First, we split it into two equations:
Now, I'm not sure if you're familiar with D notation, but:
And the second equation:
The Initial Steps
So the roots of the polynomial are
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 A and B are constants.
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.