Answer to Question #170782 in Differential Equations for Samprada Kale

Question

(x^2D^2-xD+1)y=xlogx

Answer 1

$x^2 y''-xy'+y=x\ln x$

This equation is linear, but non-homogeneous. So, at the first step we seek a particular solution of the homogeneous equation $x^2 y''-xy'+y=0$.

We seek a solution of the form $y=x^{\alpha}$ . Then

$0=x^2 y''-xy'+y=x^2 \alpha(\alpha-1)x^{\alpha-2} - x\alpha x^{\alpha-1} +x^{\alpha} =(\alpha-1)^2x^{\alpha}$

If $\alpha=1$ then we have a solution $y=cx$ , with $c$ be an arbitrary constant.

The second step is variation of constants. We replace the constant $c$ with a function $u(x)$. Then

$y=xu$

$y'=xu'+u$

$y''=xu''+2u'$

and the origin equation transforms to the following one:

$x^2(xu''+2u')-x(xu'+u)+xu=x\ln x$

$x^3u''+x^2u'=x\ln x$

$xu''+u'=x^{-1}\ln x$

$(xu')'=x^{-1}\ln x$

$xu'=\int \ln x \frac{dx}{x}=\frac{1}{2}\ln^2x+c_1$

$u'=\frac{1}{2}x^{-1}\ln^2x+c_1x^{-1}$

$u=\int( \frac{1}{2}x^{-1}\ln^2x+c_1x^{-1})dx=\int(\frac{1}{2}\ln^2x+c_1)d\ln x=\frac{1}{6}\ln^3 x+c_1\ln x+c_2$

$y=xu=\frac{1}{6}x\ln^3 x+c_1x\ln x+c_2x$

This is a general solution of ODE.