Find the particular solution that satisfies the initial condition

The given problem: is written in a form of first order "ordinary differential equation" or first order ODE.

 To evaluate this, we can apply variable separable differential equation  in which we express it in a form of  before using direct integration on each side.

To rearrange the problem, we move to the other to have an equation as: .

 Divide both sides by :

Applying direct integration:

Express as : 

Express in a form of  : 

To find the indefinite integral on both sides, we let:

then or   

then or

 The integral becomes: 

Apply the basic integration property: .

Apply the Law of Exponents: .

Then, the integral becomes:

Applying Power Rule of integration:

Note:

In radical form: 

Plug-in and , we get the general solution of differential equation:

Divide both sides by , we get:  .

Note: as arbitrary constant

For particular solution, we consider the initial condition  where   and .

Plug-in the values, we get:

.

 Then plug-in C =-1 on the general solution:  .

Rearrange into:

Taking the square root on both sides: