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11.2 Power Series Solutions

The power series method introduced in Section 11.1 can be applied to linear equations of any order (as well as to certain nonlinear equations), but its most important applications are to homogeneous second-order linear differential equations of the form

A (x) y'' + B (x) y' + C (x) y = 0,

$A (x) y ″ + B (x) y ′ + C (x) y = 0,$ (1)

where the coefficients A, B, and C are analytic functions of x. Indeed, in most applications these coefficient functions are simple polynomials.

We saw in Example 3 of Section 11.1 that the series method does not always yield a series solution. To discover when it does succeed, we rewrite Eq. (1) in the form

y'' + P (x) y' + Q (x) y = 0

$y ″ + P (x) y ′ + Q (x) y = 0$ (2)

with leading coefficient 1, and with P=B/A $P = B / A$ and Q=C/A. $Q = C / A .$ Note that P(x) and Q(x) will generally fail to be analytic at points where A(x) vanishes. For instance, consider the equation

x y'' + y' + x y = 0.

$x y ″ + y ′ + x y = 0.$ (3)

The coefficient functions in (3) are continuous everywhere. But in the form of (2) it is the equation

y'' + 1 x y' + y = 0

$y ″ + \frac{1}{x} y ′ + y = 0$ (4)

with P(x)=1/x $P (x) = 1 / x$ not analytic at x=0 $x = 0$ .

The point x=a $x = a$ is called an ordinary point of Eq. (2)—and of the equivalent Eq. (1)—provided that the functions P(x) and Q(x) are both analytic at x=a. $x = a .$ Otherwise, x=a $x = a$ is a singular point. Thus the only singular point of Eqs. (3) and (4) is x=0. $x = 0 .$ Recall that a quotient of analytic functions is analytic wherever the denominator is nonzero. It follows that, if A(a)≠0 $A (a) \neq 0$ in Eq. (1) with analytic coefficients, then x=a $x = a$ is an ordinary point. If A(x), B(x), $A (x), B (x),$ and C(x) are polynomials with no common factors, then x=a $x = a$ is an ordinary point if and only if A(a)≠0 $A (a) \neq 0$ .

Example 1

The point x=0 $x = 0$ is an ordinary point of the equation

x y'' + (sin x) y' + x 2 y = 0,

$x y^{″} + (\sin x) y^{′} + x^{2} y = 0,$

despite the fact that A(x)=x $A (x) = x$ vanishes at x=0. $x = 0 .$ The reason is that

P (x) = sin x x = 1 x (x - x 3 3 ! + x 5 5 ! - \dots) = 1 - x 2 3 ! + x 4 5 ! - \dots

$P (x) = \frac{\sin x}{x} = \frac{1}{x} (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots) = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \dots$

is nevertheless analytic at x=0 $x = 0$ because the division by x yields a convergent power series.

Example 2

The point x=0 $x = 0$ is not an ordinary point of the equation

y'' + x 2 y' + x 1/2 y = 0.

$y^{″} + x^{2} y^{′} + x^{1/2} y = 0.$

For while P(x)=x2 $P (x) = x^{2}$ is analytic at the origin, Q(x)=x1/2 $Q (x) = x^{1/2}$ is not. The reason is that Q(x) is not differentiable at x=0 $x = 0$ and hence is not analytic there. (Theorem 1 of Section 11.1 implies that an analytic function must be differentiable.)

Example 3

The point x=0 $x = 0$ is an ordinary point of the equation

(1 - x 3) y'' + (7 x 2 + 3 x 5) y' + (5 x - 13 x 4) y = 0

$(1 - x^{3}) y ″ + (7 x^{2} + 3 x^{5}) y ′ + (5 x - 13 x^{4}) y = 0$

because the coefficient functions A(x), B(x), $A (x), B (x),$ and C(x) are polynomials with A(0)≠0 $A (0) \neq 0$ .

Theorem 2 of Section 3.1 implies that Eq. (2) has two linearly independent solutions on any open interval where the coefficient functions P(x) and Q(x) are continuous. The basic fact for our present purpose is that near an ordinary point a, these solutions will be power series in powers of x−a. $x - a .$ A proof of the following theorem can be found in Chapter 3 of Coddington, An Introduction to Ordinary Differential Equations (Dover Publications, 1989).

Example 4

Determine the radius of convergence guaranteed by Theorem 1 of a series solution of

(x 2 + 9) y'' + x y' + x 2 y = 0

$(x^{2} + 9) y ″ + x y ′ + x^{2} y = 0$ (6)

in powers of x. Repeat for a series in powers of x−4 $x - 4$ .

Solution

This example illustrates the fact that we must take into account complex singular points as well as real ones. Because

P (x) = x x 2 + 9 and Q (x) = x 2 x 2 + 9,

$P (x) = \frac{x}{x^{2} + 9} and Q (x) = \frac{x^{2}}{x^{2} + 9},$

the only singular points of Eq. (6) are ±3i. $\pm 3 i .$ The distance (in the complex plane) of each of these from 0 is 3, so a series solution of the form ∑cnxn $\sum c_{n} x^{n}$ has radius of convergence at least 3. The distance of each singular point from 4 is 5, so a series solution of the form ∑cn(x−4)n $\sum c_{n} {(x - 4)}^{n}$ has radius of convergence at least 5 (see Fig. 11.2.1).

Example 5

Find the general solution in powers of x of

(x 2 - 4) y'' + 3 x y' + y = 0.

$(x^{2} - 4) y ″ + 3 x y ′ + y = 0.$ (7)

Then find the particular solution with y(0)=4, y'(0)=1 $y (0) = 4, y ′ (0) = 1$ .

Solution

The only singular points of Eq. (7) are ±2, $\pm 2,$ so the series we get will have radius of convergence at least 2. (See Problem 35 for the exact radius of convergence.) Substitution of

y = \sum n = 0 \infty c n x n, y' = \sum n = 1 \infty n c n x n - 1, and y'' = \sum n = 2 \infty n (n - 1) c n x n - 2

$y = \sum_{n = 0}^{\infty} c_{n} x^{n}, y ′ = \sum_{n = 1}^{\infty} n c_{n} x^{n - 1}, and y ″ = \sum_{n = 2}^{\infty} n (n - 1) c_{n} x^{n - 2}$

in Eq. (7) yields

\sum n = 2 \infty n (n - 1) c n x n - 4 \sum n = 2 \infty n (n - 1) c n x n - 2 + 3 \sum n = 1 \infty n c n x n + \sum n = 0 \infty c n x n = 0.

$\sum_{n = 2}^{\infty} n (n - 1) c_{n} x^{n} - 4 \sum_{n = 2}^{\infty} n (n - 1) c_{n} x^{n - 2} + 3 \sum_{n = 1}^{\infty} n c_{n} x^{n} + \sum_{n = 0}^{\infty} c_{n} x^{n} = 0.$

FIGURE 11.2.1.

Radius of convergence as distance to nearest singularity.

