5.1 Solutions about Ordinary Points

INTRODUCTION

In Section 3.3 we saw that solving a homogeneous linear DE with constant coefficients was essentially a problem in algebra. By finding the roots of the auxiliary equation we could write a general solution of the DE as a linear combination of the elementary functions xk, xkeαx, xkeαxcos βx, and xkeαxsin βx, k a nonnegative integer. But as pointed out in the introduction to Section 3.6, most linear higher-order DEs with variable coefficients cannot be solved in terms of elementary functions. A usual course of action for equations of this sort is to assume a solution in the form of infinite series and proceed in a manner similar to the method of undetermined coefficients (Section 3.4). In this section we consider linear second-order DEs with variable coefficients that possess solutions in the form of power series.

5.1.1 Review of Power Series

Recall from calculus that a power series in x − a is an infinite series of the form

.

Such a series is also said to be a power series centered at a. For example, the power series is centered at a = −l. In this section we are concerned mainly with power series in x; in other words, power series such as that are centered at a = 0. The following list summarizes some important facts about power series.

  • Convergence A power series is convergent at a specified value of x if its sequence of partial sums {SN(x)} converges; that is, if limN→∞ SN(x) = limN→∞ exists. If the limit does not exist at x, the series is said to be divergent.
  • Interval of Convergence Every power series has an interval of convergence. The interval of convergence is the set of all real numbers x for which the series converges.
  • Radius of Convergence Every power series has a radius of convergence R. If R > 0, then a power series converges for |x − a| < R and diverges for |x − a| > R.If the series converges only at its center a, then R = 0. If the series converges for all x, then we write R = ∞. Recall that the absolute-value inequality |x − a| < R is equivalent to the simultaneous inequality a − R < x < a + R. A power series may or may not converge at the endpoints a − R and a + R of this interval. FIGURE 5.1.1 shows four possible intervals of convergence for R > 0.
    Interval 1 is titled: (a) (a minus R, a + R), series diverges at both endpoints. The interval is graphed on a line. It starts from a left parenthesis placed at a minus R, goes to the right, and ends at a right parenthesis placed at a + R. The center of the interval is labeled a. Interval 2 is titled: (b) [a minus R, a + R], series converges at both endpoints. The interval is graphed on a line. It starts from a left bracket placed at a minus R, goes to the right, and ends at a right bracket placed at a + R. The center of the interval is labeled a. Interval 3 is titled: (c) (a minus R, a + R], series converges at right endpoint and diverges at left endpoint. The interval is graphed on a line. It starts from a left parenthesis placed at a minus R, goes to the right, and ends at a right bracket placed at a + R. The center of the interval is labeled a. Interval 4 is titled: (d) [a minus R, a + R), series converges at left endpoint and diverges at right endpoint. The interval is graphed on a line. It starts from a left bracket placed at a minus R, goes to the right, and ends at a right parenthesis placed at a + R. The center of the interval is labeled a.

    FIGURE 5.1.1 Intervals of convergence for R > 0

  • Absolute Convergence Within its interval of convergence a power series converges absolutely. In other words, if x is a number in the interval of convergence and is not an endpoint of the interval, then the series of absolute values |cn(x − a)n| converges.
  • Ratio Test Convergence of power series can often be determined by the ratio test. Suppose that cn ≠ 0 for all n, and that

    If L < 1 the series converges absolutely, if L > 1 the series diverges, and if L = 1 the test is inconclusive. For example, for the power series (x − 3)n/2nn the ratio test gives

    The series converges absolutely for |x − 3| < 1 or |x − 3| < 2 or 1 < x < 5. This last interval is referred to as the open interval of convergence. The series diverges for |x − 3| > 2, that is, for x > 5 or x < 1. At the left endpoint x = 1 of the open interval of convergence, the series of constants ((−1)n/n) is convergent by the alternating series test. At the right endpoint x = 5, the series (1/n) is the divergent harmonic series. The interval of convergence of the series is [1, 5) and the radius of convergence is R = 2.

  • A Power Series Defines a Function A power series defines a function f(x) = cn(x − a)n whose domain is the interval of convergence of the series. If the radius of convergence is R > 0, then f is continuous, differentiable, and integrable on the interval (a − R, a + R). Moreover, f′(x) and ∫ f (x) dx can be found by term-by-term differentiation and integration. Convergence at an endpoint may be either lost by differentiation or gained through integration. If is a power series in x, then the first two derivatives are and . Notice that the first term in the first derivative and the first two terms in the second derivative are zero. We omit these zero terms and write

    . (1)

    These results are important and will be used shortly.

