2nd Order homogeneous diff equation

**CorruptioN** · January 28th 2010, 09:38 AM

Hi Guys,

I'm having a bit of trouble with a diff equation, I've never dealt with anything quite as complicated as it before and have no idea what to do.

I've been given the equation $m x\prime \prime (t) + b x\prime (t) + k x(t) = 0$ , and also the general solution $x(t) = A\cos(\omega t + \varphi)e^{-\alpha t}$

What i have to do is find $\omega\ and\ \varphi$ in terms of k, b and m. I've never dealt with a 2nd order diff equation where x(t) isn't simply $e^{kt}$ so I don't really have a clue what to do. I tried calculating the derivatives of x and substituting them into the diff equation, but i couldn't seem to rearrange the thing to look anything like the general solution of x.

Any help would be greatly appreciated,

Thanks!

EDIT: What i have to do is find $\omega\ and\ \alpha$

**Aryth** · January 28th 2010, 11:19 AM

You're not trying to find a general solution, as you were already given it. You're looking for a way to solve for $\omega$ and $\phi$

First, you take the derivatives:

$x(t) = Ae^{-\alpha t}[\cos{(\omega t + \phi)}]$

$x'(t) = Ae^{-\alpha t}[-\omega\sin{(\omega t + \phi)} -\alpha\cos{(\omega t + \phi)}]$

$x''(t) = -A\omega^2\cos{(\omega t + \phi)}e^{-\alpha t} + A\omega\alpha\sin{(\omega t + \phi)}e^{-\alpha t} +$ $A\alpha\omega\sin{(\omega t + \phi)}e^{-\alpha t} + A\alpha^2\cos{(\omega t + \phi)}e^{-\alpha t}$

$x''(t) = Ae^{-\alpha t}[\alpha^2\cos{(\omega t + \phi)} + 2\alpha\omega\sin{(\omega t + \phi)} - \omega^2\cos{(\omega t + \phi)}]$

One of the first things you can conclude is that $Ae^{-\alpha t}$ cannot possibly be zero, or else it will render your general solution to be zero, which is obviously not true, this makes the resulting equation:

$m[\alpha^2\cos{(\omega t + \phi)} + 2\omega\alpha\sin{(\omega t + \phi)} - \omega^2\cos{(\omega t + \phi)}] +$ $b[-\omega\sin{(\omega t + \phi)} - \alpha\cos{(\omega t + \phi)}] + k\cos{(\omega t + \phi)} = 0$

Now we group all of the terms together:

$(m\alpha^2 - m\omega^2 - b\alpha + k)\cos{(\omega t + \phi)} + (2m\omega\alpha - b\omega)\sin{(\omega t + \phi)} = 0$

Now, we know that cosine and sine are linearly independent, which means that the only way that a linear combination of these two functions can add up to zero is when the coefficients of each one are both equal to zero, and so:

$m\alpha^2 - m\omega^2 - b\alpha + k = 0$

$-m\omega^2 = -m\alpha^2 + b\alpha - k$

$\omega^2 = \frac{m\alpha^2 - b\alpha + k}{m}$

Now, before we go any further, we must find alpha, which is solved in the next coefficient:

$2m\omega\alpha - b\omega = 0$

$2m\omega\alpha = b\omega$

$\alpha = \frac{b}{2m}$

Now we see alpha squared:

$\alpha^2 = \frac{b^2}{4m^2}$

Plug and simplify:

$\omega^2 = \frac{\frac{b^2}{4m} - \frac{b^2}{2m} + \frac{kb}{2m}}{m}$

Which is:

$\omega^2 = \frac{\frac{b^2}{4m} - \frac{2b^2}{4m} + \frac{2kb}{4m}}{m}$

$\omega^2 = \frac{b(k - b)}{4m^2}$

So, omega equals:

$\omega = \frac{\sqrt{b(k-b)}}{2m}$

Somebody else will have to do phi, I have to go to class.

**CorruptioN** · January 28th 2010, 11:43 AM

Wow, thank you for doing all of that! I didn't even know what linearly independent was so I had no chance of rearranging :P

EDIT: Oops, don't need phi, just needed alpha and omega, so looks like you got everything! Thanks!

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