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Math Help - Inverse Laplace

  1. #1
    Member Danshader's Avatar
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    Inverse Laplace

    Well i got this question from my tutorial. Given the answers as well with it.Still couldn't understand how my lecturer got the answer.

    Using Laplace Transform methods, solve the following ode:

     tV'' + 2V' + tV = 0

    subject to the following boundary conditions:

    V(0) = 1, V( \pi ) = 1

    Hint: You may assume that   L^{-1} \{ tan^{-1} \left( \frac{1}{s} \right) \} = \frac {sin t}{t}


    This is what i did:

     - \frac{d}{ds} \{  s^2V(s) - sv(0) - v'(0) \} + 2\{ sV(s) - v(0) \} - \frac{d}{ds}V(s) = 0

     V'(s)(-s^2 - 1) - v(0) = 0

     V(s) = \int \frac{1}{s^2 +1} ds

    letting  u = \frac{1}{s}

     V(s) = \int \frac{1}{1+u^2} du

     V(s) = tan^{-1} \frac{1}{s} + C

    Then by inversing laplace

     V(t) = \frac{sin t}{t} +L^{-1} C

    the answer given was  V(t) = \frac{sint}{t} + \frac{\pi}{t}

    i don't know how did  L^{-1} C becomes  \frac{\pi}{t}
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  2. #2
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    Quote Originally Posted by Danshader View Post
    Well i got this question from my tutorial. Given the answers as well with it.Still couldn't understand how my lecturer got the answer.

    Using Laplace Transform methods, solve the following ode:

     tV'' + 2V' + tV = 0

    subject to the following boundary conditions:

    V(0) = 1, V( \pi ) = 1

    Hint: You may assume that  L^{-1} \{ tan^{-1} \left( \frac{1}{s} \right) \} = \frac {sin t}{t}


    This is what i did:

     - \frac{d}{ds} \{ s^2V(s) - sv(0) - v'(0) \} + 2\{ sV(s) - v(0) \} - \frac{d}{ds}V(s) = 0

     V'(s)(-s^2 - 1) - v(0) = 0

     V(s) = \int \frac{1}{s^2 +1} ds

    letting  u = \frac{1}{s}

     V(s) = \int \frac{1}{1+u^2} du

     V(s) = tan^{-1} \frac{1}{s} + C

    Then by inversing laplace

     V(t) = \frac{sin t}{t} +L^{-1} C

    the answer given was  V(t) = \frac{sint}{t} + \frac{\pi}{t}

    i don't know how did  L^{-1} C becomes  \frac{\pi}{t}
    The answer given is wrong as far as I can see.
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