8.2 Section 2 Background .1 Cosmology .11 Math Section 8.4

#### Book 9 Computational Mathematics Chapter 8 Ordinary Differential Equation Initial Value Problems

Section 3 Solution by undetermined coefficients

This method can also be used to solve linear  differential equations with variable coefficients. Its application is explained by the example of the second order equation

(3.1)

with the initial conditions y(0) = y0, y'(0) = y'0? Assuming that the coefficient function in (3.1) can be expanded in power series

search for the solution of the given equation in the form

(3.2)

where the cn must be determined. Differentiating (3.2) twice with respect to x:

and substituting these series into (3.1), one finds

(3.3)

Expanding the products of the series and equating the coefficients of equal powers of x on the left- and right-hand sides, one arrives at the system of equations

(3.4)

where denotes a linear function of Consecutive equations of the system (3.4) contain one additional unknown. The coefficients c0 and c1 are determined by the initial conditions, while all remaining
ones are obtained from (2.12). It can be shown that, if the series

converge for |x| < R, then the series obtained converges in the same interval.

Note: If the initial conditions are given for x = x0, then it is advisable to make the substitution x - x0 = t, thus reducing the problem to the one considered above.

Example 1: Find the solution of the equation

(3.5)

for the initial conditions y0) = 0, y'(0) = 1.

Solution: Expand the coefficients of the given equation in power series.

. .

Find the solution of (3.5) in the form of the power series

Then

Substituting these series into (3.5) and equating the coefficients of equal powers of x, one finds the system for the determination of the coefficients ci:

By the initial conditions, c0 = 0, c1 = 1.It is readily seen that c2n+1 = 0, n = 1, 2, . . . , so that

The solution of the problem has thus the form

Note:. Sometimes, when solving a differential equation by the method of undetermined coefficients, one may find an expression for the coefficients of a
series in a general form, as is illustrated by

Example 5: Find the power series solution of the equation

(3.6)

for the initial conditions y(0)=5, y'(0)=2.

Solution: Look for the solution in the form of the series

,

when

Substituting these series into(3.6) and equating coefficients of equal powers of x on both sides leads to the system:

(3.7)

The initial conditions yield c0 = 5 and c1 = 2.  The solution of (3.7) is

Separating coefficients with odd and even subscripts:

one finds

Substituting for cl and c2, one finds

Separate the series into even and odd powers of x:

Next, find the domain of convergence of these series. For the first series, the limit of the absolute value of the ratio of consecutive terms is

for the second series:

It is seen that both series converge for all values of x, whence the solution of (3.7) can be written in the form:

Exercises

Using undetermined coefficients, find the solutions of the equations :