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Initial value differential

  1. Aug 28, 2013 #1
    Problem: Find the solution to the initial Value problem (differential equations)?

    y'=-y+e^-2x
    y(0) = 3


    Attempt: y' -y = e^(-2x) ----- (1)

    Linear equation of first order of form y' + y p(x) = q(x)
    p(x)= -1
    q(x) = e^(-2x)

    Find the integrating factor e^∫p(x) dx = e^∫ - dx = e^-x

    Multiply equation (1) by the integrating factor e^-x
    y' e^-x - y e^-x = e^(-3x) ----- (2)

    The left-hand side of (2) is the derivative of y times the integrating factor or (y e^-x)'
    (y e^-x)' = e^(-3x) ----- (3)

    Integrate both sides of (3)
    y e^-x = ∫ e^(-3x)
    y e^-x = (-1/3) e^(-3x) + C
    multiply everything by e^x

    y = (-1/3) e^(-2x) + C
    y(0)=3
    3 = (-1/3) e^0 + C
    C = 3+1/3 = 10/3

    y = (-1/3) e^(-2x) + (10/3) e^x


    How does this look?
     
  2. jcsd
  3. Aug 28, 2013 #2

    Curious3141

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    The first thing you should do is to substitute your solution back into the d.e. Does it solve the equation?

    Is this ##y' = -y + e^{-2x}## or is this ##y' - y = e^{-2x}## your equation?

    Because in the problem statement, you gave the former, but you solved the latter. The solution for the latter is correct, but if you made a simple algebraic error in bringing the ##y## over, then the solution is obviously wrong.
     
    Last edited: Aug 28, 2013
  4. Aug 28, 2013 #3

    LCKurtz

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    Isn't C included in the "everything"?
     
  5. Aug 28, 2013 #4

    Curious3141

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    Yes, it should be. The reason it didn't affect the answer is because ##e^0 = 1##.
     
  6. Aug 29, 2013 #5

    vanhees71

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    Why not using the standard way of solving linear equations, i.e., first solving the homogeneous equation
    [tex]y'=-y[/tex]
    and then using the ansatz of the variation of the constant?
     
  7. Aug 29, 2013 #6

    Curious3141

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    The integrating factor is the usual elementary method that's taught for questions like this, I think.

    I was about to suggest using a Laplace transform, which gives a quick algebraic solution in an initial value problem like this, but thought better of it, simply because it's unlikely the thread starter has covered it yet.
     
    Last edited: Aug 29, 2013
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