Question

@Mathematics Hi,could someone help me with this differential equation: x^(2)*y''=(y')^2 ?

Answer 1

\[x^2y''=(y')^2\quad,\quad z=y'\]\[x^2z'=z^2\]\[\frac{z'}{z^2}=\frac{1}{x^2}\]\[\left(-\frac{1}{z}\right)'=\frac{1}{x^2}\]\[-\frac{1}{z}=-\frac{1}{x}+C_1=\frac{-1+C_1x}{x}\]\[z=y'=\frac{x}{1-C_1x}=\frac{1}{C_1}\frac{C_1x-1+1}{1-C_1x}=\frac{1}{C_1}\left(-1+\frac{1}{1-C_1x}\right)\]\[y=\frac{1}{C_1}\left(-x-\frac{1}{C_1}\ln{(1-C_1x)}\right)+C_2\]