  • Identity Property If cn(x − a)n = 0, R > 0, for all numbers x in the interval of convergence, then cn = 0 for all n.
  • Analytic at a Point A function f is said to be analytic at a point a if it can be represented by a power series in x − a with either a positive or an infinite radius of convergence. In calculus it is seen that infinitely differentiable functions, such as ex, cos x, sin x, ln (1 + x), and so on, can be represented by Taylor series

    You might remember some of the following power series representations.

    (2)

    These Taylor series centered at a = 0, also called Maclaurin series, show that the five functions in (2) are analytic at a = 0.

    The results in (2) can also be used to obtain power series representations of other functions. For example, if we wish to find the Maclaurin series representation of, say, , we need only replace x in the Maclaurin series for ex with x2:

    Similarly, to obtain a Taylor series representation of ln x centered at a = 1 we replace x by x − 1 in the Maclaurin series for ln(1 + x):

    The interval of convergence for the power series representation of is the same as ex in (2), that is, (−∞, ∞). But the interval of convergence of the Taylor series of ln x is now (0, 2]; this interval is the interval (−1, 1] given in (2) shifted 1 unit to the right.

  • Arithmetic of Power Series Power series can be combined through the operations of addition, multiplication, and division. The procedures for power series are similar to the way in which two polynomials are added, multiplied, and divided—that is, we add coefficients of like powers of x, use the distributive law and collect like terms, and perform long division. For example, using the series in (2),

    Since the power series for ex and sin x converge for |x| < ∞, the product series converges on the same interval. Problems involving multiplication or division of power series can be done with minimal fuss using a computer algebra system.

Shifting the Summation Index

For the remainder of this section, as well as this chapter, it is important that you become adept at simplifying the sum of two or more power series, each series expressed in summation (sigma) notation, to an expression with a single Σ. As the next example illustrates, combining two or more summations as a single summation often requires a reindexing, that is, a shift in the index of summation.

EXAMPLE 1 Adding Two Power Series

Write

as one power series.

Important.

SOLUTION

In order to add the two series, it is necessary that both summation indices start with the same number and that the powers of x in each series be “in phase”; that is, if one series starts with a multiple of, say, x to the first power, then we want the other series to start with the same power. Note that in the given problem, the first series starts with x0, whereas the second series starts with x1. By writing the first term of the first series outside of the summation notation,

we see that both series on the right side start with the same power of x, namely, x1. Now to get the same summation index we are inspired by the exponents of x; we let k = n − 2 in the first series and at the same time let k = n + 1 in the second series. The right side becomes

(3)

Remember, the summation index is a “dummy” variable; the fact that k = n − 2 in one case and k = n + 1 in the other should cause no confusion if you keep in mind that it is the value of the summation index that is important. In both cases k takes on the same successive values k = 1, 2, 3, . . . when n takes on the values n = 3, 4, 5, . . . for k = n − 2 and n = 0, 1, 2, . . . for k = n + 1. We are now in a position to add the series in (3) term by term:

(4) ≡

If you are not convinced of the result in (4), then write out a few terms on both sides of the equality.

5.1.2 Power Series Solutions

A Definition

Suppose the linear second-order differential equation

a2(x)y″ + a1(x)y′ + a0(x)y = 0 (5)

is put into standard form

y″ + P(x)y′ + Q(x)y = 0 (6)

by dividing by the leading coefficient a2(x). We make the following definition.

DEFINITION 5.1.1 Ordinary and Singular Points

A point x0 is said to be an ordinary point of the differential equation (5) if both P(x) and Q(x) in the standard form (6) are analytic at x0. A point that is not an ordinary point is said to be a singular point of the equation.

Every finite value of x is an ordinary point of y″ + (ex)y′ + (sin x) y = 0. In particular, x = 0 is an ordinary point, since, as we have already seen in (2), both ex and sin x are analytic at this point. The negation in the second sentence of Definition 5.1.1 stipulates that if at least one of the functions P(x) and Q(x) in (6) fails to be analytic at x0, then x0 is a singular point. Note that x = 0 is a singular point of the differential equation y″ + (ex)y′ + (ln x)y = 0, since Q(x) = ln x is discontinuous at x = 0 and so cannot be represented by a power series in x.

