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ODE by Laplace transform

  1. Jan 12, 2016 #1
    • Member warned about posting without the homework template
    1. The problem statement, all variables and given/known data
    Use Laplace transform to solve the following ODE

    2. Relevant equations
    xy'' + y' + 4xy = 0, y(0) = 3, y'(0) = 0

    3. The attempt at a solution
    [itex]L(xy'') = -\frac{dL(y'')}{ds}[/itex]

    [itex]L(4xy) = -\frac{4dL(y)}{ds}[/itex]

    [itex]L(y'') = s²L(y) - sy(0) - y'(0) = s²L(y) -3s[/itex]

    [itex]L(y') = sL(y) - sy(0) - y(0) = sL(y) - 3[/itex]

    [itex]-\frac{d(s²L(y)-3s)}{ds} + sL(y)-\frac{4dL(y)}{ds} =0[/itex]

    [itex]-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0[/itex]

    [itex]-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]

    [itex]-\frac{dL(y)(s²+4)}{ds} + sL(y) =0[/itex] (1)

    [itex]\frac{dL(y)}{L(y)} = \frac{sds}{s²+4}[/itex]

    Integrating both sides

    [itex]ln(L(y)) = \frac{ln(s²+4)}{2} + c[/itex]

    [itex]L(y) = c\sqrt{s²+4}[/itex]

    which won't lead me to the right answer.

    I realized that if at (1) I use [itex]\frac{L(y)(s² + 4)}{ds} + sL(y) =0[/itex] instead I'll reach the right answer according to wolfram, but I can't figure out what I'm doing wrong to end up with that negative sign.

    http://www.wolframalpha.com/input/?i=xy''+++y'+++4xy+=+0,+y(0)+=+3,+y'(0)+=+0
     
    Last edited: Jan 12, 2016
  2. jcsd
  3. Jan 12, 2016 #2

    Samy_A

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    Something went wrong from:
    [itex]-\frac{d(s²L(y))}{ds} + 3 + sL(y) - 3 -\frac{4dL(y)}{ds} =0[/itex]
    to
    [itex]-\frac{s²L(y)}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]

    Check what you did with the ##-\frac{d(s²L(y))}{ds}## term.
     
  4. Jan 12, 2016 #3
    It's supposed to be
    [itex]-\frac{d(s²L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]
    fixed it on the original post.
     
  5. Jan 12, 2016 #4

    Samy_A

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    Ok, so now continue your derivation from that.
     
  6. Jan 12, 2016 #5
    [itex]-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds}[/itex]
    [itex]-2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0[/itex]
    [itex] - \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0[/itex]
    [itex] \frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0[/itex]
    [itex] \frac{d(L(y))(s²+4)}{ds} + sL(y) =0[/itex]

    Thanks, I initially solved as if s² wasn't part of the derivative.
     
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