now, if  and  are any two different roots of the transcendental equations A Jn(x) + B x J 'n (x) = 0, where A and B are constants, then A Jn() + B  J 'n () = 0, (6.41) A Jn() + B  J 'n () = 0. Multiplying the first equation by Jn(), the second by Jn() and subtracting, we find  Jn() J 'n ()   Jn() J 'n () = 0 if A B  0. 1 Therefore  xJ n (x)J n (x)dx  0 0 In case A = 0, B  0 or A  0, B = 0, the result is also easily proved. Expansion of functions in terms of Bessel functions: If we now suppose that we can expand an arbitrary function f(x) in the form f(x) =  a i J n (i x) (6.42) i where the sum is taken over the positive roots of the equation (6.41), then we can determine the coefficients ai as follows: Multiply both sides of equation (6.42) by x Jn (i x) and integrate with respect to x from 0 to 1; then 171 Advanced Engineering Mathematics 1 Chapter 6 - Special Functions 1  xf (x)J n (i x)dx =  a i  xJ n (i x)J n ( jx)dx , i 0 from which it follows that aj = where C = [J n ( )]2 2 2B  2 2 2 0 11  x f(x) J n (λ j x) dx , C0 2 2 (6.43) 2 [B  + A  B n ]. Because of its similarity to a Fourier series, a series of the type (6.42) is called a Fourier-Bessel series. We can show that if f (x) satisfies the Dirichlet conditions then at every point of continuity of f (x) in the interval [0, l] the series (6.42) converges to 1 f(x). At any point of discontinuity the series (6.42) converges to [f(x+0) + f(x0)]. 2 §6.3.6 Generating function for Jn(x): x (t  1 ) The function e 2 t =   J n (x) t n (6.44) n  is called the generating function for Bessel functions of the first kind of integral order. §6.3.7 Functions related to Bessel functions: 1. Hankel functions of the first and second kinds are defined respectively by  H (1) n  J n  x   i Yn  x  ,     H (2) J x i Y x .      n n n (6.45) 2. Modified Bessel functions: The modified Bessel function of the first kind of order n is defined as n In(x) = i ni/2 Jn( ix) = e Jn(ix) (6.46) If n is an integer In(x) = In(x), but if n is not an integer, In(x) and In(x) are linearly independent. The modified Bessel function of the second kind of order n is defined as    I n (x)  In (x)  n  0,1, 2,3, 2  sin n     K n (x)   (6.47)  I (x) I (x)    p p  lim  n  0,1,2,3,  p  n 2  sin p    172 Advanced Engineering Mathematics Chapter 6 - Special Functions These functions satisfy the differential equation 2 2 2 x y" + x y'  (x + n ) y = 0 and the general solution of this equation is y = A In(x) + B Kn(x), or if n  0, l, 2, 3,  y = R In(x) + S In(x), where A, B, R and S are constants. Equality (6.46) is very useful for deducing properties of the modified Bessel function In(x) from those of the Bessel function Jn(x). For instance, when n is an integer it follows from equation (6.28) that I n(x) = In(x) (6.48) and from equations (6.31) to (6.36) respectively that d n n [x I n (x)] = x In1(x), dx d n n [x In (x)] = x In+1(x), dx (6.49) (6.50) x I'n (x) = x In1(x)  n In(x), (6.51) x I'n (x) = n In(x) + x In+1(x), 1 I'n (x) = [In1(x) + In+1(x)], 2 2n In (x) = In1(x)  In+1(x), x I'0 (x) = I1(x). (6.52) (6.53) (6.54) (6.55) 3. Ber. Bei. Ker. Kei functions: Kelvin introduced two new functions Bern (x) and Bein (x) which are respectively the 3/2 real and imaginary parts of Jn (i x) where 2 3/2 3i/4 3/2 i =e = [1  i], i.e. Jn( i x) = Bern(x) + i Bein(x) 2 In a similar way the functions Kern (x) and Kein (x) are defined to be respectively the 2 i/4 ni/2 i/2 real and imaginary parts of e [1 + i], = Kn( i x), where i = e 2 i.e. e in/2 Kn( i x) = Kern(x) + i Kein(x). 2 2 2 The functions are useful in connection with the equation x y"+x y'  (i x + n ) y = 0, 173 Advanced Engineering Mathematics Chapter 6 - Special Functions which arise in electrical engineering and other fields. The general solution of the equation is y = A Jn(i 3/2 x) + B Kn( i x). The four functions Ber, Bei, Ker and Kei are used more often in electrical engineering than they are in physics or chemistry. Exercises 6 1) Evaluate each of the following: (7) (3)(3 / 2) a) b) 2(4)(3) (9 / 2) (6) 5 1 e) ( ) / ( ) d) 4(3) 2 2 [Answers: a) 30, 2)  3 x a)  x e 0  e st  dx = 6 , 0 1 g)  (x n x)3dx =  0 6 2x b)  x e  dt  ,s>0, s t d)  3/2 c) (3/8) , b) 16/105, Prove that 0  e)  e  x3 0 Evaluate each of the following: a) B(3, 5), 4 b) 3 d)  x (1  x) dx , 0 3/2 c) (3/8) , 45 , dx = 8 1 1 dx = ( ) , 3 3 e) 3/4, f) 4/3].  c) 1 3 a B(3/2, 2), f)  (n x) 4 dx = 24 , 0 C) B(1/3, 2/3), e)  x 4 a 2  x 2 dx . 0 [Answers: a) 1/105, b) 4/15, c) 2/ 3 , d) 1/280, 6 e) a /32 ]. 4) Prove that q 1 p 1 B(p, ql) = B (p, q) = B(pl, q). p  q 1 p  q 1 5) Prove that a B (a, b) = (b  1) B(a + l, b  1). 6) Show that B (a, b) = B(a, b  1)  B(a + l, b  1). / 2 n 1 m 1 nm ].[ ]/2[ +1] 7) Show that  cos m x sin n x dx = [ 2 2 2 0 174  x  xe dx = 3 . 0 3 . 128 3) 1 1 3 5 c) ( )( )( ) 2 2 2 6(8 / 3) f) 5(2 / 3) Advanced Engineering Mathematics Chapter 6 - Special Functions [( 1 )]2 dx 4 8) Show that  = 4  0 3  cos x   x 2 dx  = , 4 2 2 0 1 x  9) Show that  dx  0 1 x 4 =  2 2 . 1 p m 10) Show that  x p 1 (1  x m )q 1 dx = 1 B( , q), p , q , m > 0. [Hint: put x = y]. m m 0 11) Show that (a) (a + ½) = 2 21/ 2  2s (2a), a > 0. 12) Prove that / 2 / 2 b)  sin 4 x cos5 x dx = 6 a)  sin xdx = 5/32, 0  c)  cos 4 x dx = 0 0 2 3 , 8 d)  x 0 13) Show that / 2 / 2 0 0 3 8  x 3 dx = 8 . 315 16 . 9 3 m m  sin x dx =  cos x dx = 1.3.5 (m  1)   2.4.6 m 2 =  2.4.6 (m  1)  1.3.5 m if m is an even positive integer if m is an odd positive integer 14) Show that / 2 a)  sin 3 x cos 2 x dx = 2/15, 0 / 2 c)  0 tan x dx = / 2, 15) Show that a) J 3 (x)  2 2 8  sin x dx = 35/64, b) 0  xdx  d)   , 6 3 3  1 x 0 2  sin x  x cos x   , x x   3 e)  0 3x  x b) J (x) 3 2 16) Show that a)  x 4 J1 (x)dx = x J2(x)  2 x J3(x) + C. 2 3 dx  2 1 c)  x 3J 0 (x)dx = 2 J0(l)  3 J1(1), 0 175  . 2  x sin x  cos x   . x x   b)  x 3J3 (x)dx =  x J2(x)  5 x J1(x)  15 x J0(x) + 15  J 0 (x)dx , 3 2 Advanced Engineering Mathematics Chapter 6 - Special Functions d)  x 2 J 0 (x)dx = x J1(x) + x J0(x)   J 0 (x)dx . 2  (3  x 2 )sin x  3x cos x  17) Show that a) J 5 (x)   , 2   x  2 x   2  3x sin x  (3  x 2 )cos x  b) J  5 (x)   .  x  2 x2  18) Show that  J 0 (x)sin xdx = x J0(x) sin x  x J1(x) cos x + C. 1 19) Prove that a) J "n (x) = [ Jn2(x)  2 Jn(x) + Jn+2(x)] 4 1 b) J "n (x) = [Jn3(x)  3 Jn1(x) + 3Jn+1(x)  Jn+3(x) ] 8 J (x) 1 1 1 1 J1(x)  J2(x)  J3(x) + 20) Prove that a)  3 dx =   J 0 (x)dx . 3 15 15x 15x 15 x 2 3 b)  J1 ( 3 x )dx = 6 3 x J1 ( 3 x )  3 x 2 J 0 ( 3 x )  C . 176 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations CHAPTER 7 Partial Differential Equations § 7.1 Classification of equations: An equation containing partial derivatives of unknown functions ul, u2,, un is said to be partial differential equation. A partial differential equation is said to be th th of the n order if it contains at least one partial derivative of the n order and no partial derivatives of order higher than n. A partial differential equation is called linear if it is linear in the unknown functions and in their derivatives; it is called quasi-linear if it is linear in the highest order derivatives of the unknown functions. 2u 2 Thus, for instance, the equation = 5x – 3y , is a partial differential xy equation of order two, or a second order partial differential equation. 2u 2u The equation + f(x, y) = 2u is a linear partial differential equation 2 2 x y of the second order with respect to u. u  2 u u  2 u 2 + The equation + u = 0, is quasi -linear partial x x 2 y y 2 differential equation of the second order with respect to the unknown function u. 2  u  2  u  The equation   +   = u, is neither linear nor quasi-linear with respect to u.  x   y  By a solution of an equation containing partial derivatives we mean a system of functions which, when we put in the equation in place of the unknown functions, turns the equation into an identity in the independent variables. The general solution is a solution, which contains a number of arbitrary independent functions equal to the order of the equation. A particular solution is one which can be obtained from the general solution by particular choice of the arbitrary functions. 3 2 For example; as seen by substitution, u = 5 x y – 3 x y + F(x) + G(y) is a solution of 3 2 the following partial differential equation 177 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations 2u 2 = 5x – 3y. Because it contains two arbitrary independent functions F(x) and xy 3 G(y), it is the general solution. If in particular F(x) = 2 cos x, G(y) = 2y – 7, we 3 2 3 obtain the particular solution u = 5 x y – 3 x y + 2 cos x + 2y – 7. 