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Find the general solution to this ODE (with generalization)

  1. Jun 28, 2008 #1
    [tex]y''+xy'+y=0[/tex]

    PS:
    Is there a way to prove that one has found the "most" general solution by only using a particular method (e.g., integrating factors), with the possiblity of obtaining a different general solution using a different method?

    For example, using integrating factor in the ODE

    y' + x²y = x

    exp(x³/3) y' + exp(x³/3) x²y = exp(x³/3) x
    d(exp(x³/3) y) = exp(x³/3)x dx

    Integrate both sides:

    [tex]exp(x^3/3) y = \int \exp(x^3/3)x\ dx[/tex] or

    [tex]y = \frac{\int \exp(x^3/3)x\ dx}{\exp(x^3/3)}[/tex]

    The integral on the LHS does not have an elementary antiderivative, so how do one prove that this is the "most general" solution since there's a possibility of obtaining a different general solution only in terms of elementary solution using a different method?
     
  2. jcsd
  3. Jun 28, 2008 #2

    tiny-tim

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    Hi mathwizarddud! :smile:

    Define w = exp(x³/3) y.

    Then dw/dx = xp(x³/3) x.

    If there are two solutions for y, then w1 and w2, then d(w1 - w2)/dx = 0, and so w1 = w2 + constant,

    so y1 = y2 + constant*exp(-x³/3), as expected. :smile:
     
  4. Jun 28, 2008 #3
    Hello!

    I don't see how this is true:

    "Define w = exp(x³/3) y.

    Then dw/dx = xp(x³/3) x."

    dw/dx should be exp(x³/3) y' + x²exp(x³/3) y.
    Note that y = y(x).
     
  5. Jun 28, 2008 #4

    tiny-tim

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    Yes, and y' + x² y = x,

    so dw/dx = exp(x³/3) (y' + x² y) = exp(x³/3) x. :smile:
     
  6. Jun 28, 2008 #5
    I believe there is some existence & uniqueness theorem for ODEs that could tell you whether a solution is "general" or not.
     
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