Chapter 6 · Section 6.1

6.1 Power Series Solutions About an Ordinary Point

Up to now every second-order linear equation we solved had either constant coefficients or the very special Cauchy–Euler form. In both cases the solutions were built from familiar functions: exponentials, sines, cosines, powers and logarithms. Most equations with variable coefficients are not so kind. A simple-looking equation such as

y^{''} + x^{2} y = 0

has no solution that can be written with elementary functions at all. Yet equations of this type are everywhere in engineering: the deflection of a column whose stiffness changes along its length, the temperature in a fin with a tapered cross-section, waves in a medium whose properties vary with position. We need a method that does not depend on guessing the right function.

The idea of this section is simple. If we cannot guess the solution, we look for it as an infinite polynomial, a power series $y = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ , and let the differential equation tell us the coefficients one after another. To do this confidently you need three things: a short review of power series, a reliable technique for shifting the index of a sum, and a theorem that tells you when the method works and for which $x$ the series converges.

A short review of power series

Definition 6.1.1 (Power series, radius of convergence)

A power series about the point $x_{0}$ (the center) is an infinite sum

n = 0 \sum \infty a_{n} (x - x_{0})^{n} = a_{0} + a_{1} (x - x_{0}) + a_{2} (x - x_{0})^{2} + \dots .

For every power series there is a number $R$ with $0 \leq R \leq \infty$ , the radius of convergence, such that the series converges (absolutely) for $∣ x - x_{0} ∣ < R$ and diverges for $∣ x - x_{0} ∣ > R$ . At the two endpoints $x = x_{0} \pm R$ anything can happen.

The most practical way to find $R$ is the ratio test. Look at the ratio of two consecutive nonzero terms. If

n \to \infty lim \frac{a _{n + 1} ( x - x _{0} ) ^{n + 1}}{a _{n} ( x - x _{0} ) ^{n}} = L ∣ x - x_{0} ∣,

then the series converges when $L ∣ x - x_{0} ∣ < 1$ , so $R = 1/ L$ (with $R = \infty$ when $L = 0$ and $R = 0$ when $L = \infty$ ). If only every second coefficient is nonzero (a series with only even powers, say), apply the same idea to consecutive nonzero terms; we will do exactly this in the examples.

Three familiar series will keep showing up. They all converge for every $x$ except the last one, which needs $∣ x ∣ < 1$ :

e^{x} = n = 0 \sum \infty \frac{x ^{n}}{n !}, cos x = k = 0 \sum \infty \frac{( - 1 ) ^{k} x ^{2 k}}{( 2 k )!}, sin x = k = 0 \sum \infty \frac{( - 1 ) ^{k} x ^{2 k + 1}}{( 2 k + 1 )!}, \frac{1}{1 - x} = n = 0 \sum \infty x^{n} .

Theorem 6.1.1 (Working with power series)

Suppose $f (x) = \sum_{n = 0}^{\infty} a_{n} (x - x_{0})^{n}$ has radius of convergence $R > 0$ . Then on $∣ x - x_{0} ∣ < R$ :

Term-by-term differentiation. $f$ is differentiable and $f^{'} (x) = n = 1 \sum \infty n a_{n} (x - x_{0})^{n - 1}, f^{''} (x) = n = 2 \sum \infty n (n - 1) a_{n} (x - x_{0})^{n - 2},$ and these series also have radius $R$ .
Coefficients. $a_{n} = \frac{f ^{(n)} ( x _{0} )}{n !}$ ; in other words the series is the Taylor series of $f$ .
Identity principle. If $\sum_{n = 0}^{\infty} c_{n} (x - x_{0})^{n} = 0$ for every $x$ in some open interval around $x_{0}$ , then $c_{n} = 0$ for every $n$ .
Algebra. Two series with the same center can be added term by term, and a series can be multiplied by a polynomial term by term; the result converges at least where both pieces converge.

Point 3 is the engine of the whole method: after substituting a series into a differential equation we will collect everything into one series equal to zero, and then set every coefficient to zero. Point 2 explains why the lower limits in $f^{'}$ and $f^{''}$ start at $1$ and $2$ : the constant term disappears when differentiated once, and the linear term disappears when differentiated twice.

Definition 6.1.2 (Analytic function)

A function $f$ is analytic at $x_{0}$ if it equals a power series $\sum a_{n} (x - x_{0})^{n}$ with positive radius of convergence on some open interval around $x_{0}$ .