3 2 A singular solution is one which cannot be obtained from the general solution by particular choice of the arbitrary functions. A boundary-values problem involving a partial differential equation seeks all solutions of a partial differential equation which satisfy conditions called boundary conditions. We shall be primarily interested in linear equations of the second order in one unknown function. The general linear partial differential equation (will be abbreviated to P.D.E.) of order two in two independent variables has the form u 2u 2u 2u u A +B +D +C +E + F u = G, (7.1) 2 2 x     x y y x y where A, B, C, D, E, F and G; may depend on x and y but not on u. If G = 0 the equation is called homogeneous, while if G  0 it is called non-homogeneous. Because of the nature of the solutions of the equation (7.1) is often classified as 2 elliptic, hyperbolic or parabolic according as B – 4 A C = is less than, greater than or equal to zero respectively < 0  i.e. B2  4AC = 0 > 0  EllipticEquation, Parabolic Equation, Hyperbolic Equation. Example 7.1: Determine whether each of the following partial differential equations is linear or non-linear, state the order of each equation, and name the dependent and independent variables. a) 2 u 2  u . It is linear of order 2, dependent variable u, independent = a 2 y x variables x and y. 178 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations 3 2 2  z 2  z =y b) x . It is linear of order 3, dependent variable z, independent 3 2 y x variables x and y. c)  2u 2u  2u + + = 0. It is linear of order 2, dependent variable  u = 0. i.e. 2 2 2 y x z 2 u, independent variables x, y and z. 2  u  2  u  d)  x  +  y  = 1. It is nonlinear of order 1, dependent variable u,   independent variables x and y. Example 7.2: Classify each of the following equations as elliptic, hyperbolic or parabolic.  2u 2u  2u + + = 0, A = B = C = 1. a) 2 2 2 x y z 2 Then B – 4AC = 3 < 0, and the equation is elliptic. 2 2u 2  u 2 =a , A = a , B = 0 , C = –1. b) t 2 x 2 2 2 Then B – 4AC = 4 a > 0, and the equation is hyperbolic. 2 u 2 2  u , A = a , B = 0 , C = 0. =a c) t x 2 2 Then B – 4AC = 0 and the equation is parabolic.  2u u 2u 2u u +3 +2 + – 3 + 2 u = 3x – 2y, d) 4 x xy y 2 y x 2 2 A = 4, B = 3, C = l. Then B – 4AC = 9 – 16 = – 7 < 0, and the equation is elliptic.  2u 2u 2 u +x + 3y = 0 , A = y, B = 0, C = x. e) y 2 2  x x y 2 Then B – 4AC = – 4xy. In the region x y > 0, the equation is elliptic. In the region x y < 0, the equation is hyperbolic. If x y = 0, the equation is parabolic. 179 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations §7.2 Basic types of equations of Mathematical Physics: The basic equations of Mathematical Physics are the following second order partial differential equation. I) Wave equation: 2 2u 2  u =a t 2 x 2 This equation describes the motion of a beam, which can vibrate longitudinally in the x-direction. The variable u(x, t) is the longitudinal displacement from the equilibrium position of the cross section at x. The constant 2 a = g E/, where g is the acceleration due to the gravity, E is the modulus of elasticity (stress divided by strain) and depend on the properties of the beam,  is the density (mass per unit volume). This equation is applicable also to the small transverse vibrations of a taut, flexible string, such as a violin string, initially on the x-axis and set into motion. The function u(x, t) is the displacement of any point x of the string at time t. 2 The constant a = T/ where T is the tension in the string (constant), and  is the mass per unit length of the string (constant). It is assumed that no external forces act on the string but it vibrates only due to its elasticity. This equation is the simplest of the class of hyperbolic equations, II. Fourier equation for heat conduction: 2u u . =k 2 t x u 2 = k  u. t One-dimensional heat conduction. Three-dimensional heat conduction. This equation describes the flow of heat. Its derivation is based on Newton’s low. Also this equation describes the filtration of liquids and gasses in a porous medium (e.g. the filtration of oil and gas in subterranean sandstones). This equation is the 180 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations simplest of the class of parabolic equation. The constant k, called the diffusivity, is equal to k / , where the thermal conductivity k, the specific heat  and the density  are assumed constants. III Laplace’s equation: 2  u=0 This equation occurs in many fields. In the theory of electric and magnetic fields u represents the magnetic or electric potential respectively. For this reason the equation is often called the potential equation. In the theory of heat conduction, u is the steady state temperature, i.e. the temperature after a long time has elapsed, and is equivalent u = 0, in the heat conduction equation. to putting t §7.3 Formation of partial differential equation by elimination: Partial differential equations can often be formed by elimination of either arbitrary constant or arbitrary functions. Example 7.3: Eliminate the arbitrary functions f and g from y = f(x – at) + g(x + at) Solution: It is clear that y = f (x – at) + g(x + at), and x 2 y = f (x – at) + g(x + at). 2 x (1) y = –a f (x – at) + a g(x + at), and t 2 y 2 2 = a f (x – at) + a g(x + at). (2) 2 t 2 2 y 2  y = a partial differential equation of the second From (1) and (2) we get a 2 2 x t Similarly order (see wave equation). Example 7.4: Eliminate the arbitrary function f from z = f( x ). y 181 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Solution: It is clear that z 1 x z z z = f '( ) , = 0. = – x2 f '( x ) . So x + y y x y y y x y y Example 7.5: t Eliminate A and  from z = a e sin x. Solution: Differentiating partially with respect to x, we get z t = A  e cos x, x 2z 2 t = –A  e sin x. and x 2 (1) Differentiating partially with respect to t, we get z t = A  e sin x, t 2z 2 t = A  e sin x. and t 2 2z 2z + = 0. From (1) and (2) we get 2 2 x t (2) th As we know, every ordinary differential equation of the n order may be regarded as derived from a solution containing n arbitrary constants. It might be th supposed that every partial differential equation of the n order was similarly derivable from a solution containing n arbitrary functions. However, this is not true. In general, it is impossible to express the elimination of n arbitrary functions as a partial differential equation of order n. An equation of higher order is required, and the result is not unique. Example 7.6: –t Show that u(x, t) = e sin x, is a solution to the boundary-value problem u  2 u = , u(0 , t) = u( 2 , t) = 0, u(x , 0) = sin x.   t x 2 Solution: 182 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations –t From u(x, t) = e –t u (0, t) = e sin x, we have –t 0 sin 0 = 0, u( 2 , t) = e sin 2 = 0, u(x, 0) = e sin x = sin x.  i.e. the boundary conditions are satisfied. Also u u  2u 2 –t –t –t = –  e sin x. = –  e sin x, =  e cos x, t x x 2 Then substituting into the differential equation, we have u  2u 2 –t = –  e sin x =  , which is an identity. 2 t x § 7.4 Methods of finding solutions: There are many methods by which partial differential equation can be solved. The following theorems are of fundamental importance. §7.4:1 General solution: Theorem 7.1: Superposition principle: If u1, u2, ..., un are solutions of a linear homogeneous partial differential equation, then cl u1+ c2 u2+...+ cnun, where cl, c2, ..., cn are constants is also a solution. Theorem 7.2: The general solution of a linear non-homogeneous partial differential equation is obtained by adding a particular solution of the non-homogeneous equation to the general solution of the homogeneous equation. If A, B, C, D, E and F in equation (7.1) are constants, then the general solution of homogeneous equation can be found by assuming that u = e ax + by , where a and b are constants to be determined. We can sometimes find general solutions by using the methods of ordinary differential equation. Example 7.7: 2 2u a) Solve the equation =xy . xy 2 b) Find the particular solution for which u(x, 0) = x , u(1,y) = cos y. 183 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Solution: a) Write the equation as 2   u  = x y . x  y  Then integrating with respect to x, we find u 1 2 2 = x y + f(y), y 2 (1) where f(y) is an arbitrary function. Integrating (1) with respect to y we get 2 3 u = 1 x y +  f(y) dy + g(x), 6 (2) where g(x) is an arbitrary function. The result (2) can be written 2 3 u = u(x, y) = 1 x y + h(y) + g(x), 6 (3) which has two arbitrary functions and is therefore a general solution, 2 b) Since u(x, 0) = x , we have from (3) 2 Thus 2 x = h (0) + g(x), or g(x) = x – h (0) 2 3 2 u = 1 x y + h(y) + x – h (0). 6 (4) (5) Since u(l , y) = cos y, we have from (5) 3 cos y = 1 y + h(y) + 1 – h (0), 6 3 or h(y) = cos y – 1 y – 1 + h (0), 6 (6) Thus using (6) in (5) we find the required solution 2 3 2 3 u = 1 x y + x + cos y – 1 y – 1. 6 6 Example 7.8: Solve t 2u u + 3 = x. xt x Solution: We can write this equation in the form  u [t + 3 u] = x x t Integrating with respect to x, we get 2 u + 3 u = 1 x + f (t) t 2 t x2 u u f (t) +3 = + . t 2t t t or 184 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations This is a linear equation having integrating factor 3 3  dt 3 3  n t  n t t e =e =e =t . 2 2 2  3 Then [ t u ] = 1 t x + t f(t). 2 t Integrating, we get 3 2 2 3 2 3 t u = 1 t x +  t f(t) dt + h(x) = 1 t x + g(t) + h(x), 6 6 the required solution. Example 7.9: 2u  2u 2u +2 Find solutions of –3 =0 2 2  x  y y x Solution: ax + by Assume u = e 2 2 . Substituting in the given equation, we find ax + by (a + 2 a b – 3b ) e =0 or 2 2 a + 2 a b – 3b = 0 . Then (a – b)(a + 3b) = 0 and a = b or a = –3b. If a = b, e b(y – 3x) value of b. If a = –3b, e b(x + y) is a solution for any is a solution for any value of b. Since the equation is linear and homogenous, sum of these solutions is a solution. The general solution found by addition is given by u = f ( x + y) + g(y – 3x). Now, we shall obtain some special types of the partial differential equations. §7.4:2 D'Alembert methods: Let the equation to be solved is 1  2  2 = x 2 c2 t 2 (7.2) We change to new variables u = x – ct and v = x + ct. 1 i.e. x = 1 (u + v), t = [v – u]. 