First, we split it into two equations:

Now, I'm not sure if you're familiar with D notation, but:

Basically:

So therefore:

And the second equation:

The Initial Steps

1. and

2. Hence

So the roots of the polynomial are

Hence and

3. From 1 and 2 we deduce that for any quadratic function of D with constant coefficients that: . This equals 0 if k is a root of .

First we deal with the Homogeneous Equation, for the solution we expect:

1. Two linearly independent solutions, and .

2. Two arbitrary constants and .

3. The general solution will be .

First, we see that a possible solution is

From the initial work, we can see that both roots are in fact complex and in the form or , therefore, the general solution is of the form:

We have the roots:

So, therefore we arrive at the solution:

Where:

and

Where A and B are constants.

Finally:

Nonhomogeneous Equation

The general solution of this is:

Where is the general equation we had above and is a particular solution of the differential equation.

Ok, to solve this we need to go back to the formula:

Got to go to class. If anyone else would like to solve it they can.