Polynomials, $e^{x}$ , $sin x$ and $cos x$ are analytic at every point. A rational function $N (x) / D (x)$ (with common factors cancelled) is analytic at every point where $D (x_{0}) \neq = 0$ . Functions such as $x$ , $ln x$ and $∣ x ∣$ are not analytic at $x = 0$ .

Shifting the index of summation, step by step

This is the technical skill on which everything else depends, so we take it slowly. The good news: it is pure bookkeeping, and it always follows the same pattern.

The index is a dummy variable

The letter used for the index does not matter:

n = 0 \sum \infty a_{n} x^{n} = k = 0 \sum \infty a_{k} x^{k} = a_{0} + a_{1} x + a_{2} x^{2} + \dots .

Both sums stand for the same list of terms. The index is like the variable of integration in $\int_{0}^{1} t^{2} d t = \int_{0}^{1} s^{2} d s$ .

Why we need to shift

Suppose $y = \sum_{n = 0}^{\infty} a_{n} x^{n}$ . When we substitute into an equation we get sums such as

y^{''} = n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} and y = n = 0 \sum \infty a_{n} x^{n} .

To add them and use the identity principle we need, in each sum, the same power of $x$ in the general term and the same starting value of the index. In $y^{''}$ the power is $x^{n - 2}$ ; in $y$ it is $x^{n}$ . Shifting the index fixes this.

The four moves

Let us rewrite $n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2}$ so that the general term contains $x^{k}$ .

Name the new index. We want the exponent to be the new index, so set $k = n - 2.$
Solve for the old index. $n = k + 2$ .
Replace every $n$ in the general term (not only in the exponent!): $n (n - 1) a_{n} x^{n - 2} ⟶ (k + 2) (k + 1) a_{k + 2} x^{k} .$
Change the limits. The old sum started at $n = 2$ , so the new one starts at $k = 2 - 2 = 0$ . When $n \to \infty$ , also $k \to \infty$ .

The result is

n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} = k = 0 \sum \infty (k + 2) (k + 1) a_{k + 2} x^{k} .

Since the index is a dummy, we may now rename $k$ back to $n$ if we like: $\sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n}$ .

Do not take this on faith; write out the first terms of both sides and compare:

Old index	Term $n (n - 1) a_{n} x^{n - 2}$	New index	Term $(k + 2) (k + 1) a_{k + 2} x^{k}$
$n = 2$	$2 \cdot 1 \cdot a_{2} x^{0}$	$k = 0$	$2 \cdot 1 \cdot a_{2} x^{0}$
$n = 3$	$3 \cdot 2 \cdot a_{3} x^{1}$	$k = 1$	$3 \cdot 2 \cdot a_{3} x^{1}$
$n = 4$	$4 \cdot 3 \cdot a_{4} x^{2}$	$k = 2$	$4 \cdot 3 \cdot a_{4} x^{2}$

Same terms, same order. A shift only relabels the terms; it never changes the sum. A good habit: whenever you shift, check the first term of the old and new sums.

Absorbing a power of x first

Often a sum is multiplied by a power of $x$ , as in $x y^{'}$ . First move that power inside the sum, then decide whether a shift is needed:

x n = 1 \sum \infty n a_{n} x^{n - 1} = n = 1 \sum \infty n a_{n} x^{n} .

Here no shift is needed: the power is already $x^{n}$ . Notice also that the $n = 0$ term of $\sum n a_{n} x^{n}$ would be $0 \cdot a_{0} = 0$ , so we may let this sum start at $n = 0$ without changing it. Adding or removing terms that are zero is a perfectly legal way to line up starting values.

Shifting the other way

If a sum contains $x^{n + 1}$ , the move is the same with the opposite sign. For $\sum_{n = 0}^{\infty} a_{n} x^{n + 1}$ : set $k = n + 1$ , so $n = k - 1$ , the general term becomes $a_{k - 1} x^{k}$ , and the start $n = 0$ becomes $k = 1$ :

n = 0 \sum \infty a_{n} x^{n + 1} = k = 1 \sum \infty a_{k - 1} x^{k} .

Memory aid: if the exponent goes down by some amount, the subscript goes up by the same amount, and so does the starting value (and vice versa).

