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Math Help - IVP With Laplace

  1. #1
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    IVP With Laplace

    y"+4y'+29y = <delta>(t-2); y(0) = 0 ; y'(0) = 0
    .....
    Do laplace of both sides and end up with...
    e^-2s * inv laplace [ 1 / (s^2 +4s +29) ]
    I tried to comp the square, but it didnt seem to work out quite like i needed...

    1 / ( (s+2)^2 + 25)

    it looks nice bec 25 is a square too, but i dont know what to do if i did that... i could +4 -4 to top, but i still feel like i am stuck? any help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Smac View Post
    y"+4y'+29y = <delta>(t-2); y(0) = 0 ; y'(0) = 0
    .....
    Do laplace of both sides and end up with...
    e^-2s * inv laplace [ 1 / (s^2 +4s +29) ]
    I tried to comp the square, but it didnt seem to work out quite like i needed...

    1 / ( (s+2)^2 + 25)

    it looks nice bec 25 is a square too, but i dont know what to do if i did that... i could +4 -4 to top, but i still feel like i am stuck? any help
    You should have:

    Y(s)=\frac{e^{-2s}}{s^2+4s+29}

    Then:

    y(t)=u(t-2) \mathcal{L}^{-1} \left[  \frac{1}{s^2+4s+29} \right]

    CB
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  3. #3
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    yeah thats what i have, but when i did the CTS for the inverse laplace, i got stuck and idk where to go from there.
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  4. #4
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    Quote Originally Posted by Smac View Post
    yeah thats what i have, but when i did the CTS for the inverse laplace, i got stuck and idk where to go from there.
    Complete the square, use a shift theorem and recognise a standard transform.
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