Polynomial Coefficients

We shall be interested primarily in the case in which (5) has polynomial coefficients. A polynomial is analytic at any value x, and a rational function is analytic except at points where its denominator is zero. Thus, if a2(x), a1(x), and a0(x) are polynomials with no common factors, then both rational functions P(x) = a1(x)/a2(x) and Q(x) = a0(x)/a2(x) are analytic except where a2(x) = 0.

It follows, then, that x = x0 is an ordinary point of (5) if a2(x0) ≠ 0, whereas x = x 0 is a singular point of (5) if a2(x0) = 0.

For example, the only singular points of the equation (x2 − 1)y″ + 2xy′ + 6y = 0 are solutions of x2 − 1 = 0 or x = ±1. All other finite values* of x are ordinary points. Inspection of the Cauchy–Euler equation ax2y″ + bxy′ + cy = 0 shows that it has a singular point at x = 0. Singular points need not be real numbers. The equation (x2 + 1)y″ + xy′ − y = 0 has singular points at the solutions of x2 + 1 = 0; namely, x = ±i. All other values of x, real or complex, are ordinary points.

We state the following theorem about the existence of power series solutions without proof.

THEOREM 5.1.1 Existence of Power Series Solutions

If x = x0 is an ordinary point of the differential equation (5), we can always find two linearly independent solutions in the form of a power series centered at x0; that is, y =. A series solution converges at least on some interval defined by |x − x0| < R, where R is the distance from x0 to the closest singular point.

A solution of the form y = is said to be a solution about the ordinary point x0. The distance R in Theorem 5.1.1 is the minimum value or lower bound for the radius of convergence. For example, the complex numbers 1 ± 2i are singular points of (x2 − 2x + 5)y″ + xy′ − y = 0, but since x = 0 is an ordinary point of the equation, Theorem 5.1.1 guarantees that we can find two power series solutions centered at 0. That is, the solutions look like y = cnxn, and, moreover, we know without actually finding these solutions that each series must converge at least for | x | < , where R = is the distance in the complex plane from 0 to either of the numbers 1 + 2i or 1 − 2i. See FIGURE 5.1.2. However, the differential equation has a solution that is valid for much larger values of x; indeed, this solution is valid on the interval (−∞, ∞) because it can be shown that one of the two solutions is a polynomial.

Three points and two lines are graphed on an x y coordinate plane. The coordinate of the three points are as follows: (0, 0), (1, 2i), and (1, negative 2i). The points are marked with dots. Two lines connects the point (0, 0) to the points (1, 2i) and (1, negative 2i). Each line is labeled: sqrt(5).

FIGURE 5.1.2 Distance from ordinary point 0 to singular points

All the power series solutions will be centered at 0.

In the examples that follow, as well as in Exercises 5.1, we shall, for the sake of simplicity, only find power series solutions about the ordinary point x = 0. If it is necessary to find a power series solution of an ODE about an ordinary point x0 ≠ 0, we can simply make the change of variable t = x − x0 in the equation (this translates x = x0 to t = 0), find solutions of the new equation of the form y = cn tn, and then resubstitute t = x − x0.

Finding a power series solution of a homogeneous linear second-order ODE has been accurately described as “the method of undetermined series coefficients,” since the procedure is quite analogous to what we did in Section 3.4. In brief, here is the idea: We substitute y = cnxn into the differential equation, combine series as we did in Example 1, and then equate all coefficients to the right side of the equation to determine the coefficients cn. But since the right side is zero, the last step requires, by the identity property in the preceding bulleted list, that all coefficients of x must be equated to zero. No, this does not mean that all coefficients are zero; this would not make sense, because after all, Theorem 5.1.1 guarantees that we can find two linearly independent solutions. Example 2 illustrates how the single assumption that y = cnxn = c0 + c1x + c2x2 + … leads to two sets of coefficients so that we have two distinct power series y1(x) and y2(x), both expanded about the ordinary point x = 0. The general solution of the differential equation is y = C1y1(x) + C2y2(x); indeed, if y1(0) = 1, y′1(0) = 0, and y2(0) = 0, y′2(0) = 1, then C1 = c0 and C2 = c1.

EXAMPLE 2 Power Series Solutions

Solve y″ − xy = 0.

