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Thread: Solve differnetial eqation laplace transforms

  1. #1
    Member zangestu888's Avatar
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    Solve differnetial eqation laplace transforms

    Question is Solve using laplace transform i need someone to clarify the notation i dont get the liimits

    y'' − y' − 6y = 0 for {0 < t < 2 ; e^t for t > 2}, y(0) = 3, y0(0) = 4

    i dont get whats after the brackets
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by zangestu888 View Post
    Question is Solve using laplace transform i need someone to clarify the notation i dont get the liimits

    y'' − y' − 6y = 0 for {0 < t < 2 ; e^t for t > 2}, y(0) = 3, y0(0) = 4

    i dont get whats after the brackets
    The right hand side of the equation is a piece wise defined function which is 0 for $\displaystyle 0<t<2$ and $\displaystyle e^t$ for $\displaystyle t>2$, so you could write this as:

    $\displaystyle y''-y'-6y=u(t-2) e^t $

    where $\displaystyle u $is Heavyside's unit step (note it does not matter that is apprears to now be defined for $\displaystyle t<0$, that part of the solution is not relevant).

    CB
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  3. #3
    Super Member Random Variable's Avatar
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    I'll give it a try.


    $\displaystyle g(t) = \left\{\begin{array}{cc}0,&\mbox{ if } 0 \le t < 2\\e^{t}, & \mbox{ if } t \ge 2\end{array}\right. $

    You want to put $\displaystyle g(t) $ in the form $\displaystyle u(t-2)f(t-2) $.

    $\displaystyle f(t) = e^{t+2}= e^{2}e^{t} $ because then $\displaystyle f(t-2) = e^{t} $


    $\displaystyle y'' - y' -6y = u(t-2)f(t-2) $

    $\displaystyle \mathcal{L} \{y''-y'-6y\} = \mathcal{L}\{u(t-2)f(t-2)\} $

    $\displaystyle \mathcal{L}\{y'' \} - \mathcal{L} \{y'\} - 6 \mathcal{L}\{y\} = e^{-2s} \frac {e^{2}}{s-1} $

    $\displaystyle s^{2} Y(s) -sy(0) -y'(0) - sY(s) + y(0) - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1} $

    $\displaystyle s^{2} Y(s) -3s - 4 - sY(s) + 3 - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1} $

    $\displaystyle Y(s) (s^{2}-s-6) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1 $

    $\displaystyle Y(s) \big((s+2)(s-3)\big) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1 $

    $\displaystyle Y(s) = e^{-2s} \frac {e^{2}}{(s-1)(s+2)(s-3)} + \frac {3s+1}{(s+2)(s-3)} $

    $\displaystyle Y(s) = e^{-2s} \Big(\frac {\text{-}e^{2}}{6} \frac {1}{s-1} + \frac {e^{2}}{15} \frac {1}{s+2} + \frac{e^{2}}{10} \frac{1}{s-3}\Big) + \frac {1}{s+2} + \frac {2}{s-3} $

    EDIT: $\displaystyle y(t)= u(t-2) \Big(\frac {\text{-}e^{2}}{6}e^{t-2} + \frac {e^{2}}{15} e^{-2(t-2)} + \frac{e^{2}}{10} e^{3(t-2)} \Big) + e^{-2t} + 2 e^{3t} $

    so

    $\displaystyle y(t) = \left\{\begin{array}{cc}e^{-2t}+2e^{3t},&\mbox{ if } 0 \le t < 2\\\frac{-e^{t}}{6} + \frac{e^{-2t+6}}{15} + \frac {e^{3t-4}}{10} +e^{-2t} + 2 e^{3t} , & \mbox{ if } t \ge 2\end{array}\right. $


    I'm much more confident this time because I get the same answer when I break up the problem into two differential equations with different initial conditions.
    Last edited by Random Variable; Jul 14th 2009 at 11:01 PM. Reason: correcting major error
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  4. #4
    Member zangestu888's Avatar
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    thanks!! i did it also from scratch i got the same answer! except took me a while with stupied mistakes
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by zangestu888 View Post
    thanks!! i did it also from scratch i got the same answer! except took me a while with stupied mistakes
    I edited my post above.
    Last edited by Random Variable; Jul 14th 2009 at 11:05 PM.
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