Answer
\[y^2=x^2\left[(\ln cx)^2-1\right]\]
Work Step by Step
\[\frac{dy}{dx}=\frac{x\sqrt{x^2+y^2}+y^2}{xy}\]
\[\frac{dy}{dx}=\sqrt{\frac{x^2}{y^2}+1}+\frac{y}{x}\]
\[\frac{dy}{dx}=\sqrt{\frac{1}{\left(\frac{y}{x}\right)^2}+1}+\frac{y}{x}\;\;\;\ldots (1)\]
Substitute $\: y=Vx$ ____(2)
Differentiate (2) with respect to $x$
\[\frac{dy}{dx}=V+x\frac{dV}{dx} \;\;\;\ldots (3)\]
From (1),(2) and (3)
\[V+x\frac{dV}{dx}=\sqrt{\frac{1}{V^2}+1}+V\]
\[x\frac{dV}{dx}=\frac{\sqrt{1+V^2}}{V}\]
Separating variables
\[\frac{V}{\sqrt{1+V^2}}dV=\frac{1}{x}dx\]
Integrating,
\[\int\frac{V}{\sqrt{1+V^2}}dV=\int\frac{1}{x}dx+\ln c\]
$\ln c$ is constant of integration
\[\sqrt{1+V^2}=\ln |x|+\ln |c|\]
\[\sqrt{1+V^2}=\ln |cx|\]
\[1+V^2=(\ln |cx|)^2\]
From (2)
\[1+\frac{y^2}{x^2}=(\ln |cx|)^2\]
\[\frac{y^2}{x^2}=(\ln |cx|)^2-1\]
\[y^2=x^2\left[(\ln |cx|)^2-1\right]\]
Hence general solution of (1) is \[y^2=x^2\left[(\ln cx)^2-1\right]\].
