Answer
$y=\frac{x^4}{4}+2x^2+4$
Work Step by Step
Find the initial condition:
$y(0)=4+\int_0^02t\sqrt{y(t)}dt$
$y(0)=4+0$
$y(0)=4$
Construct a differential equation by differentiating the integral equation:
$\frac{d}{dx}(y(x))=\frac{d}{dx}(4+\int_0^x2t\sqrt{y(t)}dt)$
$\frac{dy}{dx}=0+2x\sqrt{y(x)}$
$\frac{dy}{dx}=2x\sqrt{y}$
Solve for $y$ by separating variables:
$\frac{1}{2\sqrt{y}}dy=xdx$
$\int \frac{1}{2\sqrt{y}}dy=\int xdx$
$\sqrt{y}=\frac{x^2}{2}+C$
Substitute the initial condition:
$\sqrt{4}=\frac{0^2}{2}+C$
$2=0+C$
$C=2$
We have now:
$\sqrt{y}=\frac{x^2}{2}+2$
$y=(\frac{x^2}{2}+2)^2$
$y=\frac{x^4}{4}+2x^2+4$
Thus, the solution is $y=\frac{x^4}{4}+2x^2+4$.