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Second Order DE

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]y'' - 4y = 0[/tex] when y = 1, y' = -1, x = 0

    2. The attempt at a solution

    [tex]y'' - 4y = 0[/tex]

    [tex]m^2 - 4 = 0[/tex]

    [tex]m = 2, m = -2[/tex]

    Substituting: [tex]y = 1, x = 0[/tex]

    [tex]1 = C1 + C2[/tex]

    [tex]C1 = -C2 + 1[/tex]

    Substituting: [tex]y' = -1, x = 0[/tex]

    [tex]-1 -C2 + 1 + C2[/tex]

    [tex]C2 = C2[/tex]

    Therefore: [tex]y = -C2e^2^x + C2e^-^2^x[/tex]

    Obviously I went wrong with my substitutions... Can I get a hint?
     
  2. jcsd
  3. Feb 23, 2007 #2
    Try solving the initial value as a system of equations, maybe this will give you a better approach than substitution.
     
  4. Feb 24, 2007 #3
    since m=+/-2
    y is of the form
    y=A(e^-2x)+B(e^2x)
    y`=-2A(e^-2x)+2B(e^2x)
    Use the given conditions. Make two equations... You know the rest.
    Hope this helps.
    Abdullah

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    Last edited: Feb 24, 2007
  5. Feb 24, 2007 #4
    thanks guys, I got it
     
  6. Feb 25, 2007 #5

    HallsofIvy

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    It would be a really smart idea to write out the general solution here:
    y= C1e2x+ C2e-2x
    Otherwise no one knows what the rest of this means!

    Okay, C1e0+ C2e0= C1+ C2= 1

    Isn't there an "=" missing here?
    If y(x)= C1e2x+ C2e-2x
    then y'(x)= 2C1e2x- 2C2e-2x
    so y'(0)= 2C1- 2C2= -1

    Surely that's not what you meant to say!

    You have C1+ C2= 1 and 2C1- 2C2= -1. Solve those two equations.
     
    Last edited: Feb 25, 2007
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