Can someone help with an idea of how to solve the following differential equation and the results obtained:
$\displaystyle \frac{d}{dx}[2x^{2}y']=2y$
Thanks for any help!
My first thought would be to go ahead and do the derivative on the left side: $\displaystyle 2x^2y''+ 4xy'= 2y$ so that $\displaystyle 2x^2y''+ 4xy'- 2y= 0$. 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 $\displaystyle y= x^n$, $\displaystyle y'= nx^{n-1}$, and $\displaystyle y''= n(n-1)x^{n-2}$. Putting those into the equation $\displaystyle 2n(n-1)x^n+ 4nx^n- 2x^n= [2n(n-1)+ 4n- 2]x^n= 0$. In order that this be true for all x, we must have $\displaystyle 2n(n-1)+ 4n- 2= 2n^2+ 2n- 2= 0$. That quadratic equation has roots $\displaystyle x= - \frac{1}{2}\pm\frac{\sqrt{5}}{2}$. The general solution would be $\displaystyle Ax^{-1/2+\sqrt{5}/2}+ Bx^{-1/2- \sqrt{5}/2}$ for constants A and B.