The change of variable, $u= \ln(x),$ converts an "Euler type equation" (also known as an "equipotential equation") to a differential equation with constant coefficients. If ax^2\frac{d^2y}{dx^2}+ bx\frac{dy}{dx}+ cy= 0 then, with u= \ln(x), \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{x}\frac{dy}{du} and \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right)= -\frac{1}{x^2}\frac{dy}{du}+ \frac{1}{x}\frac{d}{dx}\frac{dy}{du}= -\frac{1}{x^2}\frac{dy}{du}+ \frac{1}{x^2}\frac{d^2y}{du^2}.So ax^2\frac{d^2y}{dx^2}+ bx\frac{dy}{dx}+ cy= a\frac{d^2y}{du^2}- a\frac{dy}{du}+ b\frac{dy}{du}+ cy= a\frac{d^2y}{du^2}+ (b- a)\frac{dy}{du}+ cy= 0.
The characteristic equation for that constant-coefficients equation is the same as for the Euler-type equation so both have the same characteristic values. In particular, if the characteristic equation has a double root, r, then the constant-coefficients equation has the general solution y(u)= Ae^{ru}+ Bue^{ru}. Since $u= \ln(x)$ the general solution in terms of $x$ becomes y(x)= Ae^{r \ln(x)}+ B \ln(x) e^{r \ln(x)}= A e^{\ln(x^r)}+ B \ln(x) e^{\ln(x^r)}= Ax^r+ B\ln(x) x^r.