2 c    ), transforms to 1 ( + Then it is easily verified that 2 u v x    transforms to –c + c ; t u v 185 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations  2  2  2  2 + becomes +2 and finally 2 2 2 u v   v x u 2 2 2      2  2 becomes c [ –2 ]. When these changes are made, the + 2 2 2  u  v v t u  2 = 0 ; the most general solution of this is  = f(u) + g(v), equation becomes uv f and g being arbitrary functions. In the original variables this is  = f(x–ct) + g(x + ct). (7.3) In two dimensions the equation of wave motion is  2  2 1  2 + = . 2 2 2 2 c t x y (7.4) By a method similar to D'Alembert's, it can be shown that the most general solution involving only plane waves is  = f(ℓ x + my – ct) + g(ℓ x + my + ct) 2 2 (7.5) where, as before f and g are arbitrary functions and ℓ + m = 1. In three dimensions the differential equation is 1  2  2  2  2 + + , = 2 2 2 2 2 c t z x y (7.6) and the must general solution involving only plane waves is  = f(ℓ x + m y + n z – ct) + g(ℓ x + m y + n z + ct), 2 2 2 (7.7) in which ℓ + m + n = 1. § 7.4:3 Separation of variables: In this method it is assumed that a solution can be expressed as a product of unknown functions each of which depends on only one of the independent variables. The success of the method hinges on being able to write the resulting equation so that one side depends only on one variable while the other side depends on the remaining variables so that each side must be a constant. By repetition of this tile unknown can then be determined. Superposition of these solutions can actual solution. Let it be required to find the solution of the equation 186 then be used to find the Advanced Engineering Mathematics 2 2  u 2u =a , 2 2 t x Chapter 7- Partial Differential Equations (7.8) this satisfies the boundary-value conditions u(0, t) = 0 , u(ℓ , t) = 0, u | = g(x). u(x, 0) = f(x), t t=0 (7.9) (7.10) We shall seek a particular solution ( not identically equal to zero) of equation (7.8) that satisfies the boundary conditions (7.9), in the form of the product of two functions X(x) and T(t), of which the function X depend only on x, and the function T depend only on t. i.e. u(x, t) = X(x) T(t) (7.11) Substituting into equation (7.8), we get 2 X(x) T(t) = a X(x) T(t), 2 and dividing both sides by a X(x) T(t) we get: T" X" . = 2 X a T (7.12) The left hand side of this equation is a function that does not depend on x, the right hand side is a function that does not depend on t. Equation (7.12) is possible only when the left and right hand sides are not dependent either on x or on t, that is are equal to a constant number. We denote it by –, where  > 0 (later on we will consider the case  < 0). T" X" = = – . Thus 2 X a T From these equations we get two equations: X +  X = 0, (7.13) T + a  T = 0. (7.14) 2 The general solutions of these equations are X(x) = A cos  x + B sin  x, (7.15) T(t) = C cos a  t + D sin a  t, (7.16) where A, B, C and D are arbitrary constants. Substituting into equation (7.11), we get u(x, t) = (A cos  x+ B sin  x) (C cos a  t+ D sin a  t). (7.17) Now choose the constants A and B so that the conditions (7.9) are satisfied. 187 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Since T (t)  0 (otherwise we would have u(x, t)  0, which contradicts the hypothesis), the function X(x) must satisfy the conditions (7.9); that is, we must have X (0) = 0, X (ℓ) = 0. Putting the value x = 0 and x = ℓ into (7.15), we obtain 0 = A.1 + B.0, 0 = A cos  ℓ + B sin  ℓ. It follows that A = 0, B sin  ℓ = 0, B  0 (otherwise we would have u  0). Consequently, we must have sin  ℓ = 0 , whence  = n/ℓ, (n = 1, 2, 3, ) (7.18) (we do not take the value n = 0, since then we would have X  0 and u  0), and so we have xn = B sin nx , n = 1, 2, 3,   (7.19) These values of X are called eigenvalues of the given boundary-value problem. The functions Xn(x) corresponding to them are called eigenfunctions. 2 Now, if in place of –, we took (– =) k , then equation (7.13), would take the form: 2 X – k X = 0. The general solution is kx X(x) = A e –kx +Be . A non-zero solution in this form cannot satisfy the boundary conditions (7.9). Knowing that  = n ; we can write  T (t) = C cos an t + D sin an t, (n = 1, 2, 3, ) (7.20)   For each value of n, hence for every value of , we put the expressions (7.19) and (7.20) into (7.11) and obtain a solution of equation (7.8) that satisfies the boundary conditions (7.9). We denote this solution by un(x, t). un(x, t) = sin nx [C cos ant + D sin ant ].    (7.21) For each value of n we can take the constants C and D and thus write Cn and Dn (the constant B is included in Cn and Dn). Since equation (7.8) is linear and homogeneous, the sum of the solutions is also a solution, and therefore the function represented by  the series u(x, t) =  u n (x, t) = n 1  sin ant ] sin nx . =  [Cn cos ant + Dn    n 1 188 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Solution (7.22) should satisfy the initial condition (7.10). Put t = 0 in (7.22), we get  (7.23) f(x) = u(x, 0) =  Cn sin nx .  n 1 If the function f(x) is such that in the interval (0, ℓ) it may be expanded in a Fourier series, the condition (7.23), will be fulfilled if we put  Cn = 2  f (x) sin nx dx.   0 (7.24) Series (7.22) will be a solution of equation (7.8) only if the coefficients Cn and Dn are such that this series converges and that the series resulting from a double term-byterm differentiation with respect to x and to t converge as well. Differentiate the terms of (7.22) with respect to t and let t = 0. From the second condition of (7.10), we get the equality  g(x) =  Dn an sin nx .   n 1 We define the Fourier coefficients of this series:  Dn = 2  g(x) sin nx dx. an  0 (7.25) Thus, we have proved that the series (7.22), where the coefficients Cn and Dn are defined by formulas (7.24) and (7.25), is a function u(x , t), which is the solution of equation (7.8) and satisfies the boundary and initial conditions (7.9) and (7.10). Example 7.l0: a) Solve the boundary-value problem –3y u u u (0, y) = 8 e , by the method of separation of variables. =4 , x y –3y b) If u(0, y) = 8 e –5y +4e . Solution: Let u = X Y, then X Y = 4 X Y or X / 4 X = Y / Y. Since X depends only on x and Y depends only on y, and since x and y are independent variables, each side must be a constant, say C. Then X – 4 C x = 0, 189 Advanced Engineering Mathematics Y – C y = 0, 4Cx whose solutions are X = A e Chapter 7- Partial Differential Equations Cy ,Y=Be . C(4x+y) Therefore u(x, y) = X Y = A B e C(4x+y) =De . a) From the boundary condition, Cy u(0, y) = D e –3y =8e , which is possible if and only if D = 8 and C = – 3. –3(4x+y) Then u(x, y) = – 8 e b) Since D e C(4x+y) –12x–3y =–8e , is the required solution. , is a solution, then D1 eC1 (4x  y) and D2 eC2 (4x  y) , are solutions and by the principle of superposition so also is their sum. i.e. a solution is u(x, y) = D eC1 (4x  y) + D eC2 (4x  y) . 1 2 From the boundary condition, –3y –5y u(0, y) = D1 eC1 y + D2 eC2 y = 8 e + 4 e , which is possible if and only if D1= 8, D2 = 4 ; C1=–3, C2= – 5 or D1= 4, D2 = 8 ; C1= –5, C2= – 3. –3(4x+y) Then u(x, y) = 8 e –5(4x+y) +4e –12x–3y =8e –20x–5y +4e , is the required solution. § 7.4:4 Laplace transform methods: The Laplace transform of the partial differential equation and associated boundary conditions are first obtained with respect to one of the independent variables. We then solve the resulting equation for the Laplace transform of the required solution which is then found by talking the inverse Laplace transform. Given the function u(x, t) defined for a  x  b, t > 0, and satisfies the conditions for which L {u(x, t)} exists. Using Leibniz’s rule for differentiating under the integral sign, we get:  u d  st d dU st  u  L{ }= e dt = L {u} = , e u dt =  x x dx dx dx 0 0 d2U  2u and L { }= . 2 2 x dx 190 Advanced Engineering Mathematics Chapter 7  u u st  But L { } =  e dt = s  est u(x, t) dt – u(x , 0) = t t 0 0 = s U(x, s) – u(x, 0) = s U – u(x, 0), Partial Differential Equations (1) where U = U(x, s) = L{u(x, t)}. Put v = u in (1) we get t 2u v u L{ |t=0= } = L { } = s V – v(x, 0) = s L {v} – 2   t t t u = s [s L {u} – u(x, 0)] – | = t t=0 2 u = s U – s u(x, 0) – |t=0. t Example 7.11: Find the solution of –3x u u =2 + u, u(x, 0) = 6 e , which is bounded for x > 0, t > 0. x t Solution: Taking the Laplace transform of the given differential equation with respect to t, we find u u } = 2L{ } + L{u}. x t dU –3x = (2s+1) U – 2 u(x, 0) = (2s+1) U – 12 e . dx L{ Hence (1) from the given boundary condition. Note that the Laplace transformation has transformed the partial differential equation into an ordinary differential equation (1). To solve equation (1) multiply both sides by the integrating factor e   (2s 1)dx = = e(2s 1)x . Then equation (1) can be written in the form d [U e(2s 1)x ] = –12 e(2s  4)x dx Integrating both sides we get e(2s  4)x U e(2s 1)x = 6 + C. s2 191 Advanced Engineering Mathematics e3x + C e(2s 1)x . or U=6 s2 Chapter 7- Partial Differential Equations Now since u(x, t) must be bounded as x, we must have U(x, s) also e3x bounded as x  and it follows that we must choose C = 0. Then U = 6 , and s2 –2t–3x so, on taking the inverse, we find u(x, t) = 6 e . This is easily checked as the required solution. Example 7.12: Solve by Laplace transforms the boundary-value problem u  2 u = , u (x, 0) = 3 sin 2x, t x 2 u (0, t) = u (1, t) = 0, where 0 < x < 1, t > 0. Solution: Taking the Laplace transform of the partial differential equation, we find d2 U d2U s U – u (x, 0) = or – s U = –3 sin 2x. 2 2 dx dx The general solution of this equation is 3 sin 2x. U = A e s x + B e s x + s  4 2 Taking the Laplace transform of those boundary conditions, we have L{u (0 , t)} = U(0 , s) = 0 and L{u(1, t)} – U(1 , s) = 0. Using the first condition U (0, s) = 0, we have A + B = 0. Using the second condition U (l, s) = 0, we have A e s + B e s = 0. Thus A = 0, B = 0 and U = 3 sin 2x, s  4 2 from which we obtain on inversion 2 u(x, t) = 3 e4 t sin 2x. (1) This problem has an interesting physical interpretation. If we consider a solid bounded by the infinite plane face x = 0 and x = 1, the equation 192 Advanced Engineering Mathematics  2u u =k , 2 t x Chapter 7- Partial Differential Equations is the equation for heat conduction in this solid where u(x , t) is the temperature at any plane face x at any time t and k is a constant called the diffusivity, which depends on the material of the solid. The boundary conditions u(0 , t) = 0 and u(1 , t) = 0, indicate that the temperatures at x = 0 and x = 1 are kept at temperature zero, while u(x , 0) = 3 sin 2x, represents the initial temperature everywhere in 0 < x < 1. The result (1) then is the temperature everywhere in the solid at time t > 0. 