We can begin the first and third summations at n=0 $n = 0$ as well, because no nonzero terms are thereby introduced. We shift the index of summation in the second sum by +2, $+ 2,$ replacing n with n+2 $n + 2$ and using the initial value n=0. $n = 0 .$ This gives

\sum n = 0 \infty n (n - 1) c n x n - 4 \sum n = 0 \infty (n + 2) (n + 1) c n + 2 x n + 3 \sum n = 0 \infty n c n x n + \sum n = 0 \infty c n x n = 0.

$\sum_{n = 0}^{\infty} n (n - 1) c_{n} x^{n} - 4 \sum_{n = 0}^{\infty} (n + 2) (n + 1) c_{n + 2} x^{n} + 3 \sum_{n = 0}^{\infty} n c_{n} x^{n} + \sum_{n = 0}^{\infty} c_{n} x^{n} = 0.$

After collecting coefficients of cn $c_{n}$ and cn+2, $c_{n + 2},$ we obtain

\sum n = 0 \infty [(n 2 + 2 n + 1) c n - 4 (n + 2) (n + 1) c n + 2] x n = 0.

$\sum_{n = 0}^{\infty} [(n^{2} + 2 n + 1) c_{n} - 4 (n + 2) (n + 1) c_{n + 2}] x^{n} = 0.$

The identity principle yields

(n + 1) 2 c n - 4 (n + 2) (n + 1) c n + 2 = 0,

${(n + 1)}^{2} c_{n} - 4 (n + 2) (n + 1) c_{n} + 2 = 0,$

which leads to the recurrence relation

c n + 2 = ( n + 1 ) c n 4 ( n + 2 )

$c_{n + 2} = \frac{(n + 1) c_{n}}{4 (n + 2)}$ (8)

for n≧0. $n ≧ 0 .$ With n=0, 2, $n = 0, 2,$ and 4 in turn, we get

c 2 = c 0 4 \cdot 2, c 4 = 3 c 2 4 \cdot 4 = 3 c 0 4 2 \cdot 2 \cdot 4, and c 6 = 5 c 4 4 \cdot 6 = 3 \cdot 5 c 0 4 3 \cdot 2 \cdot 4 \cdot 6 .

$c_{2} = \frac{c_{0}}{4 \cdot 2}, c_{4} = \frac{3 c_{2}}{4 \cdot 4} = \frac{3 c_{0}}{4^{2} \cdot 2 \cdot 4}, and c_{6} = \frac{5 c_{4}}{4 \cdot 6} = \frac{3 \cdot 5 c_{0}}{4^{3} \cdot 2 \cdot 4 \cdot 6} .$

Continuing in this fashion, we evidently would find that

c 2 n = 1 \cdot 3 \cdot 5 \dots ( 2 n - 1 ) 4 n \cdot 2 \cdot 4 \dots ( 2 n ) c 0 .

$c_{2 n} = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1)}{4^{n} \cdot 2 \cdot 4 \dots (2 n)} c_{0} .$

With the common notation

(2 n + 1)!! = 1 \cdot 3 \cdot 5 \dots (2 n + 1) = ( 2 n + 1 ) ! 2 n \cdot n !

$(2 n + 1)!! = 1 \cdot 3 \cdot 5 \dots (2 n + 1) = \frac{(2 n + 1)!}{2^{n} \cdot n!}$

and the observation that 2⋅4⋅6⋯(2n)=2n⋅n!, $2 \cdot 4 \cdot 6 \dots (2 n) = 2^{n} \cdot n!,$ we finally obtain

c 2 n = ( 2 n - 1 ) ! ! 2 3 n \cdot n ! c 0 .

$c_{2 n} = \frac{(2 n - 1)!!}{2^{3 n} \cdot n!} c_{0} .$ (9)

(We also used the fact that 4n⋅2n=23n $4^{n} \cdot 2^{n} = 2^{3 n}$ .)

With n=1, 3, $n = 1, 3,$ and 5 in Eq. (8), we get

c 3 = 2 c 1 4 \cdot 3, c 5 = 4 c 3 4 \cdot 5 = 2 \cdot 4 c 1 4 2 \cdot 3 \cdot 5, and c 7 = 6 c 5 4 \cdot 7 = 2 \cdot 4 \cdot 6 c 1 4 3 \cdot 3 \cdot 5 \cdot 7 .

$c_{3} = \frac{2 c_{1}}{4 \cdot 3}, c_{5} = \frac{4 c_{3}}{4 \cdot 5} = \frac{2 \cdot 4 c_{1}}{4^{2} \cdot 3 \cdot 5}, and c_{7} = \frac{6 c_{5}}{4 \cdot 7} = \frac{2 \cdot 4 \cdot 6 c_{1}}{4^{3} \cdot 3 \cdot 5 \cdot 7} .$

It is apparent that the pattern is

c 2 n + 1 = 2 \cdot 4 \cdot 6 \dots ( 2 n ) 4 n \cdot 1 \cdot 3 \cdot 5 \dots ( 2 n + 1 ) c 1 = n ! 2 n \cdot ( 2 n + 1 ) ! ! c 1 .

$c_{2 n + 1} = \frac{2 \cdot 4 \cdot 6 \dots (2 n)}{4^{n} \cdot 1 \cdot 3 \cdot 5 \dots (2 n + 1)} c_{1} = \frac{n!}{2^{n} \cdot (2 n + 1)!!} c_{1} .$ (10)

The formula in (9) gives the coefficients of even subscript in terms of c0 $c_{0}$ ; the formula in (10) gives the coefficients of odd subscript in terms of c1. $c_{1} .$ After we separately collect the terms of the series of even and odd degree, we get the general solution

y (x) = c 0 (1 + \sum n = 1 \infty ( 2 n - 1 ) ! ! 2 3 n \cdot n ! x 2 n) + c 1 (x + \sum n = 1 \infty n ! 2 n \cdot ( 2 n + 1 ) ! ! x 2 n + 1) .

$y (x) = c_{0} (1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{2^{3 n} \cdot n!} x^{2 n}) + c_{1} (x + \sum_{n = 1}^{\infty} \frac{n!}{2^{n} \cdot (2 n + 1)!!} x^{2 n + 1}) .$ (11)

Alternatively,

y (x) = c 0 (1 + 1 8 x 2 + 3 128 x 4 + 5 1024 x 6 + \dots) + c 1 (x + 1 6 x 3 + 1 30 x 5 + 1 140 x 7 + \dots) .

$\begin{array}{rcl} y (x) & = & c_{0} (1 + \frac{1}{8} x^{2} + \frac{3}{128} x^{4} + \frac{5}{1024} x^{6} + \dots) \\ + c_{1} (x + \frac{1}{6} x^{3} + \frac{1}{30} x^{5} + \frac{1}{140} x^{7} + \dots) . \end{array}$ (11′)

Because y(0)=c0 $y (0) = c_{0}$ and y'(0)=c1, $y ′ (0) = c_{1},$ the given initial conditions imply that c0=4 $c_{0} = 4$ and c1=1. $c_{1} = 1 .$ Using these values in Eq. (11′), the first few terms of the particular solution satisfying y(0)=4 $y (0) = 4$ and y'(0)=1 $y ′ (0) = 1$ are

y (x) = 4 + x + 1 2 x 2 + 1 6 x 3 + 3 32 x 4 + 1 30 x 5 + \dots .

