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Math Help - General Solution of this ODE

  1. #1
    Junior Member
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    General Solution of this ODE

    Hi,

    I just need to know where to start in solving this ODE. I know it has something to do with the Method of Undetermined Coefficients, but I'm really stuck. The question is as follows:

    Find the general solution of:

    x^2y'' + 2xy' + 2y = x^2 + Bx + C

    where A and B are constants.

    Thanks a lot!
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  2. #2
    Super Member 11rdc11's Avatar
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    This a Cauchy-Euler Equation

    which is

    ar^2 + (b-a)r + c =0

    so your auxiliary equation is

    r^2 + r +2 = 0

    so

    r = -\frac{1}{2} \pm \frac{\sqrt{7}}{2}i

    and your answer is

    y = Ax^{-\frac{1}{2}}\cos{\frac{\sqrt{7}}{2}} + Bx^{-\frac{1}{2}}\sin{\frac{\sqrt{7}}{2}}
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  3. #3
    Junior Member
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    From my understanding a Cauchy-Euler equation of the 2nd order has the form: Ay''+By+C=0 which is certainly not the case here. How do you arrive at that solution?
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  4. #4
    MHF Contributor

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    You "understanding" is completely wrong. Ay"+ By+ C= 0 is a special case of the "constant coefficients" equation with Ay"+ 0y'+ By= -C. The general Cauchy-Euler equation of the 2nd order is Ax^2y"+ Bxy'+ Cy= f(x)".

    You can solve the homogeneous equation as Super Member (and he certainly is!) indicates and then try a "specific solution" of the form y= ux^2+ vx+ w for u, v, and w constants to be determined.

    More generally, the substitution t= ln(x) converts a "Cauchy-Euler" equation to an equation with constant coefficients.
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