Question $4$

$1$ pts

Solve $\frac{dy}{dx}=\frac{x^2-y^2}{2xy}$

$1\frac{3y^2}{x^2}=C|x{|}^{-3}$

$x(1-3\frac{y^2}{x^2}{)}^{\frac{1}{3}}=C$

$3\frac{y^2}{x^2}=C|x{|}^{-3}$

$1\frac{3y^2}{x^2}=\frac{C}{x}$

$1\frac{3y^2}{x^2}=C|x{|}^3$