$y (x) = 4 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{3}{32} x^{4} + \frac{1}{30} x^{5} + \dots .$ (12)

Remark

As in Example 5, substitution of y=∑cnxn $y = \sum c_{n} x^{n}$ in a linear second-order equation with x=0 $x = 0$ an ordinary point typically leads to a recurrence relation that can be used to express each of the successive coefficients c2, c3, c4, … $c_{2}, c_{3}, c_{4}, \dots$ in terms of the first two, c0 $c_{0}$ and c1. $c_{1} .$ In this event two linearly independent solutions are obtained as follows. Let y0(x) $y_{0} (x)$ be the solution obtained with c0=1 $c_{0} = 1$ and c1=0, $c_{1} = 0,$ and let y1(x) $y_{1} (x)$ be the solution obtained with c0=0 $c_{0} = 0$ and c1=1. $c_{1} = 1 .$ Then

y 0 (0) = 1, y' 0 (0) = 0 and y 1 (0) = 0, y' 1 (0) = 1,

$y_{0} (0) = 1, y_{0}^{′} (0) = 0 and y_{1} (0) = 0, y_{1}^{′} (0) = 1,$

so it is clear that y0 $y_{0}$ and y1 $y_{1}$ are linearly independent. In Example 5, y0(x) $y_{0} (x)$ and y1(x) $y_{1} (x)$ are defined by the two series that appear on the right-hand side in Eq. (11), which expresses the general solution in the form y=c0y0+c1y1 $y = c_{0} y_{0} + c_{1} y_{1}$ .

Translated Series Solutions

If in Example 5 we had sought a particular solution with given initial values y(a) and y'(a), $y ′ (a),$ we would have needed the general solution in the form

y (x) = \sum n = 0 \infty c n (x - a) n;

$y (x) = \sum_{n = 0}^{\infty} c_{n} {(x - a)}^{n};$ (13)

that is, in powers of x−a $x - a$ rather than in powers of x. For only with a solution of the form in (13) is it true that the initial conditions

y (a) = c 0 and y' (a) = c 1

$y (a) = c_{0} and y ′ (a) = c_{1}$

determine the arbitrary constants c0 $c_{0}$ and c1 $c_{1}$ in terms of the initial values of y and y'. $y ′ .$ Consequently, to solve an initial value problem, we need a series expansion of the general solution centered at the point where the initial conditions are specified.

Example 6

Solve the initial value problem

(t 2 - 2 t - 3) d 2 y d t 2 + 3 (t - 1) d y d t + y = 0; y (1) = 4, y' (1) = - 1.

$(t^{2} - 2 t - 3) \frac{d^{2} y}{d t^{2}} + 3 (t - 1) \frac{d y}{d t} + y = 0; y (1) = 4, y ′ (1) = - 1.$ (14)

Solution

We need a general solution of the form ∑cn(t−1)n. $\sum c_{n} {(t - 1)}^{n} .$ But instead of substituting this series in (14) to determine the coefficients, it simplifies the computations if we first make the substitution x=t−1, $x = t - 1,$ so that we wind up looking for a series of the form ∑cnxn $\sum c_{n} x^{n}$ after all. To transform Eq. (14) into one with the new independent variable x, we note that

t 2 - 2 t - 3 = (x + 1) 2 - 2 (x + 1) - 3 = x 2 - 4, d y d t = d y d x d x d t = d y d x = y',

$\begin{array}{c} t^{2} - 2 t - 3 = {(x + 1)}^{2} - 2 (x + 1) - 3 = x^{2} - 4, \\ \frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t} = \frac{d y}{d x} = y^{′}, \end{array}$

and

d 2 y d t 2 = [d d x (d y d x)] d x d t = d d x (y') = y'',

$\frac{d^{2} y}{d t^{2}} = [\frac{d}{d x} (\frac{d y}{d x})] \frac{d x}{d t} = \frac{d}{d x} (y ′) = y ″,$

where primes denote differentiation with respect to x. Hence we transform Eq. (14) into

(x 2 - 4) y'' + 3 x y' + y = 0

$(x^{2} - 4) y ″ + 3 x y ′ + y = 0$

with initial conditions y=4 $y = 4$ and y'=1 $y ′ = 1$ at x=0 $x = 0$ (corresponding to t=1 $t = 1$ ). This is the initial value problem we solved in Example 5, so the particular solution in (12) is available. We substitute t−1 $t - 1$ for x in Eq. (12) and thereby obtain the desired particular solution

y (t) = 4 + (t - 1) + 1 2 (t - 1) 2 + 1 6 (t - 1) 3 + 3 32 (t - 1) 4 + 1 30 (t - 1) 5 + \dots .

$y (t) = 4 + (t - 1) + \frac{1}{2} {(t - 1)}^{2} + \frac{1}{6} {(t - 1)}^{3} + \frac{3}{32} {(t - 1)}^{4} + \frac{1}{30} {(t - 1)}^{5} + \dots .$

This series converges if −1<t<3. $- 1 < t < 3 .$ (Why?) A series such as this can be used to estimate numerical values of the solution. For instance,

y (0.8) = 4 + (- 0.2) + 1 2 (- 0.2) 2 + 1 6 (- 0.2) 3 + 3 32 (- 0.2) 4 + 1 30 (- 0.2) 5 + \dots,

$y (0.8) = 4 + (- 0.2) + \frac{1}{2} {(- 0.2)}^{2} + \frac{1}{6} {(- 0.2)}^{3} + \frac{3}{32} {(- 0.2)}^{4} + \frac{1}{30} {(- 0.2)}^{5} + \dots,$

so that y(0.8)≈3.8188 $y (0.8) \approx 3.8188$ .

The last computation in Example 6 illustrates the fact that series solutions of differential equations are useful not only for establishing general properties of a solution, but also for numerical computations when an expression of the solution in terms of elementary functions is unavailable.

Types of Recurrence Relation

The formula in Eq. (8) is an example of a two-term recurrence relation; it expresses each coefficient in the series in terms of one of the preceding coefficients. A many-term recurrence relation expresses each coefficient in the series in terms of two or more preceding coefficients. In the case of a many-term recurrence relation, it is generally inconvenient or even impossible to find a formula that gives the typical coefficient cn $c_{n}$ in terms of n. The next example shows what we sometimes can do with a three-term recurrence relation.

Example 7

Find two linearly independent solutions of

y'' - x y' - x 2 y = 0.

