Page 281 - Schaum's Outline of Differential Equations

264 LINEAR DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS [CHAP. 27

equation resulting from Step 1 are set equal to their counterparts on the right side of that equation. The form of
the solution in Step 5 becomes

which has the form specified in Theorem 8.4. The first two terms comprise the general solution to the associated
homogeneous differential equation while the last function is a particular solution to the nonhomogeneous equation.

INITIAL-VALUE PROBLEMS
Solutions to initial-value problems are obtained by first solving the given differential equation and then
applying the specified initial conditions. An alternate technique that quickly generates the first few terms of the
power series solution to an initial-value problem is described in Problem 27.23.

SOLUTIONS AROUND OTHER POINTS
When solutions are required around the ordinary point x 0 ^ 0, it generally simplifies the algebra if x 0 is
translated to the origin by the change of variables t = x - x 0. The solution of the new differential equation that
results can be obtained by the power series method about t = 0. Then the solution of the original equation is
easily obtained by back-substitution.

Solved Problems

27.1. Determine whether x = 0 is an ordinary point of the differential equation

Here P(x) = -x and Q(x) = 2 are both polynomials; hence they are analytic everywhere. Therefore, every value
of x, in particular x = 0, is an ordinary point.

27.2. Find a recurrence formula for the power series solution around x = 0 for the differential equation given
in Problem 27.1.
It follows from Problem 27.1 that x = 0 is an ordinary point of the given equation, so Theorem 27.1 holds.
Substituting Eqs. (27.5) through (27.7) into the left side of the differential equation, we find

Combining terms that contain like powers of x, we have

The last equation holds if and only if each coefficient in the left-hand side is zero. Thus,

In general,