BlazingFire
New member
- Joined
- Mar 12, 2009
- Messages
- 5
The question I am currently working on right now is:
\(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t); y(0) = 0, y'(0) = 0\)
I need to solve for \(\displaystyle y(t)\) using Laplace transforms for each of the following cases: when \(\displaystyle \omega_0^2 \neq \omega\) and when \(\displaystyle \omega_0^2 = \omega\)
Both \(\displaystyle \omega_0^2\) and \(\displaystyle \omega\) are constants here. So basically in short, I'll need to solve the following two equations with the initial conditions I listed already:
1. \(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t)\)
2. \(\displaystyle y'' + \omega y = \mbox{sin}(\omega t)\)
I'll start with #1 (\(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t)\)). Here's what I've done so far:
Taking the Laplace transform of both sides gives me:
\(\displaystyle \mathfrak{L}\{y''\} + + \omega_0^2 \mathfrak{L}\{y\} = \mathfrak{L}\{\mbox{sin}(\omega t)\}\)
Using the facts that \(\displaystyle \mathfrak{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)\), \(\displaystyle \mathfrak{L}\{y\} = Y(s)\), as well as \(\displaystyle \mathfrak{L}\{\mbox{sin}(\omega t)\} = \frac{\omega}{s^2 + \omega^2}\), I rewrote the above expression as:
\(\displaystyle s^2 Y(s) - sy(0) - y'(0) + \omega_0^2 Y(s) = \frac{\omega}{s^2 + \omega^2}\)
Once I apply the intial conditions and factor out \(\displaystyle Y(s)\), the 2nd and 3rd terms on the left side disappear and I am left with:
\(\displaystyle Y(s)(s^2 + \omega_0^2) = \frac{\omega}{s^2 + \omega^2}\)
Dividing both sides by \(\displaystyle (s^2 + \omega_0^2)\) gives me:
\(\displaystyle Y(s) = \frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}\)
\(\displaystyle y(t) = \mathfrak{L^{-1}}\{{\frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}\}\)
It looks like finding the inverse transform is going to involve partial fraction decomposition and will be in the form \(\displaystyle \frac{As + B}{(s^2 + \omega^2)} + \frac{Cs + D}{(s^2 + \omega_0^2)}\). I'll probably have to solve for A, B, C and D. The next step after that, as far as I've gone, is:
\(\displaystyle \omega = (As + B)(s^2 + \omega_0^2) + (Cs + D)(s^2 + \omega^2)\)
Bleh, for some reason I don't feel like finishing this up right now (I'm drained anyway due to everything else I have to do non-math related), but I know that the rest of the problem is going to involve a lot of messy algebra.
Anyway, have I done everything right so far? If I've gone wrong anywhere, please let me know, and if you for some reason want to finish up where I've left off (particularly getting the constants), even better...
\(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t); y(0) = 0, y'(0) = 0\)
I need to solve for \(\displaystyle y(t)\) using Laplace transforms for each of the following cases: when \(\displaystyle \omega_0^2 \neq \omega\) and when \(\displaystyle \omega_0^2 = \omega\)
Both \(\displaystyle \omega_0^2\) and \(\displaystyle \omega\) are constants here. So basically in short, I'll need to solve the following two equations with the initial conditions I listed already:
1. \(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t)\)
2. \(\displaystyle y'' + \omega y = \mbox{sin}(\omega t)\)
I'll start with #1 (\(\displaystyle y'' + \omega_0^2y = \mbox{sin}(\omega t)\)). Here's what I've done so far:
Taking the Laplace transform of both sides gives me:
\(\displaystyle \mathfrak{L}\{y''\} + + \omega_0^2 \mathfrak{L}\{y\} = \mathfrak{L}\{\mbox{sin}(\omega t)\}\)
Using the facts that \(\displaystyle \mathfrak{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)\), \(\displaystyle \mathfrak{L}\{y\} = Y(s)\), as well as \(\displaystyle \mathfrak{L}\{\mbox{sin}(\omega t)\} = \frac{\omega}{s^2 + \omega^2}\), I rewrote the above expression as:
\(\displaystyle s^2 Y(s) - sy(0) - y'(0) + \omega_0^2 Y(s) = \frac{\omega}{s^2 + \omega^2}\)
Once I apply the intial conditions and factor out \(\displaystyle Y(s)\), the 2nd and 3rd terms on the left side disappear and I am left with:
\(\displaystyle Y(s)(s^2 + \omega_0^2) = \frac{\omega}{s^2 + \omega^2}\)
Dividing both sides by \(\displaystyle (s^2 + \omega_0^2)\) gives me:
\(\displaystyle Y(s) = \frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}\)
\(\displaystyle y(t) = \mathfrak{L^{-1}}\{{\frac{\omega}{(s^2 + \omega^2)(s^2 + \omega_0^2)}\}\)
It looks like finding the inverse transform is going to involve partial fraction decomposition and will be in the form \(\displaystyle \frac{As + B}{(s^2 + \omega^2)} + \frac{Cs + D}{(s^2 + \omega_0^2)}\). I'll probably have to solve for A, B, C and D. The next step after that, as far as I've gone, is:
\(\displaystyle \omega = (As + B)(s^2 + \omega_0^2) + (Cs + D)(s^2 + \omega^2)\)
Bleh, for some reason I don't feel like finishing this up right now (I'm drained anyway due to everything else I have to do non-math related), but I know that the rest of the problem is going to involve a lot of messy algebra.
Anyway, have I done everything right so far? If I've gone wrong anywhere, please let me know, and if you for some reason want to finish up where I've left off (particularly getting the constants), even better...