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Thread: Inverse Laplace

  1. #1
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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:

    $\displaystyle tV'' + 2V' + tV = 0$

    subject to the following boundary conditions:

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

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


    This is what i did:

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

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

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

    letting $\displaystyle u = \frac{1}{s} $

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

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

    Then by inversing laplace

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

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

    i don't know how did $\displaystyle L^{-1} C $ becomes $\displaystyle \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:

    $\displaystyle tV'' + 2V' + tV = 0$

    subject to the following boundary conditions:

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

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


    This is what i did:

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

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

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

    letting $\displaystyle u = \frac{1}{s} $

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

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

    Then by inversing laplace

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

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

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