# Solve by using variation of parameters

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,285 This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t. $$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ and, differentiating again, $$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$
 Quote by HallsofIvy This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t. $$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$$ and, differentiating again, $$\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)$$$$= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}$$