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