Answer
$y=\sqrt{4e^x+5}$
Work Step by Step
$yy'-2e^x=0\hspace{15mm}y(0)=3$
First we want the general solution.
$yy'=2e^x$
$\int ydy=\int 2e^xdx$
$\frac{y^2}{2}=2e^x+C$
$y^2=4e^x+C$
$y=\sqrt{4e^x+C}$
Now applying our initial condition we have
$3=\sqrt{4+C}$
$C=3^2-4=5$, and so our particular solution is
$y=\sqrt{4e^x+5}$