Second-Order, Linear Equations 2

(19)

are also solutions. It is easily checked that they are linearly independent . Thus the general solution to (9) is

(20)

This solution is sometimes written in another form. Set

Note that

We can thus set

for some angle . Hence

(21)

4. Power series solutions: When your linear equation has the form

(22)

where p and are polynomials you can sometimes find solutions in the form of power series. Suppose
that you want to solve (22) on some open interval I. Choose in I, and as sume that solutions and
have power series re presentations

(23)

and

(24)

To insure that these solutions be independent, take

(25)

and

(26)

Then the Wronskian at is

which implies the linear independence of and on I. You start by plugging the power series (23) into
equation (22). Try to write the resulting equation as a single series set equal to zero. From this you should
be able to extract a recurrence relation for the coefficients . Then use the recurrence relation and (23)
solve for the . Use the same recipe to de termine the . A few points:

a. The method of series solutions is only practical in a few cases where the polynomials and are
very simple . As we saw in class, the method worked on the Airy equation

where and .

b. When solving an initial value problem , you should choose the point in (21) and (22) to be one at which
the initial values are given.

c. The power series might only converge in a small interval centered at .

5. A Cauchy-Euler (equidimensional) equation is one of the form

(28)

For an equation of this sort, you can always find a fundamtental set. Look for solutions of the form u = t^m.
Plug this into (28) to obtain

(29)

If we take t > 0, then (29) implies that

Thus u = t^m satisfies (26) if and only if m is a root of the characteristic equation (28).

a. If (28) has real , distinct roots and , then for t > 0,

(31)

and

(32)

satisfy (26). It is easily seen that, since , these two solutions are independent. Hence the general
solution to (26) is

(33)

for t > 0.
b. If (28) has one root m, then (t) = t^m satisfies (26). Reduction of order yields a second solution,

that is linearly independent of (t). Thus, for t > 0, the general solution to (26) in this case is

(34)

c. If (28) has complex roots and , then

and

are linearly independent solutions to (26) for t > 0. By the superposition principle,

(35)

and

(36)

are also satisfy (26). It is easily checked that they are linearly independent. Thus, for t > 0, the general
solution to (26) is

(37)