$y ″ - x y ′ - x^{2} y = 0.$ (15)

Solution

We make the usual substitution of the power series y=∑cnxn. $y = \sum c_{n} x^{n} .$ This results in the equation

\sum n = 2 \infty n (n - 1) c n x n - 2 - \sum n = 1 \infty n c n x n - \sum n = 0 \infty c n x n + 2 = 0.

$\sum_{n = 2}^{\infty} n (n - 1) c_{n} x^{n - 2} - \sum_{n = 1}^{\infty} n c_{n} x^{n} - \sum_{n = 0}^{\infty} c_{n} x^{n + 2} = 0.$

We can start the second sum at n=0 $n = 0$ without changing anything else. To make each term include xn $x^{n}$ in its general term, we shift the index of summation in the first sum by +2 $+ 2$ (replace n with n+2 $n + 2$ ), and we shift it by −2 $- 2$ in the third sum (replace n with n−2 $n - 2$ ). These shifts yield

\sum n = 0 \infty (n + 2) (n + 1) c n + 2 x n - \sum n = 0 \infty n c n x n - \sum n = 2 \infty c n - 2 x n = 0.

$\sum_{n = 0}^{\infty} (n + 2) (n + 1) c_{n + 2} x^{n} - \sum_{n = 0}^{\infty} n c_{n} x^{n} - \sum_{n = 2}^{\infty} c_{n - 2} x^{n} = 0.$

The common range of these three summations is n≧2, $n ≧ 2,$ so we must separate the terms corresponding to n=0 $n = 0$ and n=1 $n = 1$ in the first two sums before collecting coefficients of xn. $x^{n} .$ This gives

2 c 2 + 6 c 3 x - c 1 x + \sum n = 2 \infty [(n + 2) (n + 1) c n + 2 - n c n - c n - 2] x n = 0.

$2 c_{2} + 6 c_{3} x - c_{1} x + \sum_{n = 2}^{\infty} [(n + 2) (n + 1) c_{n + 2} - n c_{n} - c_{n - 2}] x^{n} = 0.$

The identity principle now implies that 2c2=0, $2 c_{2} = 0,$ that c3=16c1, $c_{3} = \frac{1}{6} c_{1},$ and the three-term recurrence relation

c n + 2 = n c n + c n - 2 ( n + 2 ) ( n + 1 )

$c_{n + 2} = \frac{n c_{n} + c_{n - 2}}{(n + 2) (n + 1)}$ (16)

for n≧2. $n ≧ 2 .$ In particular,

c 4 = 2 c 2 + c 0 12, c 7 = 5 c 5 + c 3 42, c 5 = 3 c 3 + c 1 20, c 8 = 6 c 6 + c 4 56 . c 6 = 4 c 4 + c 2 30,

$\begin{array}{l} c_{4} = \frac{2 c_{2} + c_{0}}{12}, & c_{5} = \frac{3 c_{3} + c_{1}}{20}, & c_{6} = \frac{4 c_{4} + c_{2}}{30}, \\ c_{7} = \frac{5 c_{5} + c_{3}}{42}, & c_{8} = \frac{6 c_{6} + c_{4}}{56} . \end{array}$ (17)

Thus all values of cn $c_{n}$ for n≧4 $n ≧ 4$ are given in terms of the arbitrary constants c0 $c_{0}$ and c1 $c_{1}$ because c2=0 $c_{2} = 0$ and c3=16c1 $c_{3} = \frac{1}{6} c_{1}$ .

To get our first solution y1 $y_{1}$ of Eq. (15), we choose c0=1 $c_{0} = 1$ and c1=0, $c_{1} = 0,$ so that c2=c3=0. $c_{2} = c_{3} = 0 .$ Then the formulas in (17) yield

c 4 = 1 12, c 5 = 0, c 6 = 1 90, c 7 = 0, c 8 = 3 1120;

$c_{4} = \frac{1}{12}, c_{5} = 0, c_{6} = \frac{1}{90}, c_{7} = 0, c_{8} = \frac{3}{1120};$

thus

y 1 (x) = 1 + 1 12 x 4 + 1 90 x 6 + 3 1120 x 8 + \dots .

$y_{1} (x) = 1 + \frac{1}{12} x^{4} + \frac{1}{90} x^{6} + \frac{3}{1120} x^{8} + \dots .$ (18)

Because c1=c3=0, $c_{1} = c_{3} = 0,$ it is clear from Eq. (16) that this series contains only terms of even degree.

To obtain a second linearly independent solution y2 $y_{2}$ of Eq. (15), we take c0=0 $c_{0} = 0$ and c1=1, $c_{1} = 1,$ so that c2=0 $c_{2} = 0$ and c3=16. $c_{3} = \frac{1}{6} .$ Then the formulas in (17) yield

c 4 = 0, c 5 = 3 40, c 6 = 0, c 7 = 13 1008,

$c_{4} = 0, c_{5} = \frac{3}{40}, c_{6} = 0, c_{7} = \frac{13}{1008},$

so that

y 2 (x) = x + 1 6 x 3 + 3 40 x 5 + 13 1008 x 7 + \dots .

$y_{2} (x) = x + \frac{1}{6} x^{3} + \frac{3}{40} x^{5} + \frac{13}{1008} x^{7} + \dots .$ (19)

Because c0=c2=0, $c_{0} = c_{2} = 0,$ it is clear from Eq. (16) that this series contains only terms of odd degree. The solutions y1(x) $y_{1} (x)$ and y2(x) $y_{2} (x)$ are linearly independent because y1(0)=1 $y_{1} (0) = 1$ and y1'(0)=0, $y_{1} ′ (0) = 0,$ whereas y2(0)=0 $y_{2} (0) = 0$ and y2'(0)=1. $y_{2} ′ (0) = 1 .$ A general solution of Eq. (15) is a linear combination of the power series in (18) and (19). Equation (15) has no singular points, so the power series representing y1(x) $y_{1} (x)$ and y2(x) $y_{2} (x)$ converge for all x.

The Legendre Equation

The Legendre equation of order α $α$ is the second-order linear differential equation

(1 - x 2) y'' - 2 x y' + α (α + 1) y = 0,

$(1 - x^{2}) y ″ - 2 x y ′ + α (α + 1) y = 0,$ (20)

where the real number α $α$ satisfies the inequality α>−1. $α > - 1 .$ This differential equation has extensive applications, ranging from numerical integration formulas (such as Gaussian quadrature) to the problem of determining the steady-state temperature within a solid spherical ball when the temperatures at points of its boundary are known. The only singular points of the Legendre equation are at +1 $+ 1$ and −1, $- 1,$ so it has two linearly independent solutions that can be expressed as power series in powers of x with radius of convergence at least 1. The substitution y=∑cmxm $y = \sum c_{m} x^{m}$ in Eq. (20) leads (see Problem 31) to the recurrence relation

c m + 2 = - ( α - m ) ( α + m + 1 ) ( m + 1 ) ( m + 2 ) c m

$c_{m + 2} = - \frac{(α - m) (α + m + 1)}{(m + 1) (m + 2)} c_{m}$ (21)

for m≧0. $m ≧ 0 .$ We are using m as the index of summation because we have another role for n to play.

