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Integrating Factor and Unintegratable Term

  1. Jan 27, 2009 #1
    I was given the following ODE to solve and it seemed simple enough. However, after you have used the integrating factor the integral is not integratable.

    y' = (1+x^2)y +x^3, y(0)=0

    Find y(1) if y(x) is the solution to the above ODE.

    So I put it in the proper form of:

    y' + (-1-x^2)y = x^3

    INT FACTOR = e^(INTEGRAL(-1-x^2) dx) = e^(-x^3/3 - x) so:

    y = e^(x/3 + x) * (INTEGRAL e^(-x/3 - x) * x^3 dx + C)

    This "(INTEGRAL e^(-x/3 - x)" is not integratable by any means I know of. So I e-mailed the Professor and this is what he told me to do.

    "I gave this problem so you would have experience with something that doesn't have an explicit solution. The solution is in terms of an integral. You can even make it a definite integral with the upper limit variable. Choose the lower limit so the initial condition works out. Since your calculator can do numerical integration, you can then evaluate y(1)."

    I am clueless on what he is talking about. I have looked in all of the DE texts I have and all say that it is fine to leave it in the form that it is in. However, none of those examples cover what to do when there are initial conditions present and you run into this problem.

    Can anyone help me with this?

    Thanks, Steve
    Last edited: Jan 27, 2009
  2. jcsd
  3. Jan 27, 2009 #2
    How are you getting this step:

    INT FACTOR = e^(INTEGRAL(-1-x^2) dx) = e^(-x/3 - x) so:
  4. Jan 27, 2009 #3
    Using the integrating factor is given by the form:

    dy/dx + A(x)y = B(x)

    INT Factor = e^(A(x) dx)

    so A(x) = (-1-x^2)

    INt Factor is given by: e^(INTEGRAL (-1-x^2 dx)

    INT Factor = e^(-x^3/3 - x)

    Does that make sense?
  5. Jan 27, 2009 #4
    Yes but e^(-x^3/3 - x) is different from e^(-x/3 - x) that you originally wrote. I guess I was hinting that you forgot the "^3" Does that make sense?
  6. Jan 27, 2009 #5
    Fixed. Thanks.

    Any idea how to the ODE?

    Thanks, Steve
  7. Jan 27, 2009 #6
    I wouldn't use integrating factor, probably method of Frobenius / power series method is what I'd jump to.
  8. Jan 28, 2009 #7
    We haven't learned that yet. Any ideas as to what the Professor said?

    Thanks for your help.

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