Answer
\[y(x)=\frac{1}{18}(81-x^2)\]
Work Step by Step
\[\frac{dy}{dx}=\frac{y-\sqrt{x^2+y^2}}{x}\]
\[\frac{dy}{dx}=\frac{y}{x}-\sqrt{1+\frac{y^2}{x^2}}\;\;\;\ldots (1)\]
Substitute $\:y=Vx \;\;\;\ldots (2)$
Differentiate (2) with respect to $x$
\[\frac{dy}{dx}=V+x\frac{dV}{dx}\;\;\;\;\ldots (3)\]
From (1), (2) and (3)
\[ V+x\frac{dV}{dx} =V-\sqrt{1+V^2}\]
\[x\frac{dV}{dx}=-\sqrt{1+V^2}\]
Separating Variables,
\[\frac{dV}{\sqrt{1+V^2}}=-\frac{dx}{x}\]
Integrating,
\[\int\frac{dV}{\sqrt{1+V^2}}=-\int\frac{dx}{x}+\ln C\]
Where $\ln C$ is constant of integration
\[\ln\left|V+\sqrt{1+V^2}\right|=-\ln |x|+\ln C\]
\[V+\sqrt{1+V^2}=\frac{C}{x}\]
From (2)
\[\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=\frac{C}{x}\]
\[y+\sqrt{x^2+y^2}=C \;\;\;\ldots (4)\]
Using initial condition $y(3)=4$
\[4+\sqrt{9+16}=9=C\]
From (4)
\[y+\sqrt{x^2+y^2}=9\]
\[\sqrt{x^2+y^2}=9-y\]
\[x^2+y^2=81+y^2-18y\]
\[18y=81-x^2\]
\[y=\frac{1}{18}(81-x^2)\]
Hence \[y(x)=\frac{1}{18}(81-x^2)\]