In terms of the arbitrary constants c0 $c_{0}$ and c1, $c_{1},$ Eq. (21) yields

c 2 = - α ( α + 1 ) 2 ! c 0, c 3 = - ( α - 1 ) ( α + 2 ) 3 ! c 1, c 4 = α ( α - 2 ) ( α + 1 ) ( α + 3 ) 4 ! c 0, c 5 = ( α - 1 ) ( α - 3 ) ( α + 2 ) ( α + 4 ) 5 ! c 1 .

$\begin{array}{l} c_{2} = - \frac{α (α + 1)}{2!} c_{0}, \\ c_{3} = - \frac{(α - 1) (α + 2)}{3!} c_{1}, \\ c_{4} = \frac{α (α - 2) (α + 1) (α + 3)}{4!} c_{0}, \\ c_{5} = \frac{(α - 1) (α - 3) (α + 2) (α + 4)}{5!} c_{1} . \end{array}$

We can show without much trouble that for m>0 $m > 0$ ,

c 2 m = (- 1) m α ( α - 2 ) ( α - 4 ) \dots ( α - 2 m + 2 ) ( α + 1 ) ( α + 3 ) \dots ( α + 2 m - 1 ) ( 2 m ) ! c 0

$c_{2 m} = {(- 1)}^{m} \frac{α (α - 2) (α - 4) \dots (α - 2 m + 2) (α + 1) (α + 3) \dots (α + 2 m - 1)}{(2 m)!} c_{0}$ (22)

and

c 2 m + 1 = (- 1) m ( α - 1 ) ( α - 3 ) \dots ( α - 2 m + 1 ) ( α + 2 ) ( α + 4 ) \dots ( α + 2 m ) ( 2 m + 1 ) ! c 1 .

$c_{2 m + 1} = {(- 1)}^{m} \frac{(α - 1) (α - 3) \dots (α - 2 m + 1) (α + 2) (α + 4) \dots (α + 2 m)}{(2 m + 1)!} c_{1} .$ (23)

Alternatively,

c 2 m = (- 1) m a 2 m c 0 and c 2 m + 1 = (- 1) m a 2 m + 1 c 1,

$c_{2 m} = {(- 1)}^{m} a_{2 m} c_{0} and c_{2 m + 1} = {(- 1)}^{m} a_{2 m + 1} c_{1},$

where a2m $a_{2 m}$ and a2m+1 $a_{2 m + 1}$ denote the fractions in Eqs. (22) and (23), respectively. With this notation, we get two linearly independent power series solutions

y 1 (x) = c 0 \sum m = 0 \infty (- 1) m a 2 m x 2 m and y 2 (x) = c 1 \sum m = 0 \infty (- 1) m a 2 m + 1 x 2 m + 1

$y_{1} (x) = c_{0} \sum_{m = 0}^{\infty} {(- 1)}^{m} a_{2 m} x^{2 m} and y_{2} (x) = c_{1} \sum_{m = 0}^{\infty} {(- 1)}^{m} a_{2 m + 1} x^{2 m + 1}$ (24)

of Legendre’s equation of order α $α$ .

Now suppose that α=n, $α = n,$ a nonnegative integer. If α=n $α = n$ is even, we see from Eq. (22) that a2m=0 $a_{2 m} = 0$ when 2m>n. $2 m > n .$ In this case, y1(x) $y_{1} (x)$ is a polynomial of degree n and y2 $y_{2}$ is a (nonterminating) infinite series. If α=n $α = n$ is an odd positive integer, we see from Eq. (23) that a2m+1=0 $a_{2 m + 1} = 0$ when 2m+1>n. $2 m + 1 > n .$ In this case, y2(x) $y_{2} (x)$ is a polynomial of degree n and y1 $y_{1}$ is a (nonterminating) infinite series. Thus in either case, one of the two solutions in (24) is a polynomial and the other is a nonterminating series.

With an appropriate choice (made separately for each n) of the arbitrary constants c0 $c_{0}$ (n even) or c1 $c_{1}$ (n odd), the nth-degree polynomial solution of Legendre’s equation of order n,

(1 - x 2) y'' - 2 x y' + n (n + 1) y = 0,

$(1 - x^{2}) y ″ - 2 x y ′ + n (n + 1) y = 0,$ (25)

is denoted by Pn(x) $P_{n} (x)$ and is called the Legendre polynomial of degree n. It is customary (for a reason indicated in Problem 32) to choose the arbitrary constant so that the coefficient of xn $x^{n}$ in Pn(x) $P_{n} (x)$ is (2n)!/[2n(n!)2]. $(2 n)! / [2^{n} {(n!)}^{2}] .$ It then turns out that

P n (x) = \sum k = 0 N ( - 1 ) k ( 2 n - 2 k ) ! 2 n k ! ( n - k ) ! ( n - 2 k ) ! x n - 2 k,

$P_{n} (x) = \sum_{k = 0}^{N} \frac{{(- 1)}^{k} (2 n - 2 k)!}{2^{n} k! (n - k)! (n - 2 k)!} x^{n - 2 k},$ (26)

where N=〚n/2〛, $N = 〚 n / 2 〛,$ the integral part of n/2. $n / 2 .$ The first six Legendre polynomials are

P 0 (x) \equiv 1, P 2 (x) = 1 2 (3 x 2 - 1), P 4 (x) = 1 8 (35 x 4 - 30 x 2 + 3), P 1 (x) = x, P 3 (x) = 1 2 (5 x 3 - 3 x), P 5 (x) = 1 8 (63 x 5 - 70 x 3 + 15 x),

$\begin{array}{l} P_{0} (x) \equiv 1, & P_{1} (x) = x, \\ P_{2} (x) = \frac{1}{2} (3 x^{2} - 1), & P_{3} (x) = \frac{1}{2} (5 x^{3} - 3 x), \\ P_{4} (x) = \frac{1}{8} (35 x^{4} - 30 x^{2} + 3), & P_{5} (x) = \frac{1}{8} (63 x^{5} - 70 x^{3} + 15 x), \end{array}$

and their graphs are shown in Fig. 11.2.2.

FIGURE 11.2.2.

Graphs y=Pn(x) $y = P_{n} (x)$ of the Legendre polynomials for n=1, 2, 3, 4, $n = 1, 2, 3, 4,$ and 5. The graphs are distinguished by the fact that all n zeros of Pn(x) $P_{n} (x)$ lie in the interval −1<x<1 $- 1 < x < 1$ .

11.2 Problems

Find general solutions in powers of x of the differential equations in Problems 1 through 15. State the recurrence relation and the guaranteed radius of convergence in each case.

