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Math Help - IVP with Laplace Transform

  1. #1
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    IVP with Laplace Transform

    Solve the initial value problem using Laplace Transforms.

    \ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by MathTragic View Post
    Solve the initial value problem using Laplace Transforms.

    \ddot{y} - 6 \dot{y} + 9y = te^{3t}, y(0) = 1, \dot{y}(0) = 0
    Apply the Laplace transform to both sides of the equation to get

    s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}

    Apply the initial conditions and simplify to get

    s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6 \implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-3)^2}

    Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.

    Spoiler:
    For the first and third term on the RHS, use a translation theorem to make life easier. The second term on the RHS can be rewritten as \frac{1}{s-3}+\frac{3}{(s-3)^2}. Apply translation theorems when needed.
    Last edited by Chris L T521; August 30th 2009 at 10:30 PM. Reason: fixed typo
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Apply the Laplace transform to both sides of the equation to get

    s^2Y(s)-s y(0)-\dot{y}(0)-6\left(sY(s)-y(0)\right)+9Y(s)=\frac{1}{(s-3)^2}

    Apply the initial conditions and simplify to get

    s^2Y(s)-6sY(s)+9Y(s)-s+6=\frac{1}{(s-3)^2}\implies (s-3)^2Y(s)=\frac{1}{(s-3)^2}+s-6 \implies Y(s)=\frac{1}{(s-3)^4}+\frac{s}{(s-3)^2}-\frac{6}{(s-2)^2}

    Take the Inverse Laplace Transform of both sides. I leave it for you to continue from here.

    Spoiler:
    For the first and third term on the RHS, use a translation theorem to make life easier. The second term on the RHS can be rewritten as \frac{1}{s-3}+\frac{3}{(s-3)^2}. Apply translation theorems when needed.
    is the -\frac{6}{(s-2)^2} ie. the third term of the Y(s) equation supposed to be -\frac{6}{(s-3)^2} ?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    is the -\frac{6}{(s-2)^2} ie. the third term of the Y(s) equation supposed to be -\frac{6}{(s-3)^2} ?
    Ah..yes it is. Thank you for pointing out that typo. It's fixed now.
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  5. #5
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    i get the answer to be:
    y(t) = e^{3t} (2t + 1)
    is this correct?
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  6. #6
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    Quote Originally Posted by wik_chick88 View Post
    i get the answer to be:
    y(t) = e^{3t} (2t + 1)
    is this correct?
    Does it satisfy the initial conditions?

    CB
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