Решить дифференциальное уравнение : y^2-9x^2+4xyy'=0

ЛОДУ 1-го порядка:
$y^2-9x^2+4xyy'=0\\y=tx;y'=t'x+t\\t^2x^2-9x^2+4x^2t(t'x+t)=0|:x^2\\t^2-9+4tt'x+4t^2=0\\9-5t^2=\frac{4txdt}{dx}|*\frac{dx}{x(9-5t^2)}\\9-5t^2=0\\5t^2=9\\t^2=\frac{9}{5}\\t=^+_-\frac{3}{\sqrt5}\\y=^+_-\frac{3x}{\sqrt5}\\\\y=\frac{3x}{\sqrt5}\\\frac{9x^2}{5}-9x^2+\frac{36x^2}{5}=0\\0=0\\\\y=-\frac{3x}{\sqrt5}\\\frac{9x^2}{5}-9x^2+\frac{36x^2}{5}=0\\0=0$
$\frac{dx}{x}=4\frac{tdt}{9-5t^2}\\\frac{dx}{x}=-\frac{2}{5}\frac{d(9-5t^2)}{9-5t^2}\\\int\frac{dx}{x}=-\frac{2}{5}\int\frac{d(9-5t^2)}{9-5t^2}\\ln|x|=-\frac{2}{5}ln|9-5t^2|+C\\ln|x|+\frac{2}{5}ln|\frac{9x^2-5y^2}{x^2}|=C;y=^+_-\frac{3\sqrt5x}{5}$