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Thread: differential eq.

  1. #1
    Nov 2005

    differential eq.

    Let y be the solution of the initial value problem $\displaystyle y''+2y'+2y=0, y(0)=0, y'(0)=6 $

    The maximum value of $\displaystyle y $ for $\displaystyle t>0 $ is?
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    Using the characteristic equation we have:

    $\displaystyle m^{2}+2m+2=0$

    $\displaystyle m_{1}=-1+i, \;\ m_{2}=-1-i$

    This gives:

    $\displaystyle e^{-x}(C_{1}cos(x)+C_{2}sin(x))$

    Now, use your initial conditions and get $\displaystyle C_{1}=0, \;\ C_{2}=6$

    $\displaystyle y=6e^{-x}sin(x)$

    Now, what's the max value for x>0?. Graph it may help.

    Differentiate, set to 0 and solve for x.
    Last edited by galactus; Nov 24th 2008 at 05:39 AM.
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