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A function for integration factors in solving equations

�(�)= e ∫ U − −U dU If ( , )�x+ ( , )�y=0 equation is not a perfect first-order equation But It is possible to transform it to a one with multiplying by a function like �(�) which � is with respect to( or or x,y or function of them) So: Satisfies the equality: That is: (�) +� �(�) ( , )�x+�(�) ( , )�y= ( , )�x+ ( , )�y=0 = =�(�) +� dμ dμ dU = =��� , dU d d And the chain rule says � = �y= dμ dμ dU = =��� d dU d And with substituting and arrangement, the rest comes then: (�) −�(�) At last by integration. μU μ U =U − −U And μ U = e ∫ →integration with respect to U →ln � =∫ U U − −U dU is again obtained. For checking take �= and if you obtain ( )= e know is). ∫ − d − −U =��� −��� �� my verification will be correct (which as you