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Math Help - General Solution to Second Order Differntial Eq

  1. #1
    oxx
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    General Solution to Second Order Differntial Eq

    This is on a review sheet for the exam, and I have the solution, but I can't figure out how the professor got to the first step.

    The question reads: Find the general solution to:

    [(x^2)/2]y'' + 2xy' - (7/8)y = (x^4)/2 x>0


    The first line on the solution sheet reads:

    Characteristic eq: (1/2)r(r-1) + 2r - (7/8) = 0

    I understand the substituting r for y's, but I don't see how he was able to get rid of the x's and get the r(r-1) part.

    It would help a ton if someone could explain this to me.
    Thanks a lot in advance.
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  2. #2
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    Quote Originally Posted by oxx View Post
    This is on a review sheet for the exam, and I have the solution, but I can't figure out how the professor got to the first step.

    The question reads: Find the general solution to:

    [(x^2)/2]y'' + 2xy' - (7/8)y = (x^4)/2 x>0


    The first line on the solution sheet reads:

    Characteristic eq: (1/2)r(r-1) + 2r - (7/8) = 0

    I understand the substituting r for y's, but I don't see how he was able to get rid of the x's and get the r(r-1) part.

    It would help a ton if someone could explain this to me.
    Thanks a lot in advance.
    Is there a certain method you should be using to solve this DE?
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  3. #3
    oxx
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    It doesn't specify. The professor went on to get r= 1/2 and -7/2

    y = Ax^(1/2) + Bx^(-7/2)

    Then using variation of parameters his final solution was
    y=(4/105)x^4 + Ax^(1/2) + Bx^(-7/2)
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  4. #4
    MHF Contributor chisigma's Avatar
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    May be that the best approach is to 'attack' first the 'incomplete equation'...

    \frac{x^{2}}{2} y^{''} + 2 x y^{'} - \frac{7}{8} y = 0 (1)

    The 'coefficients' of this linear DE aren't constants , so that the 'standard approach' doesn't work. It is easy to verify that solutions of (1) are of the type y(x)= x^{r}... then try to find the constants  r able to solve (1)...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 8th 2010 at 12:54 PM.
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  5. #5
    oxx
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    I can't believe I couldn't figure that out. That's a lot for your help!
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