Page 290 - Schaum's Outline of Differential Equations

CHAP. 27] LINEAR DIFFERENTIAL EQUATIONS WITH VARIABLE COEFFICIENTS 273

27.23. Solve Problem 27.22 by another method.
TAYLOR SERIES METHOD. An alternative method for solving initial-value problems rests on the
assumption that the solution can be expanded in a Taylor series about the initial point ; i.e.,
X Q

(n
The terms y(x 0) and y'(x 0) are given as initial conditions; the other terms y \x 0) (n = 2,3, ...) can be obtained by
=
successively differentiating the differential equation. For Problem 27.22 we have x Q=-l, y(x Q) = y(-l) 2, and
y'(x Q) = y'(-l) = - 2. Solving the differential equation of Problem 23.22 for y", we find that

We obtain y"(x 0) = y"(—l) by substituting x 0 = — 1 into (2) and using the given initial conditions. Thus,

To obtain y"(-l), we differentiate (2) and then substitute X Q= -1 into the resulting equation. Thus,

and

(4
To obtain y \-l), we differentiate (4) and then substitute x 0= -1 into the resulting equation. Thus,

and

This process can be kept up indefinitely. Substituting Eqs. (3), (5), (7), and the initial conditions into (_/), we obtain,
as before,

One advantage in using this alternative method, as compared to the usual method of first solving the differential
equation and then applying the initial conditions, is that the Taylor series method is easier to apply when only the
first few terms of the solution are required. One disadvantage is that the recurrence formula cannot be found by the
Taylor series method, and, therefore, a general expression for the wth term of the solution cannot be obtained. Note
that this alternative method is also useful in solving differential equations without initial conditions. In such cases,
we set y(x 0) = a 0 and y'(x 0) = a 1; where a 0 and a^ are unknown constants, and proceed as before.

27.24. Use the method outlined in Problem 27.23 to solve /' - 2xy = 0; y(2) = 1, /(2) = 0.
Using Eq. (_/) of Problem 27.23, we assume a solution of the form

From the differential equation,

Substituting x = 2 into these equations and using the initial conditions, we find that