second order differential eq

**prathibha** · April 30th 2009, 10:11 PM

x^2 y" - x y' + y = 2 log x+ x^2 log x + x^2 + 1/x

**matheagle** · April 30th 2009, 10:21 PM

You can solve the homogeneous part via $y=x^m$ .

**Danny** · May 1st 2009, 03:46 AM

You can also let $u = x y' - y$ so $u' = x y''$ so your equation becomes

$xu' - u = 2 \log x+ x^2 \log x + x^2 + 1/x$

a first order linear equation. Once you solve it you then solve $xy' - y = u$ which also a first order linear equation.

**prathibha** · May 1st 2009, 11:55 PM

thank u!!

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