Section 5.2 Problem 21: Solve the initial value problem and graph the solution.

(See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation where the values of are the roots:

Since the values of are complex conjugate roots, then the general solution is where .

Thus, the general solution for our given differential equation is .

To account for both initial conditions, take the derivative of , thus,

Now, we can create our system of equations given our initial conditions:

We then solve the system of equations, which becomes easy since we already know that :

Thus, our final solution is:

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All except for b, because polynomial functions can be quadratic, cubic, witticisms and etc only non negative exponents of negative it is not a polynomial function