SOLUTION

Since there are no finite singular points, Theorem 5.1.1 guarantees two power series solutions, centered at 0, convergent for | x | < ∞. Substituting y = cnxn and the second derivative y″ − n(n − 1)cnxn−2 (see (1)) into the differential equation gives

(7)

Now we have already added the last two series on the right side of the equality in (7) by shifting the summation index in Example 1. From the result given in (4),

y″ − xy = 2c2 + [(k + 1)(k + 2)ck + 2 − ck −1] xk = 0. (8)

At this point we invoke the identity property. Since (8) is identically zero, it is necessary that the coefficient of each power of x be set equal to zero; that is, 2c2 = 0 (it is the coefficient of x0), and

(k + 1)(k + 2)ck + 2 − ck − 1 = 0,     k = 1, 2, 3, …. (9)

Now 2c2 = 0 obviously dictates that c2 = 0. But the expression in (9), called a recurrence relation, determines the ck in such a manner that we can choose a certain subset of the set of coefficients to be nonzero. Since (k + 1)(k + 2) ≠ 0 for all values of k, we can solve (9) for ck + 2 in terms of ck−1:

(10)

This formula is called a two-term recurrence relation.

This relation generates consecutive coefficients of the assumed solution one at a time as we let k take on the successive integers indicated in (10):

and so on. Now substituting the coefficients just obtained into the original assumption

y = c0 + c1x + c2x2 + c3x3 + c4x4 + c5x5 + c6x6 + c7x7 + c8x8 + c9x9 + c10x10 + c11x11 + …

we get

After grouping the terms containing c0 and the terms containing c1, we obtain y = c0y1(x) + c1y2(x), where

Since the recursive use of (10) leaves c0 and c1 completely undetermined, they can be chosenarbitrarily. As mentioned prior to this example, the linear combination y = c0y1(x) + c0y2(x) actually represents the general solution of the differential equation. Although we know from Theorem 5.1.1 that each series solution converges for |x| < ∞, this fact can also be verified by the ratio test. ≡

The differential equation in Example 2 is called Airy’s equation and is named after the English astronomer and mathematician Sir George Biddell Airy (1801–1892). Airy’s equation is encountered in the study of diffraction of light, diffraction of radio waves around the surface of the Earth, aerodynamics, and the deflection of a uniform thin vertical column that bends under its own weight. Other common forms of Airy’s equation are and See Problem 47 in Exercises 5.3 for an application of the last equation. It should also be apparent by making a sign change in (10) of Example 2 that the general solution of is where the series solutions are in this case

EXAMPLE 3 Power Series Solution

Solve (x2 + 1)y″ + xy′ − y = 0.

SOLUTION

As we have already seen on page 274, the given differential equation has singular points at x = ±i, and so a power series solution centered at 0 will converge at least for |x| < 1, where 1 is the distance in the complex plane from 0 to either i or −i. The assumption y = cnxn and its first two derivatives (see (1)) lead to

From this last identity we conclude that 2c2 − c0 = 0, 6c3 = 0, and

(k + 1)(k − 1)ck + (k + 2)(k + 1)ck + 2 = 0.

Thus,

Substituting k = 2, 3, 4, … into the last formula gives

and so on. Therefore,

The solutions are the polynomial y2(x) = x and the power series

≡

EXAMPLE 4 Three-Term Recurrence Relation

If we seek a power series solution y = cnxn for the differential equation

y″ − (1 + x)y = 0,

we obtain c2 = c0/2 and the recurrence relation

ck + 2 =      k = 1, 2, 3, . . . .

This formula is called a three-term recurrence relation.

Examination of the formula shows that the coefficients c3, c4, c5, … are expressed in terms of both c1 and c0, and moreover, the algebra required to do this becomes a little messy. To simplify life we can first choose c0 ≠ 0, c1 = 0; this yields consecutive coefficients for one solution that are expressed entirely in terms of c0. If we next choose c0 ≠ 0, c1 = 0; this yields consecutive coefficients for one solution that are expressed entirely in terms of c0 then the coefficients for the other solution are expressed in terms of c1. Using in both cases, the recurrence relation for k = 1, 2, 3, … gives

and so on. Finally, we see that the general solution of the equation is y = c0y1(x) + c1y2(x), where

and

Each series converges for all finite values of x. ≡

Nonpolynomial Coefficients

The next example illustrates how to find a power series solution about the ordinary point x0 = 0 of a differential equation when its coefficients are not polynomials. In this example we see an application of multiplication of two power series.

