# Thread: Ugly IVP Involving Laplace Transforms

1. ## Ugly IVP Involving Laplace Transforms

I say ugly because it involves omegas and too much messy algebra for my liking.

Yeah, here it is:

$y'' + \omega_0^2y = \mbox{sin}(\omega t); y(0) = 0, y'(0) = 0$

I need to solve for $y(t)$ using Laplace transforms for each of the following cases: when $\omega_0^2 \neq \omega$ and when $\omega_0^2 = \omega$

Both $\omega_0^2$ and $\omega$ are constants here. So basically in short, I'll need to solve the following two equations with the initial conditions I listed already:

1. $y'' + \omega_0^2y = \mbox{sin}(\omega t)$
2. $y'' + \omega y = \mbox{sin}(\omega t)$

I'll start with Case #1 ( $y'' + \omega_0^2y = \mbox{sin}(\omega t)$). Here's what I've done so far:

Taking the Laplace transform of both sides gives me:
$\mathfrak{L}\{y''\} + + \omega_0^2 \mathfrak{L}\{y\} = \mathfrak{L}\{\mbox{sin}(\omega t)\}$

Using the facts that $\mathfrak{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)$, $\mathfrak{L}\{y\} = Y(s)$, as well as $\mathfrak{L}\{\mbox{sin}(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$, I rewrote the above expression as:

$s^2 Y(s) - sy(0) - y'(0) + \omega_0^2 Y(s) = \frac{\omega}{s^2 + \omega^2}$

Once I apply the intial conditions and factor out $Y(s)$, the 2nd and 3rd terms on the left side disappear and I am left with:

$Y(s)(s^2 + \omega_0^2) = \frac{\omega}{s^2 + \omega^2}$

Dividing both sides by $(s^2 + \omega_0^2)$ gives me:

$Y(s) = \frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}$

It looks like finding the inverse transform is going to involve partial fraction decomposition and will be in the form $\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{s^2 + \omega_0^2}$. I'll probably have to solve for A, B, C and D. The next step after that, as far as I've gone, is:

$\omega = (As + B)(s^2 + \omega_0^2) + (Cs + D)(s^2 + \omega^2)$

And that's all I feel like doing for now, because proceeding is going to be ugly...

...bleh, anyone else want to give it a go?

2. Originally Posted by BlazingFire
I say ugly because it involves omegas and too much messy algebra for my liking.

Yeah, here it is:

$y'' + \omega_0^2y = \mbox{sin}(\omega t); y(0) = 0, y'(0) = 0$

I need to solve for $y(t)$ using Laplace transforms for each of the following cases: when $\omega_0^2 \neq \omega$ and when $\omega_0^2 = \omega$

Both $\omega_0^2$ and $\omega$ are constants here. So basically in short, I'll need to solve the following two equations with the initial conditions I listed already:

1. $y'' + \omega_0^2y = \mbox{sin}(\omega t)$
2. $y'' + \omega y = \mbox{sin}(\omega t)$

I'll start with Case #1 ( $y'' + \omega_0^2y = \mbox{sin}(\omega t)$). Here's what I've done so far:

Taking the Laplace transform of both sides gives me:
$\mathfrak{L}\{y''\} + + \omega_0^2 \mathfrak{L}\{y\} = \mathfrak{L}\{\mbox{sin}(\omega t)\}$

Using the facts that $\mathfrak{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)$, $\mathfrak{L}\{y\} = Y(s)$, as well as $\mathfrak{L}\{\mbox{sin}(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$, I rewrote the above expression as:

$s^2 Y(s) - sy(0) - y'(0) + \omega_0^2 Y(s) = \frac{\omega}{s^2 + \omega^2}$

Once I apply the intial conditions and factor out $Y(s)$, the 2nd and 3rd terms on the left side disappear and I am left with:

$Y(s)(s^2 + \omega_0^2) = \frac{\omega}{s^2 + \omega^2}$

Dividing both sides by $(s^2 + \omega_0^2)$ gives me:

$Y(s) = \frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}$

It looks like finding the inverse transform is going to involve partial fraction decomposition and will be in the form $\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{s^2 + \omega_0^2}$. I'll probably have to solve for A, B, C and D. The next step after that, as far as I've gone, is:

$\omega = (As + B)(s^2 + \omega_0^2) + (Cs + D)(s^2 + \omega^2)$

And that's all I feel like doing for now, because proceeding is going to be ugly...

...bleh, anyone else want to give it a go?
Well QickMath gives:

$\frac{\omega}{(s^2+\omega^2)(s^2+\omega_0^2)}=\fra c{\omega}{(\omega^2-\omega_0^2)(\omega_0+s^2)}-\frac{\omega}{(\omega^2-\omega_0^2)(\omega+s^2)}$

CB