Answer
$y=\frac{(x^2+4)^{2}}{4}$
Work Step by Step
Using the $FTC$ it follows:
$$y'=2x\sqrt{y}$$
$$\frac{dy}{\sqrt{y}}=2xdx$$
$$\int \frac{dy}{\sqrt{y}}=\int2xdx$$
$$2y^{\frac{1}{2}}=x^2+C$$
For $x=0$, the value of $y=4+0=4$ so:
$$2\cdot 4^{\frac{1}{2}}=0^2+C \to C=4$$
so:
$$2y^{\frac{1}{2}}=x^2+4$$
$$y^{\frac{1}{2}}=\frac{1}{2}\left(x^2+4\right)$$
$$y=\left(\frac{1}{2}\left(x^2+4\right)\right)^{2}$$
$$y=\frac{(x^2+4)^{2}}{4}$$
