`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)`