EXAMPLE 5 ODE with Nonpolynomial Coefficients

Solve y″ + (cos x)y = 0.

SOLUTION

We see x = 0 is an ordinary point of the equation because, as we have already seen, cos x is analytic at that point. Using the Maclaurin series for cos x given in (2), along with the usual assumption y = cnxn and the results in (1), we find

It follows that

and so on. This gives c2 = − c0, c3 = − c1, c4 = c0, c5 = c1, …. By grouping terms we arrive at the general solution y = c0y1(x) + c1y2(x), where

Since the differential equation has no finite singular points, both power series converge for |x| < ∞. ≡

Solution Curves

The approximate graph of a power series solution y(x) = cnxn can be obtained in several ways. We can always resort to graphing the terms in the sequence of partial sums of the series, in other words, the graphs of the polynomials SN (x) = cnxn. For large values of N, SN (x) should give us an indication of the behavior of y(x) near the ordinary point x = 0. We can also obtain an approximate solution curve by using a numerical solver as we did in Section 3.7. For example, if you carefully scrutinize the series solutions of Airy’s equation in Example 2, you should see that y1(x) and y2(x) are, in turn, the solutions of the initial-value problems

y″ − xy = 0,     y(0) = 1,     y′(0) = 0,

y″ − xy = 0,     y(0) = 0,     y′(0) = 1. (11)

The specified initial conditions “pick out” the solutions y1(x) and y2(x) from y = c0y1(x) + c1y2(x), since it should be apparent from our basic series assumption y = cnxn that y(0) = c0 and y′(0) = c1. Now if your numerical solver requires a system of equations, the substitution y′ = u in y″ − xy = 0 gives y″ = u′ = xy, and so a system of two first-order equations equivalent to Airy’s equation is

y′ = u

  u′ = xy. (12)

Initial conditions for the system in (12) are the two sets of initial conditions in (11) but rewritten as y(0) = 1, u(0) = 0, and y(0) = 0, u(0) = 1. The graphs of y1(x) and y2(x) shown in FIGURE 5.1.3 were obtained with the aid of a numerical solver using the fourth-order Runge–Kutta method with a step size of h = 0.1.

The first graph is labeled: (a) Plot of y subscript 1 (x). A curve is graphed on an x y subscript 1 plane. It starts at the approximate point (negative 10, negative 0.1), goes up and down and to the right through the approximate points (negative 9.5, 0.5), (negative 9, 0), (negative 8, negative 0.6), (negative 7, 0.6), (negative 6, negative 0.6), (negative 5, 0.8), (negative 3, negative 0.8), (negative 1, 1), (0, 1), goes up sharply and ends at the approximate point (2, 3). The second graph is labeled: (b) Plot of y subscript 2 (x). A curve is graphed on an x y subscript 2 plane. It starts at the approximate point (negative 10, negative 0.5), goes down and up and to the right through the approximate points (negative 9.9, negative 0.8), (negative 9, 0.7), (negative 8, negative 0.8), (negative 7, 0.8), (negative 5, negative 1), (negative 4, 1), (negative 2, negative 1.2), (0, 0), goes sharply up, and ends at the approximate point (2, 3.2).

FIGURE 5.1.3 Solutions of Airy’s equation

REMARKS

(i) In the problems that follow, do not expect to be able to write a solution in terms of summation notation in each case. Even though we can generate as many terms as desired in a series solution either through the use of a recurrence relation or, as in Example 5, by multiplication, it may not be possible to deduce any general term for the coefficients cn. We may have to settle, as we did in Examples 4 and 5, for just writing out the first few terms of the series.

(ii) A point x0 is an ordinary point of a nonhomogeneous linear second-order DE y″ + P(x)y′ + Q(x)y = f(x) if P(x), Q(x), and f(x) are analytic at x0. Moreover, Theorem 5.1.1 extends to such DEs—in other words, we can find power series solutions of nonhomogeneous linear DEs in the same manner as in Examples 2–5. See Problem 38 in Exercises 5.1.

5.1 Exercises Answers to selected odd-numbered problems begin on page ANS-12.

5.1.1 Review of Power Series

In Problems 1–4, find the radius of convergence and interval of convergence for the given power series.

In Problems 5 and 6, the given function is analytic at x = 0. Find the first four terms of a power series in x. Perform the multiplication by hand or use a CAS, as instructed.

