# Homework Help: 2nd order non-homgenous DE

1. Oct 31, 2012

### ElijahRockers

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

4y''+4y'+y = cos(2t), y(0)=0, y'(0)=0

2. Relevant equations

y(t)=yh+yp

3. The attempt at a solution

characteristic roots are repeated, m=-1/2, so

$y_h = A_1 e^{\frac{-t}{2}}+A_2 te^{\frac{-t}{2}}$

undetermined coefficients:

yp = Acos(2t)+Bsin(2t)

plugging into original equation and solving, I got A= -15/289 and B= 8/289

using initial conditions, I got A_1 = 15/289 and A_2 = -47/578

so

$y(t) = \frac{15}{289}e^{\frac{-t}{2}} - \frac{47}{578}te^{\frac{-t}{2}} - \frac{15}{289}sin(2t) + \frac{8}{289}cos(2t)$

should be the answer, but apparently it's wrong. I have checked over my work and can't find the fault.

Last edited: Oct 31, 2012
2. Oct 31, 2012

### Zondrina

Your process is correct. You solved L[y] = 0 and your assumption for the solution of L[y] = cos(2t) is also good. It's more than likely an arithmetic error.

3. Oct 31, 2012

### ElijahRockers

Checked it a second time. I must've typed something into the calculator wrong at the very end of the question, just my luck.
$-\frac{47}{578}$ should be $-\frac{1}{34}$ instead.