Peeling off terms

After shifting, the sums may start at different values. Write the terms of the sum that starts earlier separately until all sums start at the same index; then combine.

Example 6.1.1

Write as a single power series in $x$ :

n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} + 3 n = 0 \sum \infty a_{n} x^{n + 1} .

Solution

First sum. As shown above, with $k = n - 2$ : $n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} = k = 0 \sum \infty (k + 2) (k + 1) a_{k + 2} x^{k} .$
Second sum. With $k = n + 1$ (so $n = k - 1$ , start $k = 1$ ): $3 n = 0 \sum \infty a_{n} x^{n + 1} = k = 1 \sum \infty 3 a_{k - 1} x^{k} .$
Line up the starting values. The first sum starts at $k = 0$ , the second at $k = 1$ . Peel off the $k = 0$ term of the first sum: it is $2 \cdot 1 \cdot a_{2} x^{0} = 2 a_{2}$ .
Combine. Now both remaining sums start at $k = 1$ and contain $x^{k}$ : $2 a_{2} + k = 1 \sum \infty [(k + 2) (k + 1) a_{k + 2} + 3 a_{k - 1}] x^{k} .$
Check. The coefficient of $x^{1}$ should be $3 \cdot 2 a_{3}$ (from $n = 3$ in the first sum) plus $3 a_{0}$ (from $n = 0$ in the second). Our formula at $k = 1$ gives $3 \cdot 2 a_{3} + 3 a_{0}$ . Correct.

Common mistakes when shifting

Changing the exponent but forgetting the subscript or the factors: $n (n - 1) a_{n} x^{n - 2}$ must become $(k + 2) (k + 1) a_{k + 2} x^{k}$ , not $n (n - 1) a_{n} x^{k}$ .
Forgetting to change the lower limit.
Combining sums that start at different values, which silently drops terms. Always peel off first.

Ordinary points and singular points

We consider the homogeneous linear equation

a_{0} (x) y^{''} + a_{1} (x) y^{'} + a_{2} (x) y = 0

and its normalized form (divide by $a_{0}$ , as in Section 4.1)

y^{''} + p (x) y^{'} + q (x) y = 0, p = \frac{a _{1}}{a _{0}}, q = \frac{a _{2}}{a _{0}} .

Definition 6.1.3 (Ordinary point, singular point)

The point $x_{0}$ is an ordinary point of the equation if both $p$ and $q$ are analytic at $x_{0}$ . Otherwise $x_{0}$ is a singular point.

In this course the coefficients $a_{0}, a_{1}, a_{2}$ are almost always polynomials. Then the test is easy: after cancelling any factor common to $a_{0}, a_{1}, a_{2}$ , the singular points are exactly the zeros of $a_{0} (x)$ , including complex zeros. Every other point is ordinary.

For example, $(x^{2} - 9) y^{''} + x y^{'} + y = 0$ has singular points $x = \pm 3$ ; $(x^{2} + 1) y^{''} - y = 0$ has no real singular points, but it has the complex singular points $x = \pm i$ , and we will see that they matter.

Theorem 6.1.2 (Series solutions about an ordinary point)

Let $x_{0}$ be an ordinary point of $y^{''} + p (x) y^{'} + q (x) y = 0$ , and let $R > 0$ be the smaller of the radii of convergence of the Taylor series of $p$ and $q$ about $x_{0}$ . Then:

The equation has two linearly independent solutions $y_{1}, y_{2}$ of the form $\sum_{n = 0}^{\infty} c_{n} (x - x_{0})^{n}$ , and every solution can be written as $y = a_{0} y_{1} + a_{1} y_{2}$ ; equivalently, every solution is a power series $y = n = 0 \sum \infty a_{n} (x - x_{0})^{n}$ in which $a_{0} = y (x_{0})$ and $a_{1} = y^{'} (x_{0})$ are arbitrary and all other $a_{n}$ are determined by them.
These series converge at least for $∣ x - x_{0} ∣ < R$ .

When $a_{0}, a_{1}, a_{2}$ are polynomials with no common factor, $R$ is at least the distance from $x_{0}$ to the nearest singular point in the complex plane.

