Answer
$y=x(\sin(\ln|Cx|))$
Work Step by Step
We are given:
$$y'-x^{-1}y=x^{-1}\sqrt x^2-y^2$$
Let
$$V=\frac{y}{x}$$
then
$$\frac{dy}{dx}=V+x\frac{dV}{dx}$$
The equation becomes:
$$V+x\frac{dV}{dx}=\sqrt 1-V^2+V$$
$$x\frac{dV}{dx}=\sqrt 1-V^2$$
$$\frac{1}{\sqrt 1-V^2}dV=\frac{1}{x}dx$$
Integrating both sides:
$$\arcsin V=\ln|xC|$$
$$\rightarrow V=\sin \ln (Cx)$$
$$\frac{y}{x}=\sin \ln (Cx)$$
Hence general solution is $y=x(\sin(\ln|Cx|))$
