Answer
\[y+\sqrt{y^2+9x^2}=Cx^2\]
Work Step by Step
\[xy'-y=\sqrt{9x^2+y^2}\]
\[y'=\frac{dy}{dx}=\frac{\sqrt{9x^2+y^2}+y}{x}\]
\[\frac{dy}{dx}=\sqrt{9+\frac{y^2}{x^2}}+\frac{y}{x}\;\;\;\ldots (1)\]
Substitute $\:y=Vx\;\;\;\ldots (2)$
\[\frac{dy}{dx}=V+x\frac{dV}{dx}\;\;\;\ldots (3)\]
From (1), (2) and (3)
\[V+x\frac{dV}{dx}=\sqrt{9+V^2}+V\]
\[x\frac{dV}{dx}=\sqrt{9+V^2}\]
Separating Variables,
\[\frac{dV}{\sqrt{3^2+V^2}}=\frac{1}{x}dx\]
Integrating,
\[\int\frac{dV}{\sqrt{3^2+V^2}}=\int\frac{1}{x}dx+\ln C\]
Where $\ln C$ is constant of integration
\[\ln\left|V+\sqrt{V^2+3^2}\right|=\ln |x|+\ln C\]
\[V+\sqrt{V^2+3^2}=Cx\]
From (2)
\[\frac{y}{x}+\sqrt{\frac{y^2}{x^2}+9}=Cx\]
\[y+\sqrt{y^2+9x^2}=Cx^2\]
Hence general solution of (1)\[y+\sqrt{y^2+9x^2}=Cx^2\]
