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? 

$$\frac{ydy}{dx} - \exp^x =0$$
$$\frac{dy}{dx} = \frac{\exp^x}{y}$$
$$\int(y)dy = \int(\exp^x)dx$$
$$\frac{y^{2}}{2} = \exp^x + c $$ 
$$y^2 = 2 \exp^x + c_{}1$$ where c_{1} = 2c
$$y = \surd (2(\exp^x) + c_{1})$$ 
$$ now  y(0) = 4 $$
$$ c_{1} = 14 $$
$$so the answer is y = \surd(2(\exp^x) + 14) $$