Why complex singular points count

A power series in $x - x_{0}$ converges inside a disc in the complex plane, not only on an interval. If a singular point sits at distance $d$ from $x_{0}$ anywhere in the plane, the disc of convergence of the solution may not be able to grow past it, and its intersection with the real line is the interval $(x_{0} - d, x_{0} + d)$ . This is why a function that is perfectly smooth for all real $x$ can still have a power series with a finite radius. Example 6.1.4 shows this happening.

Two remarks. First, the theorem gives a lower bound: the actual radius can be larger. Second, if there are no singular points at all (for instance when $a_{0}$ is a nonzero constant and $a_{1}, a_{2}$ are polynomials), the series converge for every $x$ .

Example 6.1.2

Without solving, find a lower bound for the radius of convergence of power series solutions of

(x^{2} + 2 x + 5) y^{''} + x y^{'} - 3 y = 0

about (a) $x_{0} = 0$ , (b) $x_{0} = 1$ , (c) $x_{0} = - 1$ .

Solution

Singular points. They are the zeros of $a_{0} (x) = x^{2} + 2 x + 5$ : $x = \frac{- 2 \pm 4 - 20}{2} = - 1 \pm 2 i .$ There are no real singular points, so every real point is ordinary.
(a) $x_{0} = 0$ . The distance from $0$ to $- 1 + 2 i$ is $(- 1)^{2} + 2^{2} = 5$ ; by symmetry the same holds for $- 1 - 2 i$ . So $R \geq 5 \approx 2.24$ , and the series converge at least on $(- 5, 5)$ .
(b) $x_{0} = 1$ . Distance from $1$ to $- 1 + 2 i$ : $(1 - (- 1))^{2} + (0 - 2)^{2} = 8 = 22$ . So $R \geq 22 \approx 2.83$ .
(c) $x_{0} = - 1$ . Distance from $- 1$ to $- 1 + 2 i$ is $2$ , so $R \geq 2$ .

Figure 6.1.1 shows the two discs for (a) and (b). Each disc passes exactly through the singular points; its intersection with the real axis is the guaranteed interval.

Figure 6.1.1. Guaranteed discs of convergence for Example 6.1.2. The circles about

x_{0} = 0

and

x_{0} = 1

pass through the complex singular points

- 1 \pm 2 i

; the thick segments on the real axis are the intervals on which convergence is guaranteed.

The method

Here is the whole procedure. It is long the first time, but every step is mechanical.

Method in a nutshell

Check that $x_{0}$ is an ordinary point and find the guaranteed radius $R$ (distance to the nearest singular point). If $x_{0} \neq = 0$ , substitute $t = x - x_{0}$ so that you expand about $t = 0$ .
Write $y = \sum_{n = 0}^{\infty} a_{n} x^{n}$ , $y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ , $y^{''} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ and substitute.
Move every power of $x$ from the coefficients inside the sums.
Shift indices so that every sum has the same power $x^{n}$ in its general term.
Peel off terms until all sums start at the same index, then combine into one series.
Set every coefficient equal to zero (identity principle). This gives a few starting equations and a recurrence relation expressing a later coefficient in terms of earlier ones.
Use the recurrence to compute coefficients; keep $a_{0}$ and $a_{1}$ free (or set them from initial conditions $a_{0} = y (x_{0})$ , $a_{1} = y^{'} (x_{0})$ ). Look for a pattern if you can.
Write the answer as $y = a_{0} y_{1} (x) + a_{1} y_{2} (x)$ and state where it converges.

The solver below carries out steps 2 to 7 for $p (x) y^{''} + q (x) y^{'} + r (x) y = 0$ with polynomial coefficients and $p (0) \neq = 0$ . It shows the recurrence and the first terms of both basic solutions. The first preset is Example 6.1.3 below; the second is the initial-value problem of Example 6.1.4. Change the polynomials and try your own equations, then check your hand computations against it.

Worked examples

Example 6.1.3

Find the general solution of $y^{''} + 2 x y^{'} + 2 y = 0$ as a power series about $x_{0} = 0$ . Identify one of the two basic solutions in closed form.

