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I'm trying to solve this ODE :- $x (dy/dx) + y \log(x) = e^x x^{(1-1/2 \log(x))}$

I divided the equation throughout by $x$, obtaining $(dy/dx) + y\log(x)/x = e^x x^{(\log(x^{-1/2}))}$.

Then, I obtained the Integrating factor as $e^{((\log (x)^2)/2)}$.

Then, $y e^{((\log(x)^2)/2)} = \int e^{(x + ((\log(x)^2)/2))} x^{(\log(x^{-1/2}))} dx$.

I'm not sure how to proceed from here. It seems the integral on the right has to be evaluated by parts but it is quite tedious and tends to get messy. Is there a simpler way to evaluate the integral? Or is there any other approach to this differential equation?

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up vote 2 down vote accepted

You're almost there. You just need to recognize that

$$x^{\log{x^{-1/2}}} = e^{-(1/2) \log^2{x}}$$

The nasty factors cancel and the integral becomes trivial.

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Thanks a lot, Ron! Thanks to you, I don't have to go through tons of messy calculations! I don't know why I don't "see" this kind of stuff... –  Jobin Idiculla Oct 6 '13 at 14:29
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