# Homework Help: Solving A DE

1. Sep 30, 2011

### 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:
$$y_{p}=u(x)e^{-x}x^{-2}c [tex]\frac{dy}{dx}=u'(x)e^{-x}x^{-2}-u(x)e^{-x}x^{-2}-2u'(x)e^{-x}x^{-2}$$

When I shoved this back in I wound up with this for u'(t):
$$u'(x)=e^{x}x^{-2}$$

It seems...a bit strange. Did I mess up somewhere? It's a bit hard to integrate. I've gone over this several times already.

Last edited: Sep 30, 2011
2. Sep 30, 2011

### HallsofIvy

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

3. Sep 30, 2011

### vishal007win

what is u(t)?

4. Sep 30, 2011

### 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.