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Math Help - 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:

    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?
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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:

    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
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