yy' - 2e^x = 0 , y(0) = 3 Find the particular solution that satisfies the initial condition

For the given problem: yy'-2e^x=0 , we can evaluate this by applying variable separable differential equation in which we express it in a form of f(y) dy = f(x)dx .
 Then, yy'-2e^x=0 can be rearrange into yy'= 2e^x
Express y'  as (dy)/(dx):
 y(dy)/(dx)= 2e^x
Apply direct integration in the form of  int f(y) dy = int f(x)dx :
y(dy)/(dx)=2e^x
ydy= 2e^xdx
int ydy= int 2e^x dx
For the left side, we apply Power Rule integration: int u^n du= u^(n+1)/(n+1) .
int y dy= y^(1+1)/(1+1)
             = y^2/2
 For the right side, we apply basic integration property: int c*f(x)dx= c int f(x) dx and basic integration formula for exponential function: int e^u du = e^u+C on the right side.
int 2e^x dx= 2int e^x dx
                  = 2e^x+C
Combining the results for the general solution of differential equation:
y^2/2=2e^x+C
2* [y^2/2] = 2*[2e^x]+2*C     
Let 2*C= C . Just a constant.
y^2= 4e^x+C
 
 To find the particular solution we consider the initial condition y(0)=3 which implies x=0 and y =3 .
Plug them in to  y^2= 4e^x+C , we get:
3^2= 4e^0+C
9= 4*1+C
9=4+C
Then C=9-4=5 .
Plug-in C=5 iny^2= 4e^x+C , we get the particular solution:
y^2= 4e^x+5
 y = +-sqrt(4e^x+5).