Answer
$\sqrt y =\frac{1}{x} (1+x^2)^{\frac{3}{2}}+\frac{C}{x}$
Work Step by Step
Given:
$$y'+x^{-1}y=6\sqrt 1+x^2 \sqrt y$$
This equation is in form of Bernoulli equation, where $p(x)=2x^{-1}, q(x)=6\sqrt 1+x^2$ and $n=\frac{1}{2}$
$$\frac{dy}{dx}+x^{-1}y=6\sqrt 1+x^2 \sqrt y$$
Dividing both sides by $\sqrt y$
$$\frac{1}{\sqrt y}\frac{du}{dx}+\frac{2\sqrt y}{x}=6\sqrt 1+x^2$$
Let $u=\sqrt y \rightarrow \frac{du}{dx}=\frac{1}{2\sqrt y}\frac{dy}{dx}$
Then it becomes:
$$2\frac{du}{dx}+\frac{2u}{x}=6\sqrt 1+x^2$$
$$\frac{du}{dx}+\frac{u}{x}=3\sqrt 1+x^2$$
The integrating factor is:
$$I(x)=e^{\int \frac{1}{x}dx}=e^{\ln (x)}=x$$
The equation becomes:
$$\frac{d}{dx}(ux)=3x\sqrt 1+x^2$$
Integrating:
$$\frac{d}{dx}(ux)=\int 3x\sqrt 1+x^2$$
Let $1+x^2 = t \rightarrow 2x dc =dt$
$\rightarrow \int 3x\sqrt 1+x^2 dx=\frac{3}{2}\int \sqrt t dt=\frac{3}{2} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}=t^{\frac{3}{2}}=(1+x^2)^{\frac{3}{2}}+C$
where $C$ is a constant of integration
So:
$$ux=(1+x^2)^{\frac{3}{2}}+C$$
$$u=\frac{1}{x}(1+x^2)^{\frac{3}{2}}+\frac{C}{x}$$
Subtitute when $y=\sqrt y$
$$\rightarrow \sqrt y =\frac{1}{x} (1+x^2)^{\frac{3}{2}}+\frac{C}{x}$$
