Answer:

Step-by-step explanation:
The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:
A differential equation of the form:

Will have a characteristic equation of the form:

Where solutions
are the roots from which the general solution can be found.
For real roots the solution is given by:

For real repeated roots the solution is given by:

For complex roots the solution is given by:

Where:

Let's find the solution for
using the previous information:
The characteristic equation is:

So, the roots are given by:

Therefore, the solution is:

As you can see, is the same solution provided by the problem.
Moving on, let's find the derivative of
in order to find the constants
and
:

Evaluating the initial conditions:

And

Now we have found the value of the constants, the solution of the second-order IVP is:
