# ODE with Fourier Series

1. Oct 21, 2006

### Ne0

Ok we are given the ODE
$${y}^{\prime\prime}(t) + \omega^2{y(t)} = {r(t)}$$
$$r(t) = cos\omega{t}$$
$$\omega = 0.5,0.8,1.1,1.5,5.0,10.0$$
I know you can use variation of paramaters to solve for it so I start by finding the complementary solution.
$${y}^{\prime\prime}(t) + \omega^2{y(t)} = 0$$
We know solutions are of the form
$$y = \exp{(mt)}$$
so after taking derivatives and what not we get the fundamental solution
$$\cos\omega{t}, \sin\omega{t}$$
Our complementary solution is
$${y}_{c}=Acos \omega{t} + Bsin \omega{t}$$
For the particular solution we set
$${y}^{\prime\prime}(t) + \omega^2{y(t)} = cos\omega{t}$$
We then use a fourier series to expand
$$cos\omega{t}$$
Then proceed to solve for it but the problem I'm having is that I'm getting the fourier series to be zero which is strange. I know that there will be no
$${b}_{n}$$
term since cos is even but its still werid why I'm getting zero for
$${a}_{0}, {a}_{n}$$
Any help would be appreciated.

Last edited: Oct 21, 2006
2. Oct 21, 2006

### Ne0

I swear this latex thing I can't figure it out.

3. Oct 21, 2006

### Ne0

If you guys are stuck the answer in the book is:
$${y} = {c}_{1}\cos\omega{t} + {c}_{2}\sin\omega{t} + {A}(\omega)\cos\omega{t}$$

$${A}(\omega) = \frac{1}{\omega^2 - 1} {\leq} 0$$
if $$\omega^2 {\leq} 1$$

$${A}(\omega) = \frac{1}{\omega^2 - 1} {\geq} 0$$
if $$\omega^2 {\geq} 1$$

Since there was not only greater then and less then I had to use less then or equal and greater then or equal

Last edited: Oct 21, 2006
4. Oct 21, 2006

### Ne0

Can anyone help please with what I am doing wrong in the Fourier expansion?