This is called the characteristic equation of the operator (or of the differential equation). If is a root of the characteristic equation, is a solution of the differential equation. Recall that we are looking for a fundamental set consisting of two linearly independent solutions. Now a quadratic equation may have two real roots, and this would give us our two solutions (it is not hard to see that they are linearly independent).

Now, a quadratic equation may also have only one real root, or no real roots. We will have to deal with these possibilities as well. At this point it is useful to introduce the differentiation operator defined by . We can then write the operator of our differential equation as , a polynomial in . We can manipulate polynomials in just as we would manipulate ordinary polynomials. For example, we might factor . What does mean? These are operators, so they are defined by what they do to functions. The ``multiplication'' of operators is really composition: you first let the operator on the right act on the function, and then the operator on the left acts on the result. Thus for you first take , then . The characteristic polynomial is related to the way the operator acts on exponential functions: . So, for example, . Or in general, for any polynomial ,

This has the following consequences:

*If then is a solution of the homogeneous equation .**If then one solution of the non-homogeneous equation is .*

Adding constants times powers of gives us all polynomials.

Now let's apply this to where the polynomial
has only one root.
This means
where is the root. We look for a solution
of the form
for some function . The Exponential Shift
Theorem says
, so we want .
Two linearly independent solutions of this are and .
Thus a fundamental set of solutions of is
and .

- If with , . Writing we want . Clearly one solution of this is the constant . Then from we get a solution , i.e. .
- If , . Clearly one solution is , i.e. .

We have , with one root . Since the right side of the equation involves , we use the Exponential Shift Theorem with . We have , so one solution is . This gives us one solution of our differential equation: . The fundamental set of solutions of the homogeneous equation is and . Thus the general solution is . From the initial conditions we have and . Thus and , and the answer is

2002-02-07