193 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations EXERCISES 7 1) Eliminate the arbitrary functions from the following equations: 2 b) z = f(x + iy) + g(x – iy), where i = –1, a) z = f(x + ay), c) z = f(x cos  + y sin  – at) + g(x cos  + y sin  + at), 2 2 d) z = f(x – y ), [Answers: a) ax+by e) z = e z z =a , x y 2z 2z 1 2z + = 2 , c) 2 2 2 a x y t z z e) b + a = 2 a b z, x y n y f) z = x f( ). x f(ax + by), 2z 2z b) + = 0, 2 2 x y d) y z z + x = 0, x y f) x z z + y = n z.] x y 2) Eliminate the arbitrary constants from the following equations: 2 2 2 2 b) z = A ept cos qx sin ry, (where p = q + r ) a) z = A ep t cos ps , 2 2 c) z = ax + (1 – a)y + b, d) z = ax + by + a + b , e) a z + b = a x + y, z  2 z = , [Answers: a) t x 2 z z c) + = 1, x y f) z = (x – a) + (y – b) . 2z 2z 2z + = 0, b) + 2 2 2 y t x 2 z z  z  2  z  d) z = x + y +   +   , x y  x   y  2  z  2  z  f) 4z =   +   .]  x   y  2 e) z z = 1, x y 2 2 3) Show that u = f (y – 3x), where f is an arbitrary differentiable function, is a general u u +3 = 0. Find the particular solution which satisfies the solution of the equation x y condition u(0, y) = 4 sin y. [Answer: u(x, y) = 4 sin (y – 3x).] 4) Show that u(x , y) = f(2x + 5y) + g(2x – 5y) is a general solution of 2u  2u = 25 . Find a particular solution satisfying the conditions 4 2 2 y x u | = 0. [Answer: u(x, y) = sin 2x cos 5y.] u(0 , y) = u( , y) = 0, u(x , 0) = sin 2x, t t=0 194 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations 5) Determine whether each of the following partial differential equation is linear or nonlinear, state the order of each equation 4 2 2  u 2u 2u  2u 2u 2u = + a) +4 = 0, b) (x + y ) , + 2 2 4 2 2  x  y t x x y y 2 u 3u 2u 2u –4 =x , = c) u , d) x t 2 x 2 y3 u u = 12 . e) + x y u [Answers: a) Linear of order 2, b) Linear of order 4, c) Nonlinear of order 3, d) Linear of order 2, e) Nonlinear of order one.] 6) Classify each of the following equations as elliptic, hyperbolic or parabolic: u 2u 2u 2u 2u a) –2 = x + 3 y, + +2 = 5, b) 2 2 x xy   x y x y 2 2 2 u 2 u 2u  2u 2u +y +2xy = 0, = , d) x c) 2 2 2 2 x y   x y x y 2 2 2u  2u 2u = 0. +2xy e) (x – 1) + (y – 1) 2 2  x  y y x [Answers: a) Hyperbolic, b) Elliptic, c) Hyperbolic, d) Parabolic, 2 2 2 2 2 2 e) Elliptic if x + y < l, Hyperbolic if x + y > l, Parabolic if x + y = l.] 7) Show that u(x, y) = x f (2x+y) is a general solution of 2 u u x –2x = u. Find the particular solution satisfying u (1, y) = y . x y 2 [Answer: u (x, y) = x (2x + y – 2) .] –3x 8) Show that u(x, y)= 4 e cos 3y is a solution to the boundary value problem –3x 2u 2u + = 0, u(x, /2) = 0, u(x, 0) = 4 e . x 2 y 2 9) Find a partial differential equation having general solution  2u 2u 2u u = f (x – 3y) + g (2x + y). [Answer: 3 –5 –2 = 0.] 2 2  x  y x y 10) Find a partial differential equation having general solution 195 Advanced Engineering Mathematics y a) u = e f(2x – 3y), Chapter 7- Partial Differential Equations b) u = f(x – 2y) + g(2x + y). u 2u 2u u 2u –3 = 0.] [Answers: a) 3 + 2 = 2u, b) 2 –2 2 2 x y  x  y x y  2 u u 11) a) Solve x = 0. + xy y 5 4 b) Find the particular solution for which u(x, 0) = x + x – 34, u (2, y) = 3y . 6 2 4 [Answers: a) x u = f (x) + g (y), b) x u = x + x + 6y – 34 x.] 12) Find general solutions of each of the following: u u 2u 2u a) + 2 = 3u, + = 0, b) x y 2 2 x y 2u 2u  2u 2u 2u –2 = 0, c) + = , d) 2 2 2 2   x y x y x y 2u 2u  2u –3 = 0. –2 e) 2 2  x  y y x 3x [Answers: a) u = e f (y – 2x), b) u = f (x + iy) + g(x – iy), c) u = f(x + y) + g(x – y), d) u = f(x + y) + x g(x + y) e) u = f(3x + y) + g(x – y).] 13) Find general solutions of each of the following: 2 2u  2u 2u 2u 2u +2 b) –3 = + 12 t , = x sin y, a) 2 2 2 2   x y x x t y u u 4u 4u + 2 = x, c) +2 = 4, d) x y 4 3 x x y  4u 4u 4u +2 + = 16. e) 4 2 2 4 x x y y 4 [Answers: a) u = f (x – t) + g(x + t) – t , b) u = f (x + y) + g(2x + t) – 1 x sin y + 3 cos y, 4 2 2 c) u = f(y – 2x) + 1 x , 2 2 3 d) u = f(y) + x g(y) + x h(y) + T(y–2x) + 1 x y, 3 2 2 2 e) u = f(x + iy) + g(x – iy) + x h(x + iy) + x T(x – iy) + 4(x +y ) .] 196 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations 14) Solve each of the following boundary-value problems by the method of separation of variables. –x u u a) 3 + 2 = 0, u (x, 0) = 4 e , x y –5x –3x u u =2 + u, u (x, 0) = 3 e + 2 e , b) x y u  2u c) =4 , u (0, t) = 0, u (, t) = 0, u (x, 0) = 2 sin 3x – 4sin 5x. 2 t x 2 u  u u , = |x=0 = u (2, t) = 0, u (x, 0) = 8 cos 3x – 6 cos 9x . d) 2 4 4 t x x (3y – 2x)/2 [Answers: a) u = 4 e , b) u = 3 e –36t (–5x–3y) (–3x – 2y) +2e –100t , c) u = 2 e sin 3x – 4 e sin 5x, 2 2 –9 t/16 –81 t/16 cos 3x – 6 e cos 9x .] d) u = 8 e 4 4 15) Solve the boundary-value problem by the method of separation of variable 2u 2u u |t=0= f(x) if =4 , u (0, t) = u (5, t) = 0, u (x, 0) = 0, 2 2  t t x a) f (x) = 5 sin x, b) f (x) = 3 sin 2x – 2 sin 5x. [Answers: a) u = 5 sin x sin 4t, 2 b) u = 3 sin 2x sin 4t – 1 sin 5x sin 10t.] 4 5 16) Solve the boundary-value problem by the Fourier series 2u u =2 , u (0, t) = u (4, t) = 0, u (x, 0) = 25x, where 0 < x < 4, t > 0. 2 t x 2 2  em  t /8 cos m sin 1 mx.] [Answer: u (x, t) = – 200   4 m m 1 17) Show that the solution of the boundary-value problem u  2 u u u | , u(x, 0) = f (x), where 0 < x < , is given by = |x=0 = , x x= t x 2 x  2 u (x, t) = 1 a0 +  a m em t cos mx , 2 m 1 197 Advanced Engineering Mathematics 2 am =  f (x) cos m x dx, m = 0, 1, 2, 3, .  0 Chapter 7- Partial Differential Equations 18) Solve each of the following boundary value problems by using Laplace transforms. –2x u u = 3 , u (x, 0) = 4 e , a) t x –2x –4x u u = – 2u, u (x, 0) = 10 e – 6 e , b) t x u  2 u = c) , u (0, t) = u (4, t) = 0, u (x, 0) = 6 sin 1 x + 3 sinx, 2 2 t x u  2 u u u d) , = |x=0 = | , u (x, 0) = 4 sin x – 2 sin 3x. t x 2 x x x=2 –2x–6t 2 – t/4 c) u = 6 e [Answers: a) u = 4 e 2 – t sin 1 x + 3 e 2 2 – t sin x, d) u = 4 e –x–3t , b) u = –10 e 2 –9 t cos x – 2 e –4x–6t +6e , cos 3x.] 19) Solve each of the following boundary value problems by using Laplace transforms u u u = , |x=0 = u (2, t) = 0, u (x, 0) = 8 cos 3x – 6 cos 9x , a) 4 4 t x x 2u 2u u |t=0= 5 sin x, =4 , u (0, t) = u (5, t) = 0, u (x, 0) = 0, b) 2 2  t t x 2 u  u c) =4 , u (0, t) = u (3, t) = 0, u (x, 0) = 10 sin 2x – 6 sin4x. 2 t x 2 2 –9 t/16 –81 t/16 cos 3x – 6 e cos 9x , [Answers: a) u = 8 e 4 4 2 2 –16 t –64 t sin 2x – 6 e sin 4x.] b) u = 5 sin x sin 2t, c) u = 10 e 2 20) Solve each of the following boundary value problems by using Laplace transforms u 2u =2 , u (0, t) = u (5, t) = 0, u (x, 0) = 10 sin 4x – 5 sin 6x, a) 2 t x 2u 2u u | = 0, u (x, 0) = 20 sin 2x – 10 sin 5x, b) =9 , u (0, t) = u (2, t) = 0, t t=0 t 2 x 2 198 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations u  2 u = c) – 4u, u (0, t) = u (, t) = 0, u (x, 0) = 6 sin x – 4 sin 2x, t x 2 2u  2 u u d) = 16 , |x=0= 0, u (3, t) = u (x, 0) = 0, 2 2  x t x u | = 12 cos x + 16 cos 3x – 8 cos 5x, t t=0  2u u u , u (/2, t) = 0, =3 |x=0= 0, u(x, 0) = 30 cos 5x, e) 2 t  x x u u  2u f) =3 |x=0= 0, , u (/2, t) = 0, 2 t  x x u(x, 0) = 20 cos 3x – 5 cos 9x. [Answers: a) u = 10 e 2 –32 t 2 –72 t sin 4x – 5 e –5t sin 6x, –8t b) u = 20 sin 2x cos 6t – 10 sin 5x cos 15t, c) u = 6 e sin x – 4 e sin 2x, d) u = 3 cos x sin 4t + 4 cos 3x sin 12t – 2 cos 5x sin 20t,  3 5 e) u = 30 e –75t –27t cos 5x, f) u = 20 e –243t cos 3x – 5 e  2u u =2 , 0 < x < 3, t > 0, given that u (x, 0) = u (3, t) = 0, 21) Solve 2 t x cos 9x.] u (x, 0) = 54 sin 3x + 2 sin 5x – 2 sin 7x, |u (x, t)| < M, where the last condition states that u is bounded for 0 < x < 3, t > 0. 199 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Some Laplace Transforms f(t) F(s) = L {f(t)} 1 1 1 s 2 t 1 s2 3 t , (n = 0,1, 2, 3, ) n! s n 1 4 1 t  s 5 t  s3 N n 6 t p (p > 1) e te 9 t e 10 1 t 8 (s  )2 2 2 t n t t e (s  )3 (n = 1, 2, 3, ) k t 11 t e , (k > 0) 12 e e at s p 1 1 s t 7 (p  1) n! (s  )n 1 (k  1) (s  )k 1 ab (s  a)(s  b) bt 200 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations N f(t) F(s) = L{f(t)} 13 aeat  bebt ab s (s  a)(s  b) 14 sin b t 15 cos b t 16 sinh  t 17 cosh  t 18 e sin t 19 e cos t b s2  b2 s s2  b2  s2   2 s s2   2  (s  a)2  2 at sa (s  a)2  2 at 201 Advanced Engineering Mathematics Chapter 7- Partial Differential Equations Some Inverse Laplace Transforms N F(s) = L {f(t)} f(t) 1 1 s 1 2 1 s2 t 3 1 , (n = 1, 2, 3, ) n s t (n 1) (n 1)! 