Answer
$ y=4x\sin (\ln x + C)$
Work Step by Step
Given:
$$xy'=\sqrt 16x^2-y^2 +y$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x(xv'+v) =\sqrt16x^2-x^2v^2 +vx$$
$$xxv'+xv =x\sqrt 16-v^2 +vx$$
$$xv' =\sqrt16-v^2 $$
Integrating both side:
$$\int \frac{dv}{\sqrt 16-v^2}=\int \frac{1}{x}dx+C$$
$$sin^{-1}(\frac{v}{4})=\ln x + C$$
Subtitute when $y=vx \rightarrow y=\frac{y}{x} \rightarrow sin^{-1}(\frac{y}{4x})=\ln x + C$
$\rightarrow (\frac{y}{4x})=\sin (\ln x + C)$
$\rightarrow y=4x\sin (\ln x + C)$
