Answer
$\rightarrow y=(x^2-1+Ce^{-x^2})^2$
Work Step by Step
Given:
$$y'+4xy=4x^3y^{\frac{1}{2}}$$
$$\frac{dy}{dx}+4xy=4x^3y^{\frac{1}{2}}$$
This equation is in form of Bernoulli equation, where $p(x)=4x, q(x)=4x^3$ and $n=\frac{1}{2}$
Dividing both sides by $y^{\frac{1}{2}}$
$$\frac{1}{y^{\frac{1}{2}}}\frac{dy}{dx}+4xy^{\frac{1}{2}}=4x^3$$
Let $u=y^{\frac{1}{2}} \rightarrow \frac{du}{dx}=\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx}$
Then it becomes:
$$2\frac{du}{dx}+4xu=4x^3$$
$$\frac{du}{dx}+2xu=2x^3$$
The integrating factor is:
$$I(x)=e^{\int 2xdx}=e^{x^2}$$
The equation becomes:
$$\frac{d}{dx}(e^{x^2}u)=2xe^{x^2}u+e^{x^2}u\frac{du}{dx}$$
$$\frac{d}{dx}(e^{x^2}u)=2e^{x^2}x^3$$
Integrating both sides:
$$e^{x^2}u=\int 2e^{x^2}x^3$$
Let $t=x^2 \rightarrow 2x dx=dt$
$$\int 2e^{x^2}x^3 dx=\int e^ttdt$$
Continue to let $u=t, dv=e^ttdt$
$\rightarrow du=dt, v=e^t$
then
$$\int e^ttdt = te^t-\int e^t dt=te^t-e^t$$
$$\rightarrow \int 2e^{x^2}x^3 dx=x^2e^{x^2}-e^{x^2}=e^{x^2}(x^2-1)+C$$
where $C$ is a constant of integration
Then:
$$e^{x^2}u=e^{x^2}(x^2-1)+C$$
$$u=x^2-1 + \frac{C}{e^{x^2}}$$
Subtitute when $u=y^{\frac{1}{3}}$ we have:
$$\rightarrow y^{\frac{1}{2}}=x^2-1+Ce^{-x^2}$$
$$\rightarrow y=(x^2-1+Ce^{-x^2})^2$$