4 1 s 1 t 5 1 s3 t 6 1 , (k > 0) k s t (k 1) (k) 7 1 s e t 1 t te (s  )2 8 1 (s  )3 9 1 10 (s  )n , (n = 1, 2, 3, ) 1 11 12 1 t2 et 2 (s  )k , (k > 0) 1 (s  a)(s  b) 202 t (n 1) et (n 1)! t (k 1) et (k) eat  ebt ab Advanced Engineering Mathematics Chapter 7- Partial Differential Equations N F(s) = L{f(t)} f(t) 13 s (s  a)(s  b) aeat  bebt ab 1 sin bt b 14 15 16 17 s2  b2 s s2  b2 cos bt 1 sinh t  s2   2 s s2   2 cosh t 18 1 (s  a)2  2 1 at e sin t  19 sa (s  a)2  2 e cos t 203 at Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series CHAPTER 8 SOLUTION OF DIFFERENTIAL EQUATIONS IN SERIES Equation as simple as y' = 1 + x y 2 and y" = x y can not be solved by finite combinations of algebraic, exponential and trigonometric functions, and many of the equations which occur in the mathematical expression of natural phenomena can not be reduced to any of the soluble forms. We are thus led to investigate solutions, which are expressible, by infinite processes, for example, as the sum of an infinite series of elementary functions. A type of infinite series, which suggests itself, is a power series in x,  y =  an xn . n 0 §8.1 Series solutions near an ordinary point: Let us consider a linear differential equation of the second order (x) y" + (x) y' + (x) y = 0, (8.1) where (x),  (x) and  (x) are polynomials in x. The point xo is said to be ordinary point of the differential equation (8.1) if (xo)  0, otherwise a singular point. The differential equation (8.1) can be written in the form y"(x) = p(x) y' + q(x) y. (8.2) Evidently p(x), q(x) can be expanded in power series of (xxo); which converge for all x in a certain range ( p(x), q(x) analytic at x = xo ). It can be shown that if xo is an ordinary point there exist a solution of equation (8.1) of the form y = A(x) S(x) + B(x) T(x), where both S(x) and T(x) can be expressed in power series in ( x  xo). The radius of convergence of these power series is at least as large as the minimum of the radii of convergence of the series for (x) and (x). This type of equations (8.1) includes many equations of very great importance, which cannot be solved in terms of simple combinations of elementary functions. The usual procedure is to express the solution in the form of an infinite series from which tables of the value of the solution may, if desired, be computed. Thus the convergence of the series is important, not only as a basis of the validity of the process, but also as an indication of the practical value of the result, for a slowly 204 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series converging series is of little use to a computer. § 8.2 Solution developed as a Taylor series: Let us consider the differential equation given in (8.2) and assume that the coefficients p(x), q(x) are one-valued, and have derivatives of all order except possibly for certain isolated values of x. Let x = a be a value for which p(x), q(x) and all derivatives are finite: we shall obtain a solution such that y and y' have assigned finite values yo , y'o when x = a. By subsisting in (8.2) we obtain the corresponding value of the second derivative, namely y "o = p(a) y 'o + q(a) yo . Differentiating (8.2), we obtain y'" = p(x) y" + [p'(x) + q(x)]y' + q'(x) y, in addition, as yo, y 'o , y "o are known, y "'o can be obtained immediately. Continuing the process, we obtain the values of successive derivatives for x = 0, and thus we have the coefficients in the Taylor series y = yo + y 'o (x  a) + y "o n (x  a) 2 ) (x  a) + +  + y (n o 2! n! Example 8.1: To find the solution of y"  x y = 0 such that y = A, y' = B when x = 0. We have iv y" = xy, y"' = x y' + y, y = xy"+ 2y', and in general y (n) = xy (n2) + (n2) y (n3) . The origin is an ordinary point; putting x = 0 in the above, we have y "o = 0 , y "'o = yo , and in general ' y iv o = 2y o , (n  3) , we thus find that y(n) o = (n  2) yo y ov = 3y "o , 1)  2) y(3n) = (1)(4)(7)  (3n  2)A, y(3n = (2)(5)(8)(3nl)B, y(3n = 0. o o o Thus the required solution is y = A[1  x 3 4x 6 28x 9 2x 4 10x 7 80x10    ] + B[x     ] 3! 6! 9! 4! 7! 10! The ratio test proves the convergence of each series separately satisfies the differential equation. Since A and B may be regarded as ordinary constants, the above is an expression of the general solution. 205 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series When p(x), q(x) are polynomials of low degree, a more practical way of obtaining the solution is to assume a series of ascending powers of x  y = ao + a1 x + a2 x +  =  a n x n , 2 n 0 to substitute this series in the differential equation and to determine the coefficients so that the equation is identically satisfied; that is to say, so that the coefficient of each individual power of x cancels out. i.e.  y' = a1 + 2a2 x + 3a3 x + 4a4 x +  =  na n x n 1 , 2 3 n 1  y" = 2a2 + (3)(2) a3 x + (4)(3) a4 x +  =  n(n  1)a n x n  2 , 2 n 2 Substituting in the given differential equation, we get  2a2 +  (n  2)(n  1)a n  2  a n 1 ]x n = 0. n 1 n By equating to zero the coefficient of x . Therefore a2 = 0 and an+2 = a n 1 . This equation, known as a recurrence equation. (n  1)(n  2) 1 1 Since yo = A , y 'o = B , then a1= B, ao = A and a3 = A , a4 = B, (2)(3) (3)(4) 4A (2)(5)B 1 1 1 a5 = , a7 = , a8 = 0 and so on. a 2 = 0 , a6 = a3 = a4 = 6! 7! (4)(5) (5)(6) (6)(7) 1 4 28 1 2 10 80 10 Then y = A[1  x 3  x 6  x 9  ] + B[ x  x 4  x 7  x  ] . 3! 6! 9! 1! 4! 7! 10! Example 8.2: Find the solution in power series of the following differential equation y"  x y' + m y = 0. Solution: It is clear that; x = 0 is an ordinary point of this differential equation.    n 0 n 1 n 2 Let y =  a n x n , y' =  na n x n 1 , y" =  n(n  1)a n x n  2 Substituting we get 206 Advanced Engineering Mathematics   n 2  n 1 Chapter 8 - Solution Of Differential Equations In Series   n(n  1)a n x n  2   na n x n + m  a n x n = 0, n 0  hence  (n  2)(n  1)a n  2 x n +  (m  n)a n x n = 0. i.e. n 0  n 0   (n  2)(n  1)a n  2  (m  n)a n x n = 0. n 0 Equating to zero the coefficients of successive power of x, we obtain the following recurrence relation: i.e. (n+2)(n+1) an+2 + (m  n) an = 0, n = 0, l, 2, 3,  . (n  m) an+2 = an . (n  2)(n  1) We see that the coefficients of even suffix depend on ao but not on a1; those of odd suffix on a1 but not on ao, and that ao, a1, are arbitrary. Denoting by S(x) the series obtained by taking ao = 0, a1 = 1, we have m(m  2) 4 m(m  2)(m  4) 6 x  x  , S(x) = 1  m  2! 4! 6! (m  1)(m  3) 4 (m  1)(m  3)(m  5) 7 x  x  T(x) = x  m 1  3! 5! 7! a nm 0 as n. Since n  2  an (n  2)(n  1) For any finite value of m, both series converges for all values of x. The general solution is y = A S(x) + B T(x), where A, B are arbitrary constants. It will be noted that when m is an even positive integer the series, S(x) terminates and reduces to a polynomial of degree m; where m is odd, T(x) is a polynomial of degree m. Note: It is sometimes possible, by a change of dependent variable, to simplify a linear equation, or to identify it with one previously studied. The transformation most frequently employed is of the form y = u v, where y is the original, v the new dependent variable, and u a chosen function of x. 1 1 For example, if the equation y" + ( m +  x 2 )y = 0, is satisfied by a series 2 4 of ascending integral powers of x as in Example 8.2, the recurrence relation between the coefficients is (n+1)(n+2) an+2 + (m + ½) an  ¼ an2 = 0. This is a three-term recurrence relation, which is much more difficult to manipulation than one that 207 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series involves only two terms. If, however, we make the substitution y = e so that y" =  1 x2 e 4  1 x2 4 v, 2 [ v"  x v' + (¼ x  ½) v], we find that the equation in v is v"  x v' + m v = 0, which is precisely the equation considered above. § 8.3 Regular Singularity: Let us consider an equation of the form (x) y" + (x) y' + (x) y = 0 (8.1) If (x  a) is a factor of the coefficient (x), but not of both (x) and (x), the equation fails to determine a value of y" for x = a, and the procedure of (8.2) is inapplicable. We then say that x = a is a singular point or singularity of the equation. The form of a solution valid in the neighbourhood of a singularity may therefore differ from that of a solution appropriate to an ordinary point. It is important to investigate this difference. If x = a is a singular point, we can rewrite equation (8.1) in the form 2 (x  a) y" + (x  a) p(x) y' + q(x) y = 0 (8.3) The point x = a is called a regular singular point of the differential equation (8.3), if p(x) and q(x) can be expressed in convergent power series of (x  a) (by means of Taylor series of (x  a) ) If x = a is a regular singularity of the differential equation (8.3), then there always exist at least a solution of the form y = ( x  a) m   Cr (x  a)r , Co  0. r 1 This solution converges in the common range of convergence of p(x) and q(x). The substitution x  a = Z shift the singularity to the origin and so there is no loss in generality if we merely consider the equation 2 x y" + x p(x) y' + q(x) y = 0 , where   r 0 r 0 (8.4) p(x) =  pr x r , q(x) =  q r x r , and there is a regular singular point at the origin. 