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Solving differential equations using convolution, the dreaded 0=0.

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex] tx'' + (4t-2)x' + (13t-4)x = 0 [/itex]

    Use laplace transform to solve.


    3. The attempt at a solution

    I've split up the X'(s) and X(s) and integrated to get [itex] X(s) = \frac{c}{(s^2 + 4s + 13)^2} = \frac{c}{9} \left[ \frac{3}{(s+2)^2 + 9} \right]^2 [/itex]

    From this I'm guessing convolution product, using [itex] F(s) = G(s) = \frac{3}{(s+2)^2 + 9} [/itex]

    This gives [itex] f(t) = g(t) = e^{-2t}sin(3t) [/itex]

    Now I'm trying to evaluate

    [itex]x(t) = \frac{ce^{-2t}}{9} \int_0^t sin(3x)sin(3(t-x)) dx [/itex]

    which I get from the convolution formula (the e^-2x cancel out nicely).

    Here's the problem:

    I use integration by parts twice, but each time I do, no matter what I choose my u and v to be, I always end up with the integral being equal to 0... Where is my mistake???
     
  2. jcsd
  3. Nov 30, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Hard to say. But I wouldn't try to do the integral that way. sin(A)sin(B)=(cos(A-B)-cos(A+B))/2. Use that identity.
     
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