  1. sin x cos x
  2. e−x cos x

In Problems 7 and 8, the given function is analytic at x = 0. Find the first four terms of a power series in x. Perform the long division by hand or use a CAS, as instructed. Give the open interval of convergence.

In Problems 9 and 10, rewrite the given power series so that its general term involves xk.

In Problems 11 and 12, rewrite the given expression as a single power series whose general term involves xk.

In Problems 13–16, verify by direct substitution that the given power series is a particular solution of the indicated differential equation.

5.1.2 Power Series Solutions

In Problems 17 and 18, without actually solving the given differential equation, find a lower bound for the radius of convergence of power series solutions about the ordinary point x = 0. About the ordinary point x = 1.

  1. (x2 − 25)y″ + 2xy′ + y = 0
  2. (x2 − 2x + 10)y″ + xy′ − 4y = 0

In Problems 19–30, find two power series solutions of the given differential equation about the ordinary point x = 0.

  1. y″ − 3xy = 0
  2. y″ + x2y = 0
  3. y″ − 2xy′ + y = 0
  4. y″ − xy′ + 2y = 0
  5. y″ + x2y′ + xy = 0
  6. y″ + 2xy′ + 2y = 0
  7. (x − 1)y″ + y′ = 0
  8. (x + 2)y″ + xy′ − y = 0
  9. y″ − (x + 1)y′ − y = 0
  10. (x2 + 1)y″ − 6y = 0
  11. (x2 + 2)y″ + 3xy′ − y = 0
  12. (x2 − 1)y″ + xy′ − y = 0

In Problems 31–34, use the power series method to solve the given initial-value problem.

  1. (x − 1)y″ − xy′ + y = 0, y(0) = −2, y′(0) = 6
  2. (x + 1)y″ − (2 − x)y′ + y = 0, y(0) = 2, y′(0) = −1
  3. y″ − 2xy′ + 8y = 0, y(0) = 3, y′(0) = 0
  4. (x2 + 1)y″ + 2xy′ = 0, y(0) = 0, y′(0) = 1

In Problems 35 and 36, use the procedure in Example 5 to find two power series solutions of the given differential equation about the ordinary point x = 0.

  1. y″ + (sin x)y = 0
  2. y″ + exy′ − y = 0

Discussion Problems

  1. Without actually solving the differential equation (cos x)y″ + y′ + 5y = 0, find a lower bound for the radius of convergence of power series solutions about x = 0. About x = 1.
  2. How can the method described in this section be used to find a power series solution of the nonhomogeneous equation y″ − xy = 1 about the ordinary point x = 0? Of y″ − 4xy′ − 4y = ex? Carry out your ideas by solving both DEs.
  3. Is x = 0 an ordinary or a singular point of the differential equation xy″ + (sin x)y = 0? Defend your answer with sound mathematics.
  4. For purposes of this problem, ignore the graphs given in Figure 5.1.3. If Airy’s differential equation is written as y″ − xy, what can we say about the shape of a solution curve if x > 0 and y > 0? If x < 0 and y > 0? If x < 0 and y < 0?

Computer Lab Assignments

  1. (a) Find two power series solutions for y″ + xy′ + y = 0 and express the solutions y1(x) and y2(x) in terms of summation notation.

    (b) Use a CAS to graph the partial sums SN (x) for y1(x). Use N = 2, 3, 5, 6, 8, 10. Repeat using the partial sums SN (x) for y2(x).

    (c) Compare the graphs obtained in part (b) with the curve obtained using a numerical solver. Use the initial conditions y1(0) = 1, y′1(0) = 0, and y2(0) = 0, y′2(0) = 1.

    (d) Reexamine the solution y1(x) in part (a). Express this series as an elementary function. Then use (5) of Section 3.2 to find a second solution of the equation. Verify that this second solution is the same as the power series solution y2(x).

  2. (a) Find one more nonzero term for each of the solutions y1(x) and y2(x) in Example 5.

    (b) Find a series solution y(x) of the initial-value problem y″ + (cos x)y = 0, y(0) = 1, y′(0) = 1.

    (c) Use a CAS to graph the partial sums SN (x) for the solution y(x) in part (b). Use N = 2, 3, 4, 5, 6, 7.

    (d) Compare the graphs obtained in part (c) with the curve obtained using a numerical solver for the initial-value problem in part (b).

 

*For our purposes, ordinary points and singular points will always be finite points. It is possible for an ODE to have, say, a singular point at infinity.