Solving for x ? [differential equation]

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November 28th 2010, 03:57 AM
Ihaz

Solving for x ? [differential equation]

Hello,

I'm stuck with this equation and I can't figure out a way how to solve for t..
So I got the solution of a differential equation(complex roots), and now I need to find the values of t for which y=3.5

3.5 = 12e^(-αt)*cos(ω t) - 12α/ω*e^(-αt)*sin(ω t)

α and ω are constants, I tried changing it to the exponential form, cosine form but it didn't help.
In fact what I need is the time intervals for which y(t) > 3.5

Any tips or advice how to do it ?

Thanks !
November 28th 2010, 04:01 AM
Prove It

Exponential form should work...

What is your original DE?
November 28th 2010, 04:22 AM
Ihaz

d^2 y/dt^2 + 2α * dy/dt + (ω0)*y=0

But with the exponential form, I get 6 *(e^complexNum + e^conjugate) ??
November 28th 2010, 05:09 AM
Sudharaka

Quote:

Originally Posted by Ihaz

d^2 y/dt^2 + 2α * dy/dt + (ω0)*y=0

But with the exponential form, I get 6 *(e^complexNum + e^conjugate) ??

Dear Ihaz,

$\frac{d^{2}y}{dt^{2}}+2a\frac{dy}{dt}+\omega_{0}y= 0$

The solution will be, $y=Ae^{(-a+\omega)t}+Be^{(-a-\omega)t}~ where~\omega=\sqrt{a^2-\omega_{0}^{2}}~and~A,B:~arbitary~ constants$

So in this form using the boundry conditions you can find A and B. Can you give us more infromation about the question. What are the boundry conditions?
November 28th 2010, 05:21 AM
Ihaz

I already found the solution of the D.E :
12e^(-αt)*cos(ωn t) - 12α/ωn*e^(-αt)*sin(ωn t)

Now what I need is at what values of time will this be equal to 3.5 ??