(x2−1)y''+4xy'+2y=0 $(x^{2} - 1) y ″ + 4 x y ′ + 2 y = 0$
(x2+2)y''+4xy'+2y=0 $(x^{2} + 2) y ″ + 4 x y ′ + 2 y = 0$
y''+xy'+y=0 $y ″ + x y ′ + y = 0$
(x2+1)y''+6xy'+4y=0 $(x^{2} + 1) y ″ + 6 x y ′ + 4 y = 0$
(x2−3)y''+2xy'=0 $(x^{2} - 3) y ″ + 2 x y ′ = 0$
(x2−1)y''−6xy'+12y=0 $(x^{2} - 1) y ″ - 6 x y ′ + 12 y = 0$
(x2+3)y''−7xy'+16y=0 $(x^{2} + 3) y ″ - 7 x y ′ + 16 y = 0$
(2−x2)y''−xy'+16y=0 $(2 - x^{2}) y ″ - x y ′ + 16 y = 0$
(x2−1)y''+8xy'+12y=0 $(x^{2} - 1) y ″ + 8 x y ′ + 12 y = 0$
3y''+xy'−4y=0 $3 y ″ + x y ′ - 4 y = 0$
5y''−2xy'+10y=0 $5 y ″ - 2 x y ′ + 10 y = 0$
y''−x2y'−3xy=0 $y ″ - x^{2} y ′ - 3 x y = 0$
y''+x2y'+2xy=0 $y ″ + x^{2} y ′ + 2 x y = 0$
y''+xy=0 $y ″ + x y = 0$ (an Airy equation)
y''+x2y=0 $y ″ + x^{2} y = 0$

Use power series to solve the initial value problems in Problems 16 and 17.

(1+x2)y''+2xy'−2y=0 $(1 + x^{2}) y ″ + 2 x y ′ - 2 y = 0$ ; y(0)=0, $y (0) = 0,$ y'(0)=1 $y ′ (0) = 1$
y''+xy'−2y=0 $y ″ + x y ′ - 2 y = 0$ ; y(0)=1, y'(0)=0 $y (0) = 1, y ′ (0) = 0$

Solve the initial value problems in Problems 18 through 22. First make a substitution of the form t=x−a, $t = x - a,$ then find a solution ∑cntn $\sum c_{n} t^{n}$ of the transformed differential equation. State the interval of values of x for which Theorem 1 of this section guarantees convergence.

y''+(x−1)y'+y=0 $y ″ + (x - 1) y ′ + y = 0$ ; y(1)=2, y'(1)=0 $y (1) = 2, y ′ (1) = 0$
(2x−x2)y''−6(x−1)y'−4y=0 $(2 x - x^{2}) y ″ - 6 (x - 1) y ′ - 4 y = 0$ ; y(1)=0, $y (1) = 0,$ y'(1)=1 $y ′ (1) = 1$
(x2−6x+10)y''−4(x−3)y'+6y=0 $(x^{2} - 6 x + 10) y ″ - 4 (x - 3) y ′ + 6 y = 0$ ; y(3)=2, $y (3) = 2,$ y'(3)=0 $y ′ (3) = 0$
(4x2+16x+17)y''=8y $(4 x^{2} + 16 x + 17) y ″ = 8 y$ ; y(−2)=1, y'(−2)=0 $y (- 2) = 1, y ′ (- 2) = 0$
(x2+6x)y''+(3x+9)y'−3y=0 $(x^{2} + 6 x) y ″ + (3 x + 9) y ′ - 3 y = 0$ ; y(−3)=0, y'(−3)=2 $y (- 3) = 0, y ′ (- 3) = 2$

In Problems 23 through 26, find a three-term recurrence relation for solutions of the form y=∑cnxn. $y = \sum c_{n} x^{n} .$ Then find the first three nonzero terms in each of two linearly independent solutions.

y''+(1+x)y=0 $y ″ + (1 + x) y = 0$
(x2−1)y''+2xy'+2xy=0 $(x^{2} - 1) y ″ + 2 x y ′ + 2 x y = 0$
y''+x2y'+x2y=0 $y ″ + x^{2} y ′ + x^{2} y = 0$
(1+x3)y''+x4y=0 $(1 + x^{3}) y ″ + x^{4} y = 0$
Solve the initial value problem

$y'' + x y' + (2 x 2 + 1) y = 0; y (0) = 1, y' (0) = - 1.$ $y ″ + x y ′ + (2 x^{2} + 1) y = 0; y (0) = 1, y ′ (0) = - 1.$

Determine sufficiently many terms to compute y(1/2) accurate to four decimal places.

In Problems 28 through 30, find the first three nonzero terms in each of two linearly independent solutions of the form y=∑cnxn. $y = \sum c_{n} x^{n} .$ Substitute known Taylor series for the analytic functions and retain enough terms to compute the necessary coefficients.

y''+e−xy=0 $y ″ + e^{- x} y = 0$
(cosx)y''+y=0 $(\cos x) y ″ + y = 0$
xy''+(sinx)y'+xy=0 $x y ″ + (\sin x) y ′ + x y = 0$
Derive the recurrence relation in (21) for the Legendre equation.
Follow the steps outlined in this problem to establish Rodrigues’s formula

$P n (x) = 1 n ! 2 n d n d x n (x 2 - 1) n$ $P_{n} (x) = \frac{1}{n! 2^{n}} \frac{d^{n}}{d x^{n}} {(x^{2} - 1)}^{n}$

for the nth-degree Legendre polynomial.
1. Show that v=(x2−1)n $v = {(x^{2} - 1)}^{n}$ satisfies the differential equation
  
  $(1 - x 2) v' + 2 n x v = 0.$ $(1 - x^{2}) v ′ + 2 n x v = 0.$
  
  Differentiate each side of this equation to obtain
  
  $(1 - x 2) v'' + 2 (n - 1) x v' + 2 n v = 0.$ $(1 - x^{2}) v ″ + 2 (n - 1) x v ′ + 2 n v = 0.$
2. Differentiate each side of the last equation n times in succession to obtain
  
  $(1 - x 2) v (n + 2) - 2 x v (n + 1) + n (n + 1) v (n) = 0.$ $(1 - x^{2}) v^{(n + 2)} - 2 x v^{(n + 1)} + n (n + 1) v^{(n)} = 0.$
  
  Thus u=v(n)=Dn(x2−1)n $u = v^{(n)} = D^{n} {(x^{2} - 1)}^{n}$ satisfies Legendre’s equation of order n.
3. Show that the coefficient of xn $x^{n}$ in u is (2n)!/n! $(2 n)! / n!$ ; then state why this proves Rodrigues’ formula. (Note that the coefficient of xn $x^{n}$ in Pn(x) $P_{n} (x)$ is (2n)!/[2n(n!)2] $(2 n)! / [2^{n} {(n!)}^{2}]$ .)
The Hermite equation of order α $α$ is

$y'' - 2 x y' + 2 y = 0.$ $y ″ - 2 x y ′ + 2 y = 0.$
1. Derive the two power series solutions
  
