# `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` in`y^2= 4e^x+C` , we get the particular solution:

`y^2= 4e^x+5`

`y = +-sqrt(4e^x+5).`