Answer
$y=x^2-1+Ce^{-x^{2}}$
Work Step by Step
Given that:
$y'+2xy=2x^3$
Use an integration factor of:
$$\mu (x)=e^{\int P(x)dx}$$
where $P(x)=2x$
$$\mu (x)=e^{\int 2xdx}=e^{x^2}$$
Multiply the entire equation by this factor.
$$e^{x^2}(y'+2xy=2x^3)$$$$e^{x^2}y'+2e^{x^2}xy=2e^{x^2}x^3$$
Integrate each side.
$$\int e^{x^2}y'+2e^{x^2}xy=\int 2e^{x^2}x^3 dx$$
To evluate the intergral $\int 2e^{x^2}x^3 dx$
let $t=x^2 \rightarrow dt=2xdx$
The intergral become $\int te^tdt=te^t-\int e^tdt=te^t-e^t=e^t(t-1) \rightarrow \int 2e^{x^2}x^3 dx=e^{x^2}(x^2-1)$
Thus;
$$ye^{x^2}=e^{x^2}(x^2-1)+C$$ $$y=x^2-1+Ce^{-x^{2}}$$
