# Method of Variation of Parameters

1. Jun 13, 2012

### ElijahRockers

1. The problem statement, all variables and given/known data

$y''-2y'+y = \frac{e^x}{1+x^2}$

2. Relevant equations

$u_1 = -\int \frac{y_{2}g(x)}{W}dx$
$u_2 = \int \frac{y_{1}g(x)}{W}dx$
$g(x) = \frac{e^x}{1+x^2}$
W is the wronskian of y1 and y2.

3. The attempt at a solution

The characteristic equation for the homogenous solution yields a repeated root of -1, so

$y_{h} = C_{1}e^{-x} + C_{2}xe^{-x}$

When I calculated the Wronskian it simplified to

$W = e^{-2x}$

Plugging in the formula for u1 and simplifying I get

$u_{1} = -\int \frac{xe^{2x}}{1+x^2}dx$

I'm not quite sure how to go about solving this. My best shot at integration by parts didn't really seem to help. I ended up having to integrate ln(x^2 + 1)e^2x which doesn't seem any easier.

2. Jun 13, 2012

### vela

Staff Emeritus
I got W=(1-2x)e-2x.

3. Jun 13, 2012

### ElijahRockers

Hmmm, alright.. I can't seem to find my mistake

if W is
|y1 y2|
|y'1 y'2|

I got
y1 = e^-x
y2 = xe^-x
y'1 = -e^-x
y'2 = -xe^-x + e^-x = e^-x(1-x) (after factoring off the e^-x)

and y1y'2 - y'1y2 = e^(-2x)(1-x) + xe^(-2x) = e^(-2x)

4. Jun 13, 2012

### vela

Staff Emeritus
Sorry, I made a sign mistake.

5. Jun 13, 2012

### vela

Staff Emeritus
Oh, your roots are wrong. They should be +1.

6. Jun 13, 2012

### ElijahRockers

O geez... always something little.

Thanks!