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Thread: Ugly IVP Involving Laplace Transforms

  1. #1
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    Ugly IVP Involving Laplace Transforms

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

    Yeah, here it 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 Case #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:

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

    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:

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

    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?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by BlazingFire View Post
    I say ugly because it involves omegas and too much messy algebra for my liking.

    Yeah, here it 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 Case #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:

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

    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:

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

    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:

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

    CB
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