Solving a DE: Variation of Parameters & Integration Issues

[Lancelot59]

I've picked up a bit more since my last problem. I need to solve the following DE:
x^{2}\frac{dy}{dx}+x(x+2)y=e^{x}

I decided to use variation of parameters, so I re-arranged it like so:
\frac{dy}{dx}=\frac{e^{x}}{x^{2}}-(1+\frac{2}{x})y

Then solved the homogenous DE:
\frac{dy}{dx}=-(1+\frac{2}{x})y
y=e^{-x}x^{-2}c

Now for the particular solution:
y_{p}=u(x)e^{-x}x^{-2}c<br /> \frac{dy}{dx}=u&amp;#039;(x)e^{-x}x^{-2}-u(x)e^{-x}x^{-2}-2u&amp;#039;(x)e^{-x}x^{-2}<br /> <br /> When I shoved this back in I wound up with this for u&#039;(t):<br /> u&amp;#039;(x)=e^{x}x^{-2}<br /> <br /> It seems...a bit strange. Did I mess up somewhere? It&#039;s a bit hard to integrate. I&#039;ve gone over this several times already.

 

[HallsofIvy]

Looks to me like you have an algebra error somewhere. Doing exactly what you say, I get u&#039;e^{-x}= e^x or u&#039;= e^{2x}.

 

[vishal007win]

what is u(t)?

 

[Lancelot59]

u(t) is the unknown function, that when multiplied by the solution to the homogenous equation, gives you a particular solution to the DE. I forgot to put in the step where I set that part up. It should also be u(x). I'll try going over the algebra again.