  $y 1 = 1 + \sum m = 1 \infty (- 1) m 2 m α ( α - 2 ) \dots ( α - 2 m + 2 ) ( 2 m ) ! x 2 m$ $y_{1} = 1 + \sum_{m = 1}^{\infty} {(- 1)}^{m} \frac{2^{m} α (α - 2) \dots (α - 2 m + 2)}{(2 m)!} x^{2 m}$
  
  and
  
  $y 2 = x + \sum m = 1 \infty (- 1) m 2 m ( α - 1 ) ( α - 3 ) \dots ( α - 2 m + 1 ) ( 2 m + 1 ) ! x 2 m + 1 .$ $\begin{array}{l} y_{2} = x \\ + \sum_{m = 1}^{\infty} {(- 1)}^{m} \frac{2^{m} (α - 1) (α - 3) \dots (α - 2 m + 1)}{(2 m + 1)!} x^{2 m + 1} . \end{array}$
  
  Show that y1 $y_{1}$ is a polynomial if α $α$ is an even integer, whereas y2 $y_{2}$ is a polynomial if α $α$ is an odd integer.
2. The Hermite polynomial of degree n is denoted by Hn(x). $H_{n} (x) .$ It is the nth-degree polynomial solution of Hermite’s equation, multiplied by a suitable constant so that the coefficient of xn $x^{n}$ is 2n. $2^{n} .$ Show that the first six Hermite polynomials are
  
  $H 0 (x) \equiv 1, H 2 (x) = 4 x 2 - 2, H 4 (x) = 16 x 4 - 48 x 2 + 12, H 5 (x) = 32 x 5 - 160 x 3 + 120 x . H 1 (x) = 2 x, H 3 (x) = 8 x 3 - 12 x,$ $\begin{array}{l} H_{0} (x) \equiv 1, & H_{1} (x) = 2 x, \\ H_{2} (x) = 4 x^{2} - 2, & H_{3} (x) = 8 x^{3} - 12 x, \\ H_{4} (x) = 16 x^{4} - 48 x^{2} + 12, \\ H_{5} (x) = 32 x^{5} - 160 x^{3} + 120 x . \end{array}$
  
  A general formula for the Hermite polynomials is
  
  $H n (x) = (- 1) n e x 2 d n d x n (e - x 2) .$ $H_{n} (x) = {(- 1)}^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}} (e^{- x^{2}}) .$
  
  Verify that this formula does in fact give an nth-degree polynomial. It is interesting to use a computer algebra system to investigate the conjecture that (for each n) the zeros of the Hermite polynomials Hn $H_{n}$ and Hn+1 $H_{n + 1}$ are “interlaced”—that is, the n zeros of Hn $H_{n}$ lie in the n bounded open intervals whose endpoints are successive pairs of zeros of Hn+1 $H_{n + 1}$ .
The discussion following Example 4 in Section 11.1 suggests that the differential equation y''+y=0 $y ″ + y = 0$ could be used to introduce and define the familiar sine and cosine functions. In a similar fashion, the Airy equation

$y'' = x y$ $y ″ = x y$

serves to introduce two new special functions that appear in applications ranging from radio waves to molecular vibrations. Derive the first three or four terms of two different power series solutions of the Airy equation. Then verify that your results agree with the formulas

$y 1 (x) = 1 + \sum k = 1 \infty 1 \cdot 4 \cdot \dots \cdot ( 3 k - 2 ) ( 3 k ) ! x 3 k$ $y_{1} (x) = 1 + \sum_{k = 1}^{\infty} \frac{1 \cdot 4 \cdot \dots \cdot (3 k - 2)}{(3 k)!} x^{3 k}$

and

$y 2 (x) = x + \sum k = 1 \infty 2 \cdot 5 \cdot \dots \cdot ( 3 k - 1 ) ( 3 k + 1 ) ! x 3 k + 1$ $y_{2} (x) = x + \sum_{k = 1}^{\infty} \frac{2 \cdot 5 \cdot \dots \cdot (3 k - 1)}{(3 k + 1)!} x^{3 k + 1}$

for the solutions that satisfy the initial conditions y1(0)=1, y'1(0)=0 $y_{1} (0) = 1, y_{1}^{′} (0) = 0$ and y2(0)=0, y'2(0)=1, $y_{2} (0) = 0, y_{2}^{′} (0) = 1,$ respectively. The special combinations

$Ai (x) = y 1 ( x ) 3 2/3 Γ ( 2 3 ) - y 2 ( x ) 3 1/3 Γ ( 1 3 )$ $Ai (x) = \frac{y_{1} (x)}{3^{2/3} Γ (\frac{2}{3})} - \frac{y_{2} (x)}{3^{1/3} Γ (\frac{1}{3})}$

and

$Bi (x) = y 1 ( x ) 3 1/6 Γ ( 2 3 ) + y 2 ( x ) 3 - 1/6 Γ ( 1 3 )$ $Bi (x) = \frac{y_{1} (x)}{3^{1/6} Γ (\frac{2}{3})} + \frac{y_{2} (x)}{3^{- 1/6} Γ (\frac{1}{3})}$

define the standard Airy functions that appear in mathematical tables and computer algebra systems. Their graphs shown in Fig. 11.2.3 exhibit trigonometric-like oscillatory behavior for x<0, $x < 0,$ whereas Ai(x) decreases exponentially and Bi(x) increases exponentially as x→+∞. $x \to + \infty .$ It is interesting to use a computer algebra system to investigate how many terms must be retained in the y1 $y_{1}$ - and y2 $y_{2}$ -series above to produce a figure that is visually indistinguishable from Fig. 11.2.3 (which is based on high-precision approximations to the Airy functions).

FIGURE 11.2.3.

The Airy function graphs y=Ai(x) $y = Ai (x)$ and y=Bi(x) $y = Bi (x)$ .
1. To determine the radius of convergence of the series solution in Example 5, write the series of terms of even degree in Eq. (11) in the form
  
  $y 0 (x) = 1 + \sum n = 1 \infty c 2 n x 2 n = 1 + \sum n = 1 \infty a n z n$ $y_{0} (x) = 1 + \sum_{n = 1}^{\infty} c_{2 n} x^{2 n} = 1 + \sum_{n = 1}^{\infty} a_{n} z^{n}$
  
  where an=c2n $a_{n} = c_{2 n}$ and z=x2. $z = x^{2} .$ Then apply the recurrence relation in Eq. (8) and Theorem 3 in Section 11.1 to show that the radius of convergence of the series in z is 4. Hence the radius of convergence of the series in x is 2. How does this corroborate Theorem 1 in this section?
2. Write the series of terms of odd degree in Eq. (11) in the form
  
  $y 1 (x) = x (1 + \sum n = 1 \infty c 2 n + 1 x 2 n) = x (1 + \sum n = 1 \infty b n z n)$ $y_{1} (x) = x (1 + \sum_{n = 1}^{\infty} c_{2 n + 1} x^{2 n}) = x (1 + \sum_{n = 1}^{\infty} b_{n} z^{n})$
  
  to show similarly that its radius of convergence (as a power series in x) is also 2.

