Решить дефиринцал (x^2-2xy)y'=xy-y^2

Решите задачу:

$(x^2-2xy)y'=xy-y^2\\\\y'= \frac{xy-y^2}{x^2-2xy} \\\\t=\frac{y}{x}\; ,\; y=tx\; ,\; y'=t'x+t\\\\t'x+t= \frac{tx^2-t^2x^2}{x^2-2tx^2} \\\\ t'x+t=\frac{t-t^2}{1-2t} \\\\t'x= \frac{t-t^2}{1-2t}-t\\\\t'x= \frac{t-t^2-t+2t^2}{1-2t}\\\\ \frac{dt}{dx} = \frac{t^2}{x(1-2t)} \\\\\int \frac{(1-2t)\, dt}{t^2} =\int \frac{dx}{x}\\\\\int (t^{-2}-\frac{2}{t})dt=\int \frac{dx}{x}\\\\\frac{t^{-1}}{-1}-2\cdot ln|t|=ln|x|+C\\\\ -\frac{x}{y}-2\cdot ln\left | \frac{y}{x} \right |=ln|x|+C$