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Math Help - Help With Differential Equation

  1. #1
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    Help With Differential Equation

    Can someone help with an idea of how to solve the following differential equation and the results obtained:

    \frac{d}{dx}[2x^{2}y']=2y

    Thanks for any help!
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  2. #2
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    Re: Help With Differential Equation

    My first thought would be to go ahead and do the derivative on the left side: 2x^2y''+ 4xy'= 2y so that 2x^2y''+ 4xy'- 2y= 0. That's an "Euler type" equation- also called and "equi-potential" equation. The change of variable t= ln(x) will convert it to a linear equation with constant coefficients.

    A "short cut" that will work here is to try y= x^n, y'= nx^{n-1}, and y''= n(n-1)x^{n-2}. Putting those into the equation 2n(n-1)x^n+ 4nx^n- 2x^n= [2n(n-1)+ 4n- 2]x^n= 0. In order that this be true for all x, we must have 2n(n-1)+ 4n- 2= 2n^2+ 2n- 2= 0. That quadratic equation has roots x= - \frac{1}{2}\pm\frac{\sqrt{5}}{2}. The general solution would be Ax^{-1/2+\sqrt{5}/2}+ Bx^{-1/2- \sqrt{5}/2} for constants A and B.
    Thanks from Kramer
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  3. #3
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    Re: Help With Differential Equation

    Thanks a lot for reply. What would the result be if you tried taking integral on both sides? Would that not help the cause at all?
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