208 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series §8.4 The method of Frobenius: F. G. Frobenius, of Berlin (1873) assume as a trial solution C y=x   ar xr , (8.5) r 1 where the a's are constants. The index C will be determined by a quadratic equation called the Indicial Equation. The roots of this equation may be equal, different and differing by an integer, or different and differing by a quantity not an integer. These cases will have to be discussed separately. It is clear that,  y' =  (C  r)a r x (C  r 1) , r 0  y" =  (C  r)(C  r  1)a r x (C  r  2) . (8.6) r 0 Substituting in (8.4) we get    r 0  r 0 r 0  (C  r)(C  r  1)a r x (C  r) +  p r x r  (C  r)a r x (C  r) +  +  q r x r  a r x (C  r) =0 r 0 r 1 C By equating to zero the coefficients of x we have, [C(C  l) + po C + qo] ao = 0 , as ao  0 i.e. C(C  l) + Po C + qo = 0 (8.7) Equation (8.7) is called the Indicial Equation. It is quadratic in C. It has two roots C1, C2. We have four cases depending on C1, C2. Equating to zero the coefficients of x C+1 C+2 ,x , , we have [C(C + l) + po(C + l) + qo] a1 + (p1 C + q1) ao = 0  [(C + S)(C + S  1) + po(C + S) + qo ] aS + + terms which are linear in aS1, aS2,  ao = 0 (8.8) Corresponding to a root C of the Indicial Equation, we can obtain the successive coefficients a1, a2,  , etc. in terms of ao. We always obtain two solutions of (8.4) in the form (8.5) when the roots of I.E. are complex or are real but do not differ by an integer. There is clearly only one solution of the type (8.5) when the roots of the I.E. 209 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series coincide. When the roots of the I.E. differ by the integer m the larger root yield the desired solution. It may also happen that for the smaller root, the left-hand side of equation (8.8) vanishes identically. In this case we shall see that a second solution of the type (8.5) exists. Case I. Roots of I.E. unequal and differing by a quantity not an integer: Example 8.3: Consider the equation 2 x y" + (x + l) y' + 3y = 0  put  r  a r x =  a r x (C  r) , ao  0, C y=x r 1  giving r 1  y" =  (C  r)(C  r  1)a r x (C  r  2) . y' =  (C  r)a r x (C  r 1) , r 0 r 0 Substituting we get    r 0 r 0 r 0 2  (C  r)(C  r  1)a r x (C  r 1) +  (C  r)a r x (C  r) +  (C  r)a r x (C  r 1) +  + 3  a r x (C  r) = 0.  i.e.  r 1   2(C  r)(C  r  1)  (C  r) a r x (C  r 1) r 0   r 0 r 0  +  (C  r  3)a r x (C  r) = 0. r 0   (C  r)(2C  2r  1) a r x (C  r 1) +  (C  r  3)a r x (C  r) = 0. put r  l = m in the first summation we get (C  1) C(2C  1) ao x   +   (C  m  1)(2C  2m  1) a m 1x (C  m) + m 0 +  (C  r  3)a r x (C  r) = 0. r 0 i.e.  C(2C  1) (C ao x    (C  m  1)(2C  2m  1)a m 1  (C  m  3)a m  x (C  m) = 0. m 0 C1 By equating to zero the coefficients of x C C+1 ,x ,x C(2C  1) ao = 0 ,  , we have I.E. (Indicial Equation) (C + m l)(2C + 2m + l) am+1 + (C + m +3) am = 0 210 1) + Advanced Engineering Mathematics C = 0 or C =  Thus, if Chapter 8 - Solution Of Differential Equations In Series 1 . 2 C = 0, (m + l)(2m + 1) am+1 + (m + 3) am = 0, m = 0, l, 2, 3,  (m  3) am+1 =  am (m  1)(2m  1) i.e. Hence a1 = 3ao , a2 = 2ao, a3 =  2 ao , a4 = 1 ao , a5 =  1 ao and so on, replacing ao 3 7 45 by A we get: 3 4 5 2 y1 = A(l  3x + 2x  2 x + 1 x  1 x + ). 3 If C = i.e. 7 45 1 , (m + 3 )(2m + 2) am+1 + (m + 7 )am = 0. 2 2 2 (2m  7) am+1 =  am . (2m  3)(2m  2) Hence a1 =  7 ao , a2 = 21 ao, a3 =  11 ao , a4 = 1113 ao = 143 ao, 40 80 6 8980 5760 13 13 a5 =  a = a and so on, replacing ao by B we get: 6880 o 3840 o 4 5 2 3 y2 = B x [1  7 x + 21 x  11 x + 143 x  13 x +]. 40 80 6 5760 3840 Thus y = A y1 + B y2 is a solution which contains two arbitrary constants, and so may be considered the complete primitive. m+C In the series we obtained Um = am x U m 1 a m 1 (C  m  3) x. = x=  Then (C  m  1)(2C  2m  1) Um am and the limit when m  is zero, independent of the value of x. Hence both series obtained are convergent for all values of x. Case II. Roots of Indicial Equation equal: Consider the equation (8.4) and its trial solution (8.5). As before equating to C+1 zero all coefficients of x , x C+2 ,  , but do not yet put the coefficient of x equal to zero. Insert in the series (8.5) the values of the coefficients ar in terms of C, we have  y = ao x C  f r (C)x r = ao Y , fo = 1, r 0 Substituting in equation (8.4), we get 2 2 C [x D + x p(x) D + q(x)] Y = ao x [C(C  l) + po C + qo] (8.10) 211 (8.9) Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series Differentiate each side with respect to C. The order of differentiation with respect to x and C may be interchanged and so Y  2 2 [x D + x p(x) D + q(x)] = a o x C [C(C  1)  po C  q o ] = C C     = a o x C [C(C  1)  po C  q o ] ℓn x + a o x C [2C  po  1] . Now choose (8.11) C(C  l) + po C + qo = 0 , if the roots of the indicial equation are equal, then the sum of the two roots C1+ C2 = 2C = 1  po . It is clear that the right-hand side of equation (8.11) is zero,  Y  for C1 = C2 = C, and so Y and  are solutions of our differential equation   C C  C 1  Since Y = x C  f r (C)x r r 0 , fo = 1,   f (C)    Y  C1 C1 r =  + x ( nx) f (C )x then  x   r  xr .  r 1   C C  C1 r 0 r  0  C C1 (8.12)  f (C)  Note that fr(C) are rational functions, and so  r  is best calculated by C   C1 logarithmic differentiation. In general, if the Indicial Equation has two equal roots, we get two Y independent solutions by substituting this value of C in Y and . The second C solution will always consist of the product of the first solution (or a numerical multiple of it) and ℓn x, added to another series. Example 8.4: Integrate in series the equation x y" + y' + x y = 0. Solution: Let y = x C  r   a r x =  a r x (C  r) . r 1  r 1  Then y' =  (C  r)a r x (C  r 1) , y" =  (C  r)(C  r  1)a r x (C  r  2) . r 0 r 0 Substitute in the differential equation we obtain 212 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series     (C  r)(C  r  1)  (C  r) a r x (C  r 1) +  a r x (C  r  2) = 0. r 0 r 0   r 0 r 0  (C  r)2 a r x (C  r 1) +  a r x (C  r  2) = 0. i.e. 2 C ao x C1 2 C 2 + (C + 1) a1 x + (C + 2) a2 x C+1  +  (C  r) 2 a r x (C  r 1) + r 3  +  a r x (C  r  2) = 0. r 0 Thus 2 C ao x C1  2 C 2 + (C + 1) a1 x + (C + 2) a2 x C+1 + +  (C  r  3)2 a r  3  a r  x (C  r  2) = 0.   r 0 Equate coefficients of successive powers of x to zero, we get: 2 1) The Indicial Equations C = 0. 2 2) (C + l) a1 = 0, i.e. a1 = 0 3) (C + 2) a2 = 0, i.e. a2 = 0 4) (C + r + 3) ar+3 + ar = 0 2 2 Therefore a3 =  a6 =  1 (C  6) 1 2 1 (C  3) a , 2 o a3 = 1 2 i.e. ar+3 =  a4 = 0 , a , 2 o 1 (C  r  3)2 ar a5 = 0, a7 = 0 , (C  6) (C  3) 1 a6 =  ao , a9 =  (C  9) 2 (C  9) 2 (C  6)2 (C  3)2 a8 = 0, a10 = 0 , a11 = 0. Hence   x3 x6 x9     , y = xC 1  2 (C  3)2 (C  6)2 (C  3)2 (C  6)2 (C  9)2  (C  3)  is a solution if C = 0 i.e. y = [1  x3 2 3  x6 2 2 3 .6  x9 2 2 2 3 .6 .9  ] , is a solution. Differentiating (1) with respect to C we get 3 2x 6 2x 6 Y C  2x  + = Y ℓn x + x   2 C (C  3)3 (C  6) 2 (C  3) 2 (C  6)3  (C  3) 213 (1) Advanced Engineering Mathematics  3 2x9 2 (C  3) (C  6) (C  9) 2  Chapter 8 - Solution Of Differential Equations In Series 2x9 2 3 2  2 2x9 2 3 (C  3) (C  6) (C  9) (C  3) (C  6) (C  9)  2x 3 18x 6  198x 9  Y      = y ℓn x +  .   3 3 3 3 3 3  C C  0 3 .6 3 .6 .9  3  and +  Y  = Thus the complete solution a y + b    C C  0  2x 3 18x 6   x3  x6 x9 198x 9 [a + b ℓn x] 1  2  2 2  2 2 2   + b  3  3 3  3 3 3   . 3 .6 3 .6 .9 3 .6 3 .6 .9  3   3   x3  x6 x9  Y  = [a + b ℓn x]      1 i.e. a y + b   +  2 4 2 6 2  C C  0 3 3 .(2!) 3 (3!)    x3  x6 x9 1 + 2b   (1  )  (1  1  1 )   . 2 2 2 3 34 (2!) 2 37 (3!)2  3  The series are easily proved to be convergent for all values of x. Case III: Roots of Indicial Equation differing by an integer, making a coefficient of Y infinite: Let the roots of differential equation are C1 and C1 + n (where n is positive integer), and let some of the coefficients of Y become infinite when C = C1 (the smaller root). (Y)C1  n is a solution. In general fr(C) have a factor (C  C1) in the denominator. Write Y1 = ( C  C1 ) Y. Then 2 2 (8.13) C [x D + x p(x) D + q(x)] Y1 = ao x (C  C1) [C(C  1)  po C  q o ] (8.14) Differentiate each side with respect to C. The order of differentiation with respect to x and C may be interchanged and so Y  2 2 a o x C (C  C1 )[C(C  1)  po C  q o ] = [x D + x p(x) D + q(x)] 1 = C C    + a o x C [C(C  1)  po C  q o ] + a o x C (C  C1 )[2C  po  1] .  = a o x C (C  C1 )[C(C  C1 )  po C  q o ] ℓn x + (8.15) Now, since C1(C11) + po C1 + qo = 0, then if we put C = C1, in Equation (8.15) we get 214 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series  Y  2 2 =0 [x D + x p(x) D + q(x)]  1  C   C  C1 i.e.  Y1  is a solution of the differential equation (8.4).  C   C  C1 Thus we have found three solution to the differential equation (8.4) which of the  Y  second order. Possible solutions are (Y)C1  n , (Y1 )C1 ,  1  . C   C  C1 The second of these is a multiple of the first. In general, if the Indicial Equation has two roots C1 and C1 + n differing by an integer n, and if some of the coefficients of Y become infinite when C = C1, we modify the form of Y by multiplying it by (C  C1). We then get two independent solutions by putting C = C1, in the modified form of Y Y1 . The result of putting C = C1 + n in Y merely gives a (which is Y1) and C numerical multiple of that obtained by putting C = C1. Example 8.5: Solve in series the following differential equation x y"  3 y' + x y = 0. Solution: Let y = x   r 1  r 1  a r x r =  a r x (C  r) . C Then y' =  (C  r)a r x r 0 (C  r 1)  , y" =  (C  r)(C  r  1)a r x (C  r  2) . r 0 Substituting we get   r 0   r 0 r 0 r 0   (C  r)(C  r  1)  3(C  r) a r x (C  r 1) +  a r x (C  r 1) = 0.  (C  r)(C  r  4)a r x (C  r 1) +  a r x (C  r 1) = 0. i.e. C(C  4) ao x C1 C + (C + 1)(C3) a1 x +   +  (C  r)(C  r  4)a r x (C  r 1) +  a r x (C  r 1) = 0. r 0 r 3 Thus C(C  4) ao x  C1 C + (C + 1)(C3) a1 x + +   (C  r  2)(C  r  2)a r  2  a r  x (C  r 1) = 0. r 0 215 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series Equate coefficients of successive powers of x to zero we get : 1) The Indicial Equation C(C  4)ao = 0 i.e. C = 0 or C = 4. 2) (C + l)(C  3) a1 = 0 i.e. a1 = 0. 3) (C + r + 2)(C + r  2) ar+2 + ar = 0 1 i.e. ar+2 = ar . (C  r  2)(C  r  2) Therefore 0 = a1 = a3 = a5 = a7 = a9 =  = a2ml. 1 1 a2 =  ao , ao , a4 = (C  2)(C  2) (C  4)C(C  2)(C  2) 1 ao , a6 = 2 (C  6)(C  4)C(C  2) (C  2) 1 a8 = ao , 2 2 (C  8)(C  6)(C  4) C(C  2) (C  2) The roots of the Indicial equation are C = 0 or C = 4. But if we put C = 0, the coefficients a4 , a6 , a8 , become infinite. Now Y1 = (C  C1) Y = (C  0) Y = C Y = C 1 4 C 2 =x [C x  x + (C  2)(C  2) (C  4)(C  2)(C  2) 1 1 6 8 x + x ]  (C  6)(C  4)(C  2) 2 (C  2) (C  8)(C  6)(C  4) 2 (C  2)2 (C  2) Putting C = 0 we get a solution 1 1 1 1 x6  x8 + x10  = u= x4 + 6.4.2.2.2 8.6.4.4.2.2.2 10.8.6.6.4.4.2.2.2 4.2.2 4 6 8 10 2x 2 x 2 x 2 x =            2!  2  3!.1!  2  4!.2!  2  5!.3!  2  But  . Y1 (C2  4) C = Y1 ℓn x + x 1  x2  2 C  [(C  2)(C  2)] 1 1 1  4  1 x +     (C  4)(C  2)(C  2)  C  4 C  2 C  2  216 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 1 1 2 1  6  1     C  6 C  4 C  2 C  2  x  2 (C  6)(C  4)(C  2) (C  2) 1 1 2 2 1  8  1        x + . (C  8)(C  6)(C  4) 2 (C  2)2 (C  2)  C  8 C  6 C  4 C  2 C  2  +  Y  Putting C = 0 we get another solution  1  =v. C   C  0 1 1 1 1 1 6 1 1 1 2 2 1 v = u ℓn x + 1 + x 2  [   ]x + [     ]x8  4 4.2.2 4 2 2 8.6.4.4.2.2.2 8 6 4 2 2 2 4 6 1 x 1  1 1  x  x i.e. v = u ℓn x + 1 +    1     +    2 3  2  2.2!  2  3!.1!  2 8 10 1   1 1 1  1  x  1   1 1 1 1  1 1  x  1         1    +       +   4!.2!   2 3 4  2   2  5!.3!   2 3 4 5   2 3    2  4 Then the general solution is given by A u + B v. Putting C = 4 we get YC=4 = 2 u. The series are easily proved to be convergent for all values of x, except at x = 0. Case IV. Roots of Indicial Equation differing by an integer, making a coefficient of Y indeterminate: If the Indicial Equation has two roots C1, and C1 + n differing by an integer n, and if one of the coefficients of Y becomes indeterminate when C = C1, the complete primitive is given by putting C = C1 + n in Y, which then contains two arbitrary constants. The result of putting C = C1 in Y merely gives a numerical multiple of one of the series contained in the first solution. Example 8.6: With the help of power series, find the solutions of the equation x y" + (x 1) y'  y = 0 . Solution: Let y = x C   r 1  r 1  a r x r =  a r x (C  r) .  Then y' =  (C  r)a r x (C  r 1) , y" =  (C  r)(C  r  1)a r x (C  r  2) , r 0 and proceeding as usual, we get r 0   r 0 r 0  (C  r)(C  r  2)a r x (C  r 1) +  (C  r  1)a r x (C  r) = 0. 217 Advanced Engineering Mathematics C(C2) ao x C1 Chapter 8 - Solution Of Differential Equations In Series  +   a r  (C  r  1)a r 1  (C  r  1)x (C  r) = 0. r 0 Then C(C2) ao = 0 i.e. C = 0 or C = 2 (1) [ar + (C + r + l) ar+1 ] (C + r1) = 0 r = 0 : (ao + (C + 1)a1) (C  l) = 0 i.e. a1 = r = l: [a1 + (C + 2) a2] C = 0, 1 ao C 1 (2) 1 a2 C3 1 a3 i.e. a4 = C4 r = 2: [a2 + (C + 3) a3] (C + 1) = 0, i.e. a3 = r = 3: [a3 + (C + 4) a4] (C + 2) = 0, (3) (4) (5) (6) In general ar+1 =  ar/ (C + r + 1) for all values of r, except at C + r +1 = 0. Equation (1) gives C = 0 or C =2. The coefficient of [a1 + (C + 2) a2 ] in (4) vanishes when C = 0, but as there is no other term in the equation this makes a2 indeterminate instead of infinite. If C = 2 , a2 =  ½ a1 = ½ ao . Thus, if C = 0, from equations (3), (5), (6) we get 1 2 1 2 1 a1 =  ao , a3 =  a2 =  a2 , a4 =  a3 = a2 = a2 , 3 3! 3.4 4! 4 1 1 2 1 1 2 2 a5 =  a4 = a2 = a2, a6 =  a5 = a2 = a2 , a7 =  a2 , 5 3.4.5 5! 6 3.4.5.6 6! 7! This gives us 2 2 2 2 2   (Y)C=0 = ao(1  x) + a2  x 2  x 3  x 4  x 5  x 6  x 7   = 3! 4! 5! 6! 7!   2 2 2 2 2  2 = ao(1  x) + a2 x 1  x  x 2  x 3  x 4  x 5   . 4! 5! 6! 7!  3!  This contains two arbitrary constants, so it may be taken as the complete primitive. It x is clear that (Y)C=0 = (ao  2a2)(l  x) + a2 e . But we have the other solution given by C = 2. 1 2 1 2 2 1 a1 =  ao , a2 =  a1 =  ao , a3 =  a2 = ao ,  a2 , 3 4! 5 5! 3! 4 1 2 1 2 a4 =  a3 = ao , a5 =  a4 = ao 6 6! 7 7! 218 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series and so on. 2 2 2 2 2  2 (Y)C=0 = ao x 1  x  x 2  x 3  x 4  x 5   4! 5! 6! 7!  3!  x = 2 ao e 2 ao (1  x) = 2 ao ex +2 ao (x 1) . That is a constant multiple of the second series in the first solution. Example 8.7: Obtain the solutions of 2 (1  x ) y" + 2 x y' + y = 0. Solution: Proceeding as usual, we get C(C  1)ao x C2  + (C + 1) C a1 x C1 + +  (C  r  2)(C  r  1)a r  2  2(C  r)  (C  r)(C  r  1)  1 a r  x (C  r) = 0. r 0 Hence C(C  1) ao = 0, (1) (C + 1) C a1 = 0 , (2) a2(C + 2)(C + l) + ao[2C  C(C  l) + l] = 0 , (3) a3(C + 3)(C + 2) + a1[2(C + 1)  (C + 1) + 1] = 0, (4) and in general (C + r + 2)(C + r  l) ar+2 +[2(C + r)  (C + r)(C + r l) + l] ar = 0 (5) Equation (1) gives C = 0 or C = 1. The coefficient of a1 in (2) vanishes when C = 0, but as there is no other term in the equation this makes a1 indeterminate instead of infinite. If C = 1, a1 = 0. Thus, if C = 0, from equations (3), (4), (5) we get 1 1 1 1 1 a2 =  ao , a3 =  a1 , a4 =  a2 = ao , a5 =  a1, 8 40 2 2 4 1 3 a6 = ao , a7 = a1 . This gives us 80 560 1 6 1 5 3 7 1 2 1 4 1 3 (Y)C=0 = ao [1  x + x + x + ] + a1 [ x  x + x + x + ]. 8 80 40 560 2 2 This contains two arbitrary constants, so it may be taken as the complete primitive. The series may be proved convergent for |x| < l. 219 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series But we have the other solutions given by C = 1. Working out the coefficients, 1 4 3 6 1 2 (Y)C=1 = ao x[1  x + x + x + ], 40 560 2 that is, a constant multiple of the second series in the first solutions. Some cases where the method fails: If the differential equations is not in the form (8.4), where p(x) and q(x) are finite when x = 0, then the equation is said to have no regular primitives (integrals) in ascending powers of x. Let us consider the following differential equation 4 3 x y" + 2 x y'  y = 0. If we try to apply the usual method, we get for the indicial equation, ao = 0, which has no roots, as by hypothesis ao  0. Such a differential equation is said to have no regular integrals in ascending powers 1/x of x. It is easy to verify that e 1/x equation. Of course e 1/x and e are solutions to the given differential 1/x and e can be expanded in powers of (1/x) . 1 2 y=0, It is easy to see that the equation x y" + 2 x y'  2 x gives us ao = 0 , a1 = 0 and all other coefficients are also zero. If we try to apply the usual method on the differential equation 2 x y"  (l  3x) y' + y = 0. We can see that no integrals that is regular in ascending powers of x, as the one series obtainable diverges for all values of x. The following equation 4 2 3 2 3 x (1  x ) y" + 2x y'  (l  x ) y = 0 , has no integrals that are regular in either ascending or descending powers of x. This (x + 1/x) equation has the primitive : a e (x + 1/x) +be 220 . Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series Exercises 8 Solve the following differential equations in series of powers of x : 2 1. y" + y'  x y = 0 , 2 2. y' = y  x ; y(0) = 2 , y'(0) = l. [Answer: y = 2 + x  1 2 5 3 1 4 x + x + x + ] 6 8 2 [Answer: y = 1 + x + 1 2 2 3 7 4 x  x  x   ]. 2 3 12 y(0) = 1. 3. y' = x + 1 ; y(0) = 1. y 1 1 [Answer: y = 1 + x + x 3  x 4   ]. 3 3 y 4. y' = y + x e ; y(0) = 0. [Answer: y = 8. y' = 2x + cos y ; y(0) = 0. 2 1 2 1 3 1 4 x  x  x   ]. 2 6 6 1 1 [Answer: y = x + x 2  x 3  x 4   ]. 6 4 3 6. y' = x + y ; y(1) = 1. 3 4 2 [Answer: y = 1 + 2(x1) + 4(x1) + 25 (x1) + 81 (x1) +]. 3 4 2 7. y" = x y  y ; y(0) = 1, y'(0) = 2. 1 2 1 3 1 4 x  x  x   ]. 2 3 3 [Answer: y = 1 + 2x  2 2 8. y" = y' + x y  y ; y(0) = 4, y'(0) = 2. 4 2 3 [Answer: y = 4  2x + 2x  2x + 19 x + ]. 6 2 9. y' = y + x ; y(0) = 2. 10. y' = 2y + x  1 ; y(1) = A. 2 11. y' = y + x 3 2 [Answer: y = 2  2x  x ]. 2(x1)  1 x + 1 ]. [Answer: y = ( A + 1 ) e 4 ; y(0) = 1 . 2 2 4 2 3 5 4 [Answer: y = 1 + 1 x + 1 x + 1 x + 9 x + 21 x +]. 2 2 4 8 16 32 320 2 12. y' = x  y ; y(0) = 0. 3 [Answer: y = 1 x  1 x 7  3 13. (1  x) y' = 1 + x  y ; y(0) = 0. 221 7.9 2 x11   ]. 7.11.27 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series [Answer: y = x + 1 x 2  1 x 3  1 x 4   + 1.2 2.3 3.4 1 x n +, (n  1)n the series converges for 1  x  1 ]. 14. x y" + y = 0 ; y(0) = 0 , y'(0) = 1. [Answer: y = x  18. y" + x y = 0 ; y(0) = 1 , y'(0) = 0. 1 x2 + 1 x3 + 1 x4+ ]. 2 2.(1!) 3.(2!)2 4.(3!) 2 3 6 9 [Answer: y = 1  1 x + 1.4 x  1.4.7 x + ]. 3! 16. y" + 2 y' + y = 0 ; y(0) = 1 , y'(0) = 0. x 9! [Answer: y = sin x / x]. 17. y" + 1 y' + y = 0 ; y(0) = 1 , y'(0) = 0. x 6! [Answer: y = 1  12 x 2  2 1 x4  1 x 6   ]. 2 2 2 2 .4 2 .4 .6 2 2 18. y" + y cos x = 0 ; y(0) = a , y'(0) = 0. [Answer: y = a(1  1 x 2  2 x 4  9 x 6  55 x8   ]. 2! 4! 6! 8! 