12yy' - 7e^x = 0 Find the general solution of the differential equation

For the given problem: 12yy'-7e^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, 12yy'-7e^x=0 can be rearrange into 12yy'= 7e^x
Express y'  as (dy)/(dx) :
12y(dy)/(dx)= 7e^x
Apply direct integration in the form of int f(y) dy = int f(x)dx :
12y(dy)/(dx)= 7e^x
12ydy= 7e^xdx
int12ydy= int 7e^x dx
For the both side , we apply basic integration property: int c*f(x)dx= c int f(x) dx
12 int ydy= 7int e^x dx
Applying Power Rule integration: int u^n du= u^(n+1)/(n+1) on the left side.
12int y dy= 12 *y^(1+1)/(1+1)
               = (12y^2)/2
               =6y^2
Apply basic integration formula for exponential function: int e^u du = e^u+C on the right side.
7int e^x dx = 7e^x+C
Combining the results for the general solution of differential equation:
6y^2=7e^x+C
or 
(6y^2)/6=(7e^x)/6+C
y^2 = (7e^x)/6+C
y = +-sqrt((7e^x)/6+C)