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Math Help - Inital Value Problem with Laplace Transforms

  1. #1
    Member diddledabble's Avatar
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    Post Inital Value Problem with Laplace Transforms

     y" + 6y' +5y = 12e^t

    Solve with the initial conditons y(0)=-1 y'(0)=6

     L{y"}s +6L{y'}s + 5Y(s)= 12e^t

    I am stuck on this because of the 12e^t

    The other problems I think I have done right.
    Example:
    y" -y' - 2y =0 y(0)=-2 y'(0)=5

     L{y'}(s)=sY(s)-y(0) = sY(s) +2
     L{y"}(s) = s^{2}Y(s) -sy10-y'(0) = s^{2}Y(s) + s -5

     (s^{2}Y(s) +s-5) - (sY(s) +2) -2Y(s) = 0
     Y(s)(s^{2}-s-2)+s=0

    Y(s) = \frac{-s}{s^{2}-s-2}

    Y(s) = \frac{-s}{(s-2)(s+1)}
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    Use a LaPlace on the 12e^{t}

    Which is \frac{12}{p-1}

    Then, solve your equation for Y in terms of p.
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  3. #3
    Super Member Random Variable's Avatar
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     y'' +6y'+5y=12e^{t}

     s^{2}Y(s)-sy(0)-y'(0) + 6sY(s) - 6y(0) + 5Y(s) = \frac {12}{s-1}

     s^{2}Y(s) +s - 6 + 6sY(s) + 6 + 5Y(s) = \frac{12}{s-1}

     Y(s)\big(s^{2}+6s+5\big) = \frac {12}{s-1} -s

    Y(s)\Big((s+5)(s+1)\Big) = \frac{12+s- s^{2}}{s-1}

     Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}

    y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}
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  4. #4
    Member diddledabble's Avatar
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    Post You lost me.

     Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}

    You lost me when you seperated out the fractions. How did you do that?

    y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by diddledabble View Post
     Y(s) = \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{1}{s-1} -\frac{3}{4} \frac{1}{s+5} -\frac{5}{4} \frac{1}{s+1}

    You lost me when you seperated out the fractions. How did you do that?

    y(t) = e^{t} - \frac{3}{4}e^{-5t} - \frac{5}{4}e^{-t}
     \frac{12+s-s^{2}}{(s-1)(s+5)(s+1)} = \frac{A}{s-1} + \frac{B}{s+5} + \frac{C}{s+1}

     12+ s- s^2 = A(s+5)(s+1) + B(s-1)(s+1) + C (s-1)(s+5)

    You could expand and equate like terms to find A, B, and C. Or you could let s=1, -1, and -5.
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