# Diff EQ

1. Sep 17, 2006

### Warr

I took an ODE course last year, but I seem to have forgotten some stuff. I need to solve this equation:

$$\frac{d^2u}{dt^2} + {\omega}^2u = f_osin({\mu}t)$$

with the boundry conditions:

u(0) = 0, du/dt(0) = 0

When I tried to solve the homogenenous equation first, I got

$$u_g(t)=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}$$

I then differentiated and set up the system with the two boundy conditions...but I got c1+c2=0 and c1-c2=0...c1=c2=0.

This seems wrong. Any help would be appreciated

2. Sep 17, 2006

### AKG

I think those might be called initial conditions, not boundary conditions, not that it matters. Anyways, you need to find a particular solution, add it to your ug, then find the ci by looking at the initial conditions.

3. Sep 17, 2006

### Warr

Here is what I did now:

let the particular solution be of the form $$u_p=Asin({\mu}t})$$

differentiated twices gives $$u_p''=-A{\mu}^2sin({\mu}t})$$

so now from my origional diff eq I get

$$A({\omega}^2-{\mu}^2)sin({\mu}t)=f_osin({\mu}t)$$

hence

$$A = \frac{f_o}{{\omega}^2-{\mu}^2}$$

and therefore

$$u_p=\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})$$

so now the general solution is

$$u=c_1e^{i{\omega}t}+c_2e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})$$

plugging in the initial conditions from the first post, I got

$$c_1=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}$$

$$c_2=-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}$$

and therefore the entire solution to be

$$u=\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{i{\omega}t}-\frac{{\mu}f_o}{2({\mu}^2-{\omega}^2)}e^{-i{\omega}t}+\frac{f_o}{{\omega}^2-{\mu}^2}sin({\mu}t})$$

Last edited: Sep 17, 2006
4. Sep 17, 2006

### AKG

I think you're missing a factor of $i\omega$ in the denominators of c1 and c2.

5. Sep 18, 2006

### Warr

You are right. Also, is it right to say that the solution exists as long as $$|\mu| {\neq} |\omega|$$ (assuming that they are both real)?

Thanks for the help!

Last edited: Sep 18, 2006
6. Sep 18, 2006

### HallsofIvy

If you are going to use $sin(\mu t)$ as you particular solution, why not use $sin(\omega t)$ and $cos(\omega t)$ in the general solution?

Yes, the solution is what you have as long as $|\mu|\ne|\omega|$. However, if you are thinking that when they are equal, a solution does not exist, that is not necessarily so. You just need to use a more complicated function as your specific solution.