Answer
$\frac{y}{x}-\frac{y^3}{3x^3}=\ln |x|+C$
Work Step by Step
We are given:
$$\frac{dy}{dx}=\frac{x^2}{x^2-y^2}+\frac{y}{x}$$
$$x\sec^2(xy)dy+[y\sec^2(xy)+2x]dx=0$$
Assume that $V=\frac{y}{x} \rightarrow \frac{dy}{dx}=V+x\frac{dV}{dx}$
The equation becomes:
$$x\frac{dV}{dx}=\frac{1}{1-V^2}$$
$$(1-V^2)dV=\frac{1}{x}dx$$
Integrating both sides:
$$V-\frac{V^3}{3}=\ln |x|+C$$
$$\frac{y}{x}-\frac{y^3}{3x^3}=\ln |x|+C$$
The general solution is $\frac{y}{x}-\frac{y^3}{3x^3}=\ln |x|+C$
