# Help with interpretation of question. (Differential equations)

1. Apr 26, 2012

### sid9221

http://dl.dropbox.com/u/33103477/Linear%20oscilator.png [Broken]

I am having trouble understand the question, what I have done its solve the equation using the substitution $$x=e^{rx}$$

Then, I have the solution given by:

$$x(t)=c_1 e^{t(\sqrt{\gamma^2 - \omega^2 })} + c_2e^{-t(\sqrt{\gamma^2 - \omega^2)}}$$

So at x(0)=0,

$$c_1 + c_2 = 0$$

and, x'(0)=v
$$v=c_2(\sqrt{\gamma^2 - \omega^2} - \omega) + c_1(\sqrt{\gamma^2 - \omega^2}+\omega)$$

Not quite sure how to proceed as if \omega is greater than \gamma I get imaginary values and the opposite gives real, but what does that mean ?

Last edited by a moderator: May 5, 2017
2. Apr 26, 2012

### Dustinsfl

Assuming everything is correct and I see it is correct up to taking the derivative and plugging in zero (I was too lazy to take the derivative). If omega > gamma, you would have $e^{ait}$ where I am using a = radical.

What is $e^{ait}\mbox{?}$ Can you rewrite e in this form as some other transcendental?

If they are real, you just have a DE with of the form Ae+Be.
Equal you just have a constant.

Also, I see you didn't solve for you constant. You set it up but that was it.
$$\begin{bmatrix}1&1&0\\ \sqrt{\gamma^2-\omega^2}-\omega & \sqrt{\gamma^2-\omega^2}+\omega & v \end{bmatrix}$$
When you solve this system, what do you get for $c_1,c_2$?

Last edited: Apr 26, 2012
3. Apr 27, 2012

### sid9221

Okay so working out $$c_1 = -v/2\omega$$ and $$c_2 = v/2\omega$$ what does that mean for the formula I don't get what I'm suppose to interpret from these results ?