11.2 Application Automatic Computation of Series Coefficients

Repeated application of a recurrence relation to grind out successive coefficients is—especially in the case of a recurrence relation with three or more terms—a tedious aspect of the infinite series method. Here we illustrate the use of a computer algebra system not only to automate this task, but also to explore interactively the graphical effect of changing the number k of terms we include in a partial-sum approximation to the actual solution given by the full infinite series. In Example 7 we saw that the coefficients in the series solution y=∑cnxn $y = \sum c_{n} x^{n}$ of the differential equation

y'' - x y' - x 2 y = 0

$y ″ - x y ′ - x^{2} y = 0$ (1)

are given in terms of the two arbitrary coefficients c0 $c_{0}$ and c1 $c_{1}$ by

c 2 = 0, c 3 = c 1 6, and c n + 2 = n c n + c n - 2 ( n + 2 ) ( n + 1 ) for n ≧ 2.

$c_{2} = 0, c_{3} = \frac{c_{1}}{6}, and c_{n + 2} = \frac{n c_{n} + c_{n - 2}}{(n + 2) (n + 1)} for n ≧ 2.$ (2)

It would appear to be a routine matter to implement such a recurrence relation, but a twist results from the fact that a typical computer system array is indexed by the subscripts 1, 2, 3, …, rather than by the subscripts 0, 1, 2, … that match the exponents in the successive terms of a power series that begins with a constant term. For this reason we first rewrite our proposed power series solution in the form

y = \sum n = 0 \infty c n x n = \sum n = 1 \infty b n x n - 1

$y = \sum_{n = 0}^{\infty} c_{n} x^{n} = \sum_{n = 1}^{\infty} b_{n} x^{n - 1}$ (3)

where bn=cn−1 $b_{n} = c_{n - 1}$ for each n≧1. $n ≧ 1 .$ Then the first two conditions in (1) imply that b3=0 $b_{3} = 0$ and b4=16b2 $b_{4} = \frac{1}{6} b_{2}$ ; also, the recurrence relation (with n replaced with n−1 $n - 1$ ) yields the new recurrence relation

b n + 2 = c n + 1 = ( n - 1 ) c n - 1 + c n - 3 ( n + 1 ) n = ( n - 1 ) b n + b n - 2 n ( n + 1 ) .

$b_{n + 2} = c_{n + 1} = \frac{(n - 1) c_{n - 1} + c_{n - 3}}{(n + 1) n} = \frac{(n - 1) b_{n} + b_{n - 2}}{n (n + 1)} .$ (4)

Now we are ready to begin. Suppose that we want to calculate the terms through the 10th degree in (2)with the initial conditions b1=b2=1. $b_{1} = b_{2} = 1 .$ Then either the Maple commands


k := 11:          # k terms
b := array(1..k):
b[1] := 1:        # arbitrary
b[2] := 1:        # arbitrary
b[3] := 0:
b[4] := b[2]/6:
for n from 3 by 1 to k − 2 do
   b[n+2] := ((n−1) b[n] + b[n−2])/(n (n+1));
   od;

or the Mathematica commands


k = 11;            (∗ k terms ∗)
b = Table[0, {n,1,k}];
b[[1]] = 1;        (∗ arbitrary ∗)
b[[2]] = 1;        (∗ arbitrary ∗)
b[[3]] = 0;
b[[4]] = b[[2]]/6;
For[n=3, n<=k−2,
   b[[n+2]]=((n−1) b[[n]] + b[[n−2]])/(n (n+1)); n=n+1];

quickly yield the coefficients {bn} ${b_{n}}$ corresponding to the solution

y (x) = 1 + x + x 3 6 + x 4 12 + 3 x 5 40 + x 6 90 + 13 x 7 1008 + 3 x 8 1120 + 119 x 9 51840 + 41 x 10 113400 + \dots .

$\begin{array}{l} y (x) = \\ 1 + x + \frac{x^{3}}{6} + \frac{x^{4}}{12} + \frac{3 x^{5}}{40} + \frac{x^{6}}{90} + \frac{13 x^{7}}{1008} + \frac{3 x^{8}}{1120} + \frac{119 x^{9}}{51840} + \frac{41 x^{10}}{113400} + \dots . \end{array}$ (5)

You might note that the even- and odd-degree terms here agree with those shown in Eqs. (18) and (19), respectively, of Example 7.

The Matlab commands


k = 11;                     % k terms
b = 0∗(1:k);
b(1) = 1;                   % arbitrary
b(2) = 1;                   % arbitrary
b(3) = 0;
b(4) = b(2)/6;
for n = 3:k−2
   b(n+2) = ((n−1) ∗b(n) + ∗b(n−2))/(n∗ (n+1));
   end
format rat, b

give the same results, except that the coefficient b10 $b_{10}$ of x9 $x^{9}$ is shown as 73/31801 rather than the correct value 119/51840 shown in Eq. (4). It happens that

73 31801 \approx 0.0022955253 while 119 51840 \approx 0.0022955247,

$\frac{73}{31801} \approx 0.0022955253 while \frac{119}{51840} \approx 0.0022955247,$

so the two rational fractions agree when rounded to 9 decimal places. The explanation is that (unlike Mathematica and Maple) Matlab works internally with decimal rather than exact arithmetic. But at the end its format rat algorithm converts a correct 14-place approximation for b10 $b_{10}$ into an incorrect rational fraction that’s “close but no cigar.”

The Matlab commands above form the basis for the interactive display shown in Fig. 11.2.4, which graphs the actual solution (blue curve)of the differential equation (1), with initial conditions b1=b2=1, $b_{1} = b_{2} = 1,$ together with the approximate solution (black curve) consisting of the terms through the fourth degree (k=5 $k = 5$ ) in (5). The pop-up menu allows the user to vary the number of terms k and thus immediately see the graphical effect of changing the number of terms included in the series expansion. With k=10, $k = 10,$ the actual and approximate solutions are indistinguishable in this viewing window.

FIGURE 11.2.4.

Matlab interactive display. The blue curve represents the actual solution of the initial value problem y''−xy'−x2y=0, y(0)=y'(0)=1, $y ″ - x y ′ - x^{2} y = 0, y (0) = y ′ (0) = 1,$ whereas the black curve shows the partial-sum approximation of the series solution (5) with terms through the fourth degree (k=5 $k = 5$ ).

Finally, you can substitute b1=1, b2=0 $b_{1} = 1, b_{2} = 0$ and b1=0, b2=1 $b_{1} = 0, b_{2} = 1$ separately (instead of b1=b2=1 $b_{1} = b_{2} = 1$ ) in the commands shown here to derive partial sums of the two linearly independent solutions displayed in Eqs. (18) and (19) of Example 7. This technique can be applied to any of the examples and problems in this section.

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Table of Contents for 11.2 Power Series Solutions

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Table of Contents for
11.2 Power Series Solutions