TITLE: Alembert method $xy''-y'=x^2e^x$, $y_1=x^2$
QUESTION [1 upvotes]: The question asks to use the Alembert method to solve $xy''-y'=x^2e^x$, being $y_1=x^2$ a solution of the associated homogeneous equation. 
I did $y=ux^2$, 
$y'=u'x^2+2xu$,
$y''=u''x^2+4xu'+2u$
$u''x+3u'=e^x$
Making $u'=v$
$v'x+3v=e^x$
then i got $v=(e^x+c_1)/x$
How can i get $u$? Is all i have done before right? 
Thanks

REPLY [1 votes]: You have $v$ (According to Wolfram):
$$v = \frac{e^x}{x} - \frac{2e^x}{x^2} + \frac{2e^x}{x^3}$$
When using the reduction of order method (Also known as the method you described above), then you do not need the constant values.
Since $u' = v$
$$u' = \frac{e^x}{x} - \frac{2e^x}{x^2} + \frac{2e^x}{x^3}$$
Integrating we get:
$$u = \frac{e^x}{x} - \frac{e^x}{x^2}$$
Therefore, our second solution is:
$$y_2 = ux^2 = \left(\frac{e^x}{x} - \frac{e^x}{x^2}\right)x^2 = xe^x - e^x = e^x(x-1)$$
Our general solution is:
$$y = c_1x^2 + c_2(e^x(x-1))$$
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