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Linear First Order Equation

  1. Feb 5, 2006 #1

    Zem

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    I'm using an integrating factor, rho(x), to solve an equation of the form
    dy/dx + P(x)y = Q(x)

    I need to find the particular solution.
    y' = 1 + x + y + xy; y(0) = 0
    y' - y - xy = 1 + x
    dy/dx + y(-1-x) = 1 + x
    P(x) = (-1-x), Q(x) = (1 + x),
    rho(x) = e^(-x-1/2x^2)
    (Multiply both sides by rho(x))
    e^(-x - 1/2x^2)(dy/dx) + e^(-x - 1/2x^2)(1-x) = e^(-x - 1/2x^2)(1+x)

    Dx[y * e^(-x - 1/2x^2)] = e^(-x - 1/2x^2)(1+x)
    y * e^(-x - 1/2x^2) = int_e^(-x - 1/2x^2)(1+x)dx

    (Multiply both sides by the reciprocal of rho(x))
    y(x) = e^(x + 1/2x^2) * int_e^(-x - 1/2x^2)(1+x)dx

    (U substitution) This is where I get stuck. It seems that I need u substitution to find the integral on the right side
    u = e^(-x - 1/2x^2)
    du = (-x - 1) * e^(-x - 1/2x^2)
    dv = 1
    v = (1+x)

    When I set that up in uv - int_v du it looks just as bad as the original integral.

    Thanks in advance for any assistance.
     
    Last edited: Feb 5, 2006
  2. jcsd
  3. Feb 5, 2006 #2
    I don't have a clue what you did, that is some of the messiest math I have seen, I can't follow your steps at all. You do know how to find an integrating factor right? Just e raised to the integral of p(x), you multiply everything in the original equation by that and you should get a function that is the derivative of a product of y and some function of x. It kind of looks like you tried to raise e to the integral of the function you were supposed to be integrating.
     
  4. Feb 5, 2006 #3

    Zem

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    I have cleaned it up. I attempted to use LaTeX notation for e and the integral sign, but made typos and my edits didn't appear.
    Yes, that was a mistake I made in the post that I didn't make in writing it out. This is my integrating factor...
    P(x) = (-1-x), Q(x) = (1 + x),
    rho(x) = e^(-x-1/2x^2)
    (Multiply both sides by rho(x))
    e^(-x - 1/2x^2)(dy/dx) + e^(-x - 1/2x^2)(1-x) = e^(-x - 1/2x^2)(1+x)
    (Which is the product rule of...)
    y * e^(-x - 1/2x^2) = int_e^(-x - 1/2x^2)(1+x)dx

    I'll fix the original post.
     
    Last edited: Feb 5, 2006
  5. Feb 5, 2006 #4
    Ok It looks like you did it right then, and the right hand integral is just the integral of -eudu
     
  6. Feb 5, 2006 #5

    Zem

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    By -e^udu, do you mean this?
    int_e^(-x - 1/2x^2) * (-x - 1) * e^(-x - 1/2x^2)

    I am certain my ti-89 would just give me 2 integrals if I punch that integral into it. Is there a way I could make a less complex integral for Q(x) for my calculator?
     
  7. Feb 5, 2006 #6
    You had this right int_e^(-x - 1/2x^2)(1+x)dx?

    Let u = -x - 1/2x2 then du = -(1+x) so you have -eudu
     
  8. Feb 5, 2006 #7

    Zem

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    Beautiful. Thanks again!
     
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