This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us .

This tells us that the homogeneous solution is . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that . Then,

 and

Plugging into the original equation, we have

Which implies that  and . Solving via substitution,

Thus, the particular solution is  and the overall solution is the particular plus the homogeneous.

So 

We first note that the differential equation has characteristic equation 

 ,

since the roots of this characteristic polynomial are linearly independent of the forcing function   

,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form 

We first figure that the forcing function   is linearly independent  to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

We start with the assumption that the particular solution must be of the form 

 .

Then we solve the first and second derivatives with this assumption, that is, 

 and .

Then we plug in these quantities into the given equation to get:

 , which solves for .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

First solve the homogeneous portion:

Therefore,  is a repeated root thus one of the complimentary solutions  is,

Now find the remaining complimentary solution .

Now solve for  and .

Where 

and 

Therefore,

Now, combine both of the complimentary solutions together to arrive at the general solution.

The form of a guess for a particular solution is 

To compute the Wronskian first calculates the roots of the homogeneous portion.

Therefore one of the complimentary solutions is in the form,

where,

Next compute the Wronskian:

Now take the determinant to finish calculating the Wronskian.

To compute the Wronskian first calculates the roots of the homogeneous portion.

Therefore one of the complimentary solutions is in the form,

where,

Next compute the Wronskian:

Now take the determinant to finish calculating the Wronskian.

Because the inhomogeneity does not take a form we can exploit with undetermined coefficients, we must use variation of parameters. Thus, first we find the complementary solution. The characteristic equation of  is , with solutions of . This means that  and .

To do variation of parameters, we will need the Wronskian, 

Variation of parameters tells us that the coefficient in front of  is  where  is the Wronskian with the  row replaced with all 0's and a 1 at the bottom. In the 2x2 case this means that 

. Plugging in, the first half simplifies to 

and the second half becomes 

Putting these together with the complementary solution, we have a general solution of 

We know the solution consists of a homogeneous solution and a particular solution.

The auxiliary equation for the homogeneous solution is

The homogeneous solution is

The particular solution is of the form

It requires variation of parameters to solve

Solving the system gets us

Integrating gets us

So 

Our solution is