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Second order diff eq with i

  1. Mar 4, 2007 #1
    Crap, nevermind, I left b^2 out of the quadratic formula, thanks anyways.

    Here's the question:
    Find y as a function of t from the diff eq:
    y''+6y'+25y=0 with the initial conditions y(0)=8 and y'(0)=8

    I used the form r^2+6r+25=0 to solve for r and through the quadratic equation got r = -3+/-5i
    so my equation now looks like
    c1*e^(-3t)*cos(5t)+c2*e^(-3t)*sin(5t)=8
    The second part of the equation cancels and I'm left with c1=8
    Now to find y'(0) I have:
    -3*c1*e^(-3t)*cos(5t)+c1*e^(-3t)*(-5*sin(5t))+-3*c2*e^(-3t)*sin(5t)
    +c2*e^(-3t)*(5*cos(5t))
    The second and third terms in that cancel out and I'm left with
    -3*c1*e^(-3t)*cos(5t)+c2*e^(-3t)*5*cos(5t), and plugging in 0 for t and 8 for c1 I get:
    -24+5*c2=8
    5*c2=32
    c2=32/5

    So the final equation looks like
    8*e^(-3t)*cos(5t)+32/5*e^(-3t)*sin(5t)

    This isn't right and I can't figure out what I did wrong. Any help would be awesome.

    Thanks a lot.
     
  2. jcsd
  3. Mar 4, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    Does this mean you've solved the problem?
     
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