Finding the function of x in a differential equation $y=f(x)$ If the graph of $y=f(x)$ contains the point $(0,2)$, $\frac{dy}{dx}=\frac{e^{2x}}{y}$ and $f(x) > 0$ for all $x$, then $f(x) = ?$
$dy/dx = \frac{e^{2x}}y$
$y\cdot  dy = e^{2x}dx$
(The following step confuses me, because of the $dy$).
$\int y\cdot  dy = \int e^{2x}dx$
$\frac{1}2y^2 = \frac{1}2e^{2x}+C$
$y=\sqrt{e^{2x}+2C}$
or
$y=\sqrt{e^{2x}+C}$
from $y(0) = 2$ I get $C = 2^2-e^{2x}$
I can't get to the answer for $f(x) = ?$

If we have $$\frac{1}{2}y^{2} = \frac{1}{2}e^{2x} + C$$ then $$y = \sqrt{e^{2x}+C'}$$ where $C'=2C$.
For the initial value problem:
$y(0) = 2 = \sqrt{e^{2(0)}+C'} \iff C' = 4 - e^{2(0)} = 4 - 1 = 3$
So $$y(x) = \sqrt{e^{2x} + 3}$$
