Find the general solution for:

$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$

Consider the change $of$ variable:

$x=e^t,$ then $t=lnx,$ and

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{dy}{dx}=-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x^2}\frac{d^{2}y}{d{t}^2}$

therefore, the original equation becomes:

$\frac{d^{2}y}{d{t}^2}-4y=0x^2\frac{d^{2}y}{d{x}^2}+ax\frac{dy}{dx}+by=0$

with $a,b$ constant, after the change $of$ variable becomes:

$\frac{d^{2}y}{d{t}^2}+(a-1)\frac{dy}{dx}+by=0$

that $is$ a second order differential equation with constant coefficientsy$(t)=c_1{e}^{2t}+c_2{e}^{-2t}$

and going back $to$ the original variables, $we$ obtain:

$y(x)=c_1{x}^2+c_2{x}^{-2}$