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Linear ode with variable coefficient

  1. Dec 15, 2004 #1
    Does anyone have any guidance on how to solve a problem of thes form

    p''(x) - 2(1-3x^2)p'(x) = 0

    It looks very similar to a characteristic equation, but the coefficients aren't constant
  2. jcsd
  3. Dec 15, 2004 #2
    Sort of, let q(x) = p'(x) so that q'(x) = p''(x). The "new" equation is

    q'(x) - 2(1 - 3x^2)q(x) = 0,

    which can be solved with an integrating factor.

    In this case, the integrating factor is e^(2x(x^2 - 1)).

    e^(2x(x^2 - 1))q'(x) + e^(2x(x^2 - 1))(-2)(1 - 3x^2)q(x) = 0
    d/dx( e^(2x(x^2 - 1)) * q(x) ) = 0
    e^(2x(x^2 - 1)) * q(x) = C
    q(x) = Ce^(-2x(x^2 - 1))

    So p'(x) = Ce^(-2x(x^2 - 1)). But that function doesn't seem to have a "nice" antiderivative, so I don't know how to continue.
  4. Dec 15, 2004 #3
    so what?? do you need an explicit solution?? you have it


    you may have a polinomial solution using Frobenius method, but youll get the same thing (as expected).
    Last edited: Dec 15, 2004
  5. Dec 15, 2004 #4
    I did do that and I did get p'(x)=c exp(2*(x-x^3)) also....yeah I do need an explicit solution...and this is also where I got stuck because I couldn't see how to solve for p explicitly
  6. Dec 16, 2004 #5
    it is solved explicitly, you have the expresion right there...

    the integral above cant be solved in a closed form, but it is as explicit as a trig function
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