Help With Differential Equation

**Kramer** · October 26th 2012, 10:11 AM

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!

**HallsofIvy** · October 26th 2012, 10:51 AM

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.

**Kramer** · October 26th 2012, 11:00 AM

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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