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

This problem contains two questions that need to be solved for: order of the differential equation and whether it is linear or nonlinear.

To determine the order of the differential equation, look for the highest derivative in the equation.

For this particular function recall that, 

therefore the highest derivative is three which makes the equation a third ordered differential equation.

The second part of this problem is to determine if the equation is linear or nonlinear. For a differential equation to be linear two characteristics must hold true:

1. The dependent variable  and all its derivatives have a power involving one.

2. The coefficients depend on the independent variable .

Looking at the given function,

it is seen that all the variable  and all its derivatives have a power involving one and all the coefficients depend on  therefore, this differential equation is linear.

To answer this problem completely, the differential equation is a linear, third ordered equation.

By the cauchy-peano theorem, for , as long as  is continuous on a closed rectangle around our starting point, we have local existence. All three functions are continuous everywhere, so they enjoy local existence at every starting point.

We can show that the solutions to differential equations are unique by showing that  is Lipschitz continuous in y. If  is continuous, then this will suffice to show the Lipschitz continuity.

Note that the first and third equations are continuous for all y and t, but that the second is not continuous when . More concretely, when , both the equation  and the equation  would satisfy the differential equation.