2 19. (1 + x )y" + x y'  y = 0 ; y(0) = a , y'(0) = 0. [Answer: Recurrence relation (n+2) an+2 + (n  1) an = 0,  1.3.5. (2k  3) 2k 2 y = Ao x + A1[1 + 1 x +  (1)k 1 x ]. k 2 2 k! k 2 2 20. y"  x y'  y = 0 . [Answer: Recurrence relation (n+2)(n+1) an+2  (n  1) an1  an = 0, 1 4 1 5 1 6 13 7 2 y = A(1 + 1 x + x  x  x  x  ) + 2 24 20 720 2520 1 1 1 5 7 6 41 7 + B( x + x 3  x 4  x  x  x   )]. 6 12 120 360 5040 2 2 21. y"  2 x y' + 4x y = x + 2 x + 2. [Answer: Recurrence relation (n+3)(n+2) an+3  2(n  2) an = 0, n  2, 2 2 2 9 1 1 1 10 y = A(1  x 3  x 6  x  ) + B( x  x 4  x 7  x  ) + 3 45 405 6 63 567 1 1 1 1 7 1 9 1 10 + ( x 2  x3  x 4  x 6  x  x  x   )]. 3 12 45 126 405 1134 2 2 22. 2 x y"  x y' + ( x + 1 ) y = 0. [Answer: Recurrence relation (C + n + 1)(2C + 2n + 3) an+2 + an = 0, 222 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 1 1 4 1 x  x 6  ) + y = A x (1  x 2  6 168 11088 1 2 1 4 + B x( 1  x  x  1 x 6   ) ]. 28080 10 360 2 23. 3 x y" + 2 y' + x y = 0. [Answer: Recurrence relation (C + n + 3)(3C + 3n +8) an+3 + an = 0, 1 1 1 3 1 6 1/3 x 6  ) + B x ( 1  x  x ) ] y = A (1  x 3  24 2448 30 3420 24. x y" + y'  y = 0. 2 [Answer: Recurrence relation (C + n + 1) an+1  an = 0, n  0, C = 0 , 0 ; 1 1 3 y = (A + B ℓn x) (1 + x  x2  x  )  2 2 (2!) (3!) 1 1 1 1 1  2 B( x  (1  )x 2  (1   )x 3  ) ]. 2 2 3 (2!)2 (3!)2 2 28. (x  x ) y"  3y' + 2y = 0. [Answer: Recurrence relation (C + n  3)an+1  (C + n  2) an = 0, C = 4 , 0 ; 4 2 y = A (x + 2 x + 3) + B x (1 x)2 .] 2 26. (x  x) y" + 3y'  2y = x + 32 . x [Answer: see 25, Recurrence relation (C + n  3)an+1  (C + n  2) an = 0, C = 4 , 0; 4 2 y=A(x +2x+3)+B x (1 x) 2 3 2 2 + 1 x  3 .] 4 5x 2 27. 2(x + x ) y"  (x  3x ) y' + y = 0 [Answer: Recurrence relation an =  anl , C = 1 , ½ 2 3 4 5 y = (A x + B x)( l  x + x  x + x  x +) , | x | < l )]. 28. 4 x y" + 2(l  x) y'  y = 0 [Answer: Recurrence relation 2(c + n + 1) an+1 = an, 1 1 2 1 3 y = A (l  x x  x  ) + 2 3 2.1! 2 .2! 2 .3! 1 1 2 1 + B x (1  x x  x 3  ) 1.3 1.3.5 1.3.5.7 2 2 29. 2 x y"  x y' + (1  x ) y = 0 [Answer: Recurrence relation (C + n + l)(2C + 2n + 3) an+2 = an , a2n+1 = 0, 223 Advanced Engineering Mathematics C = 1, ½, y = Ax (1  Chapter 8 - Solution Of Differential Equations In Series 1 2 1 1 x  x4  x 6  ) + B]. 2.5 2.4.5.9 2.4.6.5.9.13 30. x y" + y' + x y =0 Bessel's equation of order zero. 2 [Answer: Recurrence relation (C + n + 2) an+2 + an = 0 , a2n+l = 0 , C = 0 , 0 1 2 1 1 y = (A + B ℓn x) (1  x  x4  x 6  ) + 2 2 2 2 (2.4) (2.4.6) 1 1 1 1 1 1 + B ( x2  (1  )x 4  (1   )x 6  ) 2 2 3 22 (2.4)2 (2.4.6)2 2 2 31. x y"  x y' + (x + 1) y = 0. 2 [Answer: Recurrence relation (C + n + 1) an+2 + an = 0 , a2n+l = 0, C = 1 , 1, 1 2 1 1 y = x(A + B ℓn x) (1  x  x4  x 6  ) + 22 24 (2!)2 26 (3!)2 1 1 1 1 1 1 + B x ( x2  (1  )x 4  (1   )x 6  ) ]. 2 2 3 22 (2.4)2 (2.4.6)2 32. x y"  2 y' + y = 0 [Answer: Recurrence relation (C + n  2) (C + n + 1)an+1 + an = 0, 1 1 1 5 y = (A + B ℓn x) ( x 3  x 4  x  ) + 12 48 480 1 1 1 19 3 137 3 + B (1  x  x 2  x 3  x  x  ) ]. 2 4 36 576 28800 33. x y" + 2 y' + x y = 0 [Answer: Recurrence relation (C + n + 2)(C + n + 3) an+2 + an = 0, a2n+1 = 0 , C = 0 , 1 1 1 1 1 1 y = A [1  x 2  x 4  ] + B [1  x 2  x 4  ] ]. x 2! 4! 3! 5! 2 34. x (x + l) y" + x(x + l)y'  y = 0. [Answer: Recurrence relation (C + n + 2) an+1 + (C + n )an = 0 , C = 1, 1 1 1 1 1 y = A x [1  x  x 2  x 3  ] + B x (1+x)]. 3 6 10 38. 2x y" + y'  y = x + 1. [Answer: Recurrence relation (C + n + 1) (2C + 2n + 1) an+1  an = 0 , C = 0, ½ 1 1 1 1 1 3 x  ] 2x]. y = A [1  x  x 2  x 3  ] + B x [1  x  x 2  6 90 3 30 360 224 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 3 36. (1  x ) y" + 6 x y = 0. [Answer: Recurrence relation (n + 3) an+3  (n 3)an = 0 , C = 0, ½ 1 1 1 3 y = A(1  x ) + B [x  x 4  x 7  x10  ] ]. 2 14 35 2 2 2 37. 2 x (1 + x ) y" + xy'  12 x y = 0. [Answer: Recurrence relations (2C + 2n + 3) an+2 = (2C + 2n  6)an , C = 0 , ½ 4 8 16 8 x  ] + y = A [1  2x 2  x 4  x 6  7 77 385 5 5 5 6 x  ] . + B x [1  x 2  x 4  4 32 128 38. x y" + y'  4 x y = 0 2 [Answer: (C + n + 2) an+2 = 4 an , C = 0 , 0 ; 1 1 y = (A + B ℓn x)( 1  x 2  x4  x6   )  2 2 (2!) (3!) 1 1 1 1 1  B (x 2  (1  )x 4  (1   )x 6  ) .] 2 2 3 (2!) 2 (3!) 2 39. x y" + 2y' + x y = 0. ( See problem 33 ) [Answer: (C + n + 2)(C + n + 3) an+2 + an = 0, a2n+1 = 0 , C = 0 , 1 1 1 1 1 1 1 y = A (  x  x 3  ) + B ( 1  x 2  x 4  x 6   ) = x 2! 4! 3! 5! 7! A B = cos x  sin x .] x x 2 2 40. 2 x y" + ( 3x  2x ) y'  (x + 1) y = 0 [Answer: (C + n + 2) an+1  an = 0, , C = ½ , 1 1 1 1 1 1 1 y = A (  1  x  x 3  ) + B x [1  (2x)  (2x)2  (2x)3  ] . x 2! 3! 5 5.7 5.7.9 2 2 41. 9 x y"  (x  2) y = 0 [Answer: (3C + 3n + 5)(3C + 3n + 4) an+2  an = 0, C = 1/3 , 2/3 1 2 1 1 1/3 y = A x (1  x  x4  x 6  ) + 5.6 5.6.11.12 5.6.11.12.17.18 1 2 1 1 2/3 + B x (1  x  x4  x 6  ) ]. 6.7 6.7.12.13 6.7.12.13.18.19 2 42. x y"  x2 y' + ( x  2 ) y = 0 [Answer: (C + n + 2) an+1  an = 0, C = 1 , 2, 225 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 1 1 4 1 5 1 x x  x  ) ]. y = A (  1  ) + B (x 2  x 3  4 4.5 4.5.6 x 2 2 2 43. x y" + 2 x y'  ( x + 2 x + 2) y = 0 [Answer: (C + n + 1)(C + n + 4) an+2  2 an+1  an = 0, C = 1 , 2 1 1 3  1  y = (A + B ℓn x) x 1  x  x 2  x   + 5 20  2  1  1 2 1 4 1 5 7 6  +A        1 x x x x x  ]. 2  2 8 40 720 x 44. x y"  x y'  y = 0 [Answer: (C + n) an+1  an = 0, C = 0 , 1;   1 3 11 x x3  x 4   ]. y = (A + B ℓn x) x e + B 1  x 2  12.22 12.22.32  12  48. x y" + y'  x y = 0 2 [Answer: (C + n + 2) an+2  an = 0, C = 0 , 0;   1 2 3 3 4  1  y = (A + B ℓn x) 1  x  x 4   +B   x 2  x   ]. 128  4  22.42  22  3 46. (2x + x ) y"  y'  6 x y = 0 [Answer: (2C + 2n + 1) an+2 + (C + n  3)an = 0, C = 0 , 3/2 ; 3 4 1 6 1 8 2 x + x + ) + y = A( 1 + 3x + x  5 15 65 3 2 1.3 4 1.3.5 6 1.3.5.9 8 3/2 x + x  x + )]. +B x ( 1 + x  8 8.16 8.16.24 8.16.24.32 47. 4 x y" + 2 y' + y = 0 [Answer: 2(C + n + 1)(2C + 2n + 1) an+1 + an = 0, C = 0 , ½; 2 2 y = A ( 1  x/2! + x /4!  ) + B x ( 1  x/3! + x /5!  ) = = A cos x + B sin x )]. 48. 2x(l  x) y" + (l  x)y' + 3y = 0 y= A [Answer: (2C + 2n + 1) an+1  (2C + 2n  3)an = 0, C = 0 , ½; 3 2 3 3 3 4 3 5 x  x  x  x   )] x (l  x) + B( 1  3x + 1.3 3.5 5.7 7.9 2 49. ( x  x ) y" + (1  x) y'  y = 0 2 2 [Answer: (C + n + 1) an+1  (C + n + 2) an = 0, C = 0 , 0; 2 2 2 2 2 3 2 4 y = (A + B ℓn x)(1 + 2 x + 3 x + 4 x + 5 x + )  226 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 2 3  2B(1.2 x + 2.3 x + 3.4 x +  )]. 2 50. ( x  x ) y" + (1  x) y'  y = 0 2 2 [Answer: (C + n + 1) an+1  [(C + n) + 1]an = 0, C = 0 , 0; 2 3 3 4 2 y =(A + B ℓn x)(1+x+ 2 x + 2.5 x + 2.5.10 x + ) + B(2x + x + 14 x +  )]. 4 4.9 27 4.9.16 51. x y" + (1 + x) y' + 2y = 0 2 [Answer: (C + n + 1) an+1 + (C + n + 2) an = 0, C = 0 , 0; 3 2 y =(A + B ℓn x)(12x+ 3 x  4 x +) + 3! 2! 1 1 2 1 1 1 3 1 + B(2 (2  ) x + 3 (2   ) x + 4 (2    ) x   )]. 3! 2! 2 3 2 3 4 2 2 2 52. x y" + x y' + (x  1) y = 0 [Answer: (C + n + 1) (C + n + 3) an+2 + an = 0, C = 1 , 1; 3 5 y =(A + B ℓn x)( 1 x+ 21 x + 2 12 x + ) + 2 2 .4 2 .4 .6 5 +B(1/x + 12 x  (1  1 ) 21 x + (1  1  1 ) 2 12 x 4 2 .4 4 6 2 .4 .6 2 3 2 +  )]. 2 53. x y" + x y' + ( x  4 ) y = 0 Bessel's equation of order 2. [Answer: (C + n ) (C + n + 4) an+2 + an = 0, C = 2 , 2; 2 4 6 8 x + 3 2 12 x ) + y =(A + B ℓn x)( 21 x + 31 x  3 12 2 .4 + B (x 2 2 .4.6 2 .4 .6.8 2 .4 .6 .8.10 2 4 4 31 1 1 11 x  x + x  )]. + 2+ 2 2 2 2 (2.4) (2.4.6) (2.4.6.8) 54. x ( 1  x ) y"  3 x y'  y = 0 [Answer: (C + n ) an+1 (C + n + 1) an = 0, C = 0 , 1; 2 3 4 y =(A + B ℓn x) ( x + 2x + 3x + 4x + ) + 2 3 + B(1 + x + x + x + x4 + ) = ( A + B ℓn x)( x(l  x) 2 1 ) + B(l  x) ]. 58. x ( 1  x ) y"  ( 1 + 3x ) y'  y =0 [Answer: (C + n 1) an+1 (C + n + 1) an = 0, C = 0 , 2; 2 3 4 y =(A + B ℓn x) ( 1.2 x + 2.3x + 3.4x + ... ) + 2 3 4 + B( 1 + x + 5x + 11x + 19x + )]. 2 3 56. ( x + x + x ) y" + 3 x2 y'  2 y = 0 [Answer: (C + n +1) an+2 +(C + n  1) an+1 +(C + n ) an = 0, C = 0 , 1; 2 3 4 5 )+ y =(A + B ℓn x) ( 2x + 2x  x  x + 5 x   4 227 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series 2 3 4 + B( 1  x  5x  x + 11x /3   )]. 2 57. ( 1  x ) y"  2 x y' + 2 y = 0 Legendre's equation of order unity. 2 4 6 [Answers: ( n + 1) an+2  (n  1) an = 0; y = A x + B(l  x  x /3  x /5 +  ]. 2 58. ( 1  x ) y"  2 x y' + m(m + 1) y = 0 2 Legendre's equation of order n. [Answers: (n + 2) (n + 1) an+2  ( n  m) ( n + m + 1) an = 0; 4 y = A(1  m(m + l)x /2! + m(m + l)(m + 3)(m  2)x /4!  ) + 3 5 + B( x  (m  l)(m + 2)x /3! + (m  l)(m  3)(m + 2)(m + 4) x /5!   )]. 2 59. y" + x y = 0 [Answers: (n + 4) (n + 3) an+4 + an = 0; 4 8 12 y = A (1  x /(3.4) + x /(3.4.7.8)  x /(3.4.7.8.11.12)+ ) + 5 9 13 + B(x  x /(4.5) + x /(4.8.8.9)  x /(4.8.8.9.12.13) +  ) ]. 2 60. (2 + x ) y" + x y' + (1 + x) y = 0 2 [Answers: 2(n + 3) (n + 2) an+3 + [(n + 1) + 1] an+1 + an = 0; 2 3 4 5 y = A( 1  x /4  x /12 + 5x /96 + 13x /480 + ) + 3 4 5 + B( x  x /6  x /24 + x /24 +  ) ]. 228 Advanced Engineering Mathematics Chapter 8 - Solution Of Differential Equations In Series REFERENCES (1) Advanced Mathematics for Engineers and Scientists By Murray R. Spiegel. (2) Advanced Calculus, By Murray R. Spiegel. (3) Laplace Transforms, By Murray R. Spiegel. (4) Vector Analysis, By Murray R. Spiegel. (5) Differential Equations, By Jr. Frank Ayres. (Schaum's outline series in Mathematics-McGraw-Hill book company) (6) A course in Mathematics Parts I, II, III. By Dr. Naguib G. Bakhoom. (7) Mathematics for Engineers, By Dr. Abdel-Latif El-Sedeek. (8) Mathematics for Engineering Students (1984) By Dr. A. A. Azim, Dr. M. Shams, and Dr. Labib Iskandar (9) Solved Examples for Engineering Students (1983) By Dr. A. A. Azim, Dr. M. Shams, and Dr. Labib Iskandar. 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