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Solution to 2nd order diff eq

  1. Feb 22, 2007 #1
    Never mind, I figured it out.

    Here's the question:
    Find the general solution to the homogeneous differential equation
    [​IMG]

    The solution has the form [​IMG]

    enter your answers so that [​IMG]

    I'm supposed to find f1(t) and f2(t).

    I know the form ar^2+br+c=0 but in this case it's only r^2=0 so r=0.
    Also y=c1f1(t)+c2f2(t)=c1e^(r1t)+c2e^(r2t)

    f1(t)=e^(r1t) and f1(0)=1 so f1(0)=e^(r1*0)=1
    I know that's right

    I'm stuck on f2(t)
    f2(t)=e^(r2t) and f2(2)=0 so f2(2)=e^(r2*2)=2
    I tried solving for r2
    r2*2=ln(2)
    r2=ln(2)/2=.3466
    and then plugged it into e^(rt), so it was e^(.3466*t)
    This isn't right.

    I don't know what else to try. Both r1 and r2 should be equal to 0 to satisfy r^2=0, but then f2(t)=e^(rt)=e^(0*t)=1 and that wouldn't satisfy f(2)=2.

    Can anyone tell me what else I can try?

    Thanks a lot.
     
    Last edited: Feb 22, 2007
  2. jcsd
  3. Feb 23, 2007 #2
    There is a problem with your attempt. You have repeated roots. The exponential form doesn't work. If that does not get you anywhere, you can just use general anti-derivatives (integrate both sides) to get the general form of the solution.

    This ought to give you the idea of the problem. I hope nobody tries to one-up my answer and gives it all away like you couldn't get it.
     
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