Find the particular solution that satisfies the initial condition This is how I am going about it
$yy'-e^{x}=0$  ;   $y(0)= 4$
I put it in standard form
$\frac{dy}{dx}-\frac{e^{x}}y$=0
$P(x)=e^{x}$
$Q(x)=0$ 
$I(x) = e^{\int e^{x}}$= ?
I'm not sure if I am doing it correctly but if so, what would I(x) come out to be? 

$yy'=e^{x}$,
$\implies \frac{1}{2}(y^{2})'=e^{x}$ so $(y^{2})'=2e^{x}$ 
$\implies y^{2}=2e^{x}+C$.
Now $y(0)= 4$, $4 = 2+C \implies C = 2.$
Thus the particular solution that satisfies the initial condition is $y^{2}=2e^{x}+2$.
