# Question with Diff Eq

• Sep 21st 2009, 04:56 PM
Katzenjammer
Question with Diff Eq
I haven't had Diff Eq, and in my notes for an unrelated class, part of a way to solve a solution was with the following diff eq.

(d^2)x/d(t^2) + a^2 * x = 0 | a=constant; t=0; dx/dt = 0

I really don't know what to do.

Any help is greatly appreciated.

Thanks
• Sep 21st 2009, 06:25 PM
Chris L T521
Quote:

Originally Posted by Katzenjammer
I haven't had Diff Eq, and in my notes for an unrelated class, part of a way to solve a solution was with the following diff eq.

(d^2)x/d(t^2) + a^2 * x = 0 | a=constant; t=0; dx/dt = 0

I really don't know what to do.

Any help is greatly appreciated.

Thanks

Supposing a solution of the form $\displaystyle x=e^{rt}$, we see that our auxiliary equation becomes $\displaystyle r^2+a=0$.

However, I ask this question: is $\displaystyle a$ a positive or negative constant?

I don't want to continue on until this is clarified.
• Sep 21st 2009, 07:00 PM
Katzenjammer
Quote:

Originally Posted by Chris L T521
Supposing a solution of the form $\displaystyle x=e^{rt}$, we see that our auxiliary equation becomes $\displaystyle r^2+a=0$.

However, I ask this question: is $\displaystyle a$ a positive or negative constant?

I don't want to continue on until this is clarified.

I have no idea, it was at the end of class and my Dynamics teacher said "finish this for next class."

I'm assuming it's a positive, since it's acceleration.
• Sep 21st 2009, 07:07 PM
Chris L T521
Quote:

Originally Posted by Katzenjammer
I have no idea, it was at the end of class and my Dynamics teacher said "finish this for next class."

I'm assuming it's a positive, since it's acceleration.

Quote:

Originally Posted by Chris L T521
Supposing a solution of the form $\displaystyle x=e^{rt}$, we see that our auxiliary equation becomes $\displaystyle r^2+a=0$.

However, I ask this question: is $\displaystyle a$ a positive or negative constant?

I don't want to continue on until this is clarified.

Ok, continuing on...

The solution to the characteristic equation is $\displaystyle r^2=-a\implies r=\pm i\sqrt{a}$.

So it follows that the general solution to the equation is $\displaystyle x(t)=c_1\cos\!\left(\sqrt{a}t\right)+c_2\sin\!\lef t(\sqrt{a}t\right)$

Are you sure there isn't another initial condition? Since its a second order equation, there should be one more initial condition.
• Sep 21st 2009, 07:18 PM
Katzenjammer
Just peeked back my notes, he only gave the three initial conditions.

It wouldn't surprise me if he left something out, he's known to do that. :-p