Answer
See below
Work Step by Step
Given $x\frac{dy}{dx}-y=\sqrt 4x^2-y^2\\
x\frac{dy}{dx}=\sqrt 4x^2-y^2+y$
Since $x\gt 0$, the equation becomes:
$\frac{dy}{dx}=\sqrt 4-\frac{y^2}{x^2}+\frac{y}{x}$
Let $V=\frac{y}{x}$ then $y=Vx\\\frac{dy}{dx}=V+x\frac{dV}{dx}$
The equation becomes: $V+ x\frac{dV}{dx}=\sqrt 4-V^2+V\\\rightarrow x\frac{dV}{dx}=\sqrt 4-V^2\\\rightarrow \frac{dV}{\sqrt 4-V^2}=\frac{dx}{x}$
Integrating both sides: $\int \frac{1}{\sqrt 4-V^2}dV=\int \frac{1}{x}dx$
Solve the left side: $V=2\sin t\rightarrow dV=2\cos t dt\\
\rightarrow \int \frac{1}{\sqrt 4-V^2}dV=\int\frac{1}{\sqrt 4-4\sin^2(t)}2\cos t dt\\
\rightarrow \int \frac{1}{\sqrt 4-V^2}dV=\int \frac{1}{\sqrt 1-\sin^2(t)}\cos t dt$
With $1-\sin^2(t)=\cos^2(t)$
Hence, $\int \frac{1}{\sqrt 1-\sin^2(t)}\cos t dt=\int \frac{1}{\sqrt \cos^2(t)}\cos t dt\\
\rightarrow \int 1dt=t\\
\rightarrow t=\arcsin \frac{V}{2}$
Hence, $\arcsin \frac{V}{2}=\ln|x|+\ln c$
Since $V=\frac{y}{x}$, we have: $\arcsin \frac{y}{2x}=\ln|x|+\ln c\\
\rightarrow \frac{y}{2x}=\sin(\ln(cx))$
Hence, $y(x)=2x\sin (\ln(cx))$