Solution

Ordinary point. Here $p (x) = 2 x$ and $q (x) = 2$ are polynomials, so there are no singular points. Every solution is a power series that converges for all $x$ .
Substitute. With $y = \sum_{n = 0}^{\infty} a_{n} x^{n}$ : $n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} + 2 x n = 1 \sum \infty n a_{n} x^{n - 1} + 2 n = 0 \sum \infty a_{n} x^{n} = 0.$
Move the $x$ inside. $2 x \sum n a_{n} x^{n - 1} = \sum_{n = 1}^{\infty} 2 n a_{n} x^{n}$ . Its $n = 0$ term would be zero, so we may start it at $n = 0$ .
Shift the first sum. With $k = n - 2$ , then renaming $k$ as $n$ : $n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} = n = 0 \sum \infty (n + 2) (n + 1) a_{n + 2} x^{n} .$
Combine. All three sums now contain $x^{n}$ and start at $n = 0$ : $n = 0 \sum \infty [(n + 2) (n + 1) a_{n + 2} + 2 n a_{n} + 2 a_{n}] x^{n} = 0.$
Recurrence. Every bracket must vanish. Since $2 n a_{n} + 2 a_{n} = 2 (n + 1) a_{n}$ , we can cancel the factor $n + 1$ : $a_{n + 2} = - \frac{2}{n + 2} a_{n}, n = 0, 1, 2, \dots$
Even coefficients. They are tied to $a_{0}$ : $a_{2} a_{4} a_{6} a_{8} = - \frac{2}{2} a_{0} = - a_{0}, = - \frac{2}{4} a_{2} = \frac{1}{2} a_{0}, = - \frac{2}{6} a_{4} = - \frac{1}{6} a_{0}, = - \frac{2}{8} a_{6} = \frac{1}{24} a_{0} .$ The pattern is $a_{2 k} = \frac{( - 1 ) ^{k}}{k !} a_{0}$ . (Indeed, with $n = 2 k - 2$ the recurrence reads $a_{2 k} = - a_{2 k - 2} / k$ .)
Odd coefficients. They are tied to $a_{1}$ : $a_{3} = - \frac{2}{3} a_{1}, a_{5} = - \frac{2}{5} a_{3} = \frac{4}{15} a_{1}, a_{7} = - \frac{2}{7} a_{5} = - \frac{8}{105} a_{1},$ so $a_{2 k + 1} = \frac{( - 2 ) ^{k}}{1 \cdot 3 \cdot 5 \dots ( 2 k + 1 )} a_{1}$ .
General solution. $y = a_{0} y_{1} (1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + \dots) + a_{1} y_{2} (x - \frac{2}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{105} x^{7} + \dots) .$
Closed form of $y_{1}$ . $y_{1} = \sum_{k} \frac{( - x ^{2} ) ^{k}}{k !}$ is the exponential series with $x^{2}$ replaced by $- x^{2}$ , so $y_{1} = e^{- x^{2}}$ . A direct check: $y_{1}^{'} = - 2 x e^{- x^{2}}$ , $y_{1}^{''} = (4 x^{2} - 2) e^{- x^{2}}$ , and $(4 x^{2} - 2) - 4 x^{2} + 2 = 0$ .
Convergence. For $y_{1}$ the ratio of consecutive nonzero terms is $\frac{a _{n + 2} x ^{n + 2}}{a _{n} x ^{n}} = \frac{2∣ x ∣ ^{2}}{n + 2} \to 0$ , and the same holds for $y_{2}$ . Both series converge for every $x$ , as Theorem 6.1.2 promised.

The second solution $y_{2}$ is not an elementary function. (It equals $e^{- x^{2}} \int_{0}^{x} e^{s^{2}} d s$ , known as Dawson's integral.) The power series is a perfectly good way to define and compute it.

Figure 6.1.2. Partial sums of

y_{1}

in Example 6.1.3 compared with

e^{- x^{2}}

. Each partial sum is excellent near

x = 0

; adding more terms extends the range of agreement. Because the radius is infinite, any interval can be covered by taking enough terms.

Figure 6.1.3. The two basic solutions of Example 6.1.3:

y_{1}

(even, with

y_{1} (0) = 1

y_{1}^{'} (0) = 0

) and

y_{2}

(odd, with

y_{2} (0) = 0

y_{2}^{'} (0) = 1

). Every solution is

a_{0} y_{1} + a_{1} y_{2}

Even and odd solutions

Whenever the recurrence links $a_{n + 2}$ only to $a_{n}$ (a two-term recurrence with step 2), the even coefficients depend only on $a_{0}$ and the odd ones only on $a_{1}$ . Then $y_{1}$ contains only even powers and $y_{2}$ only odd powers. This happens when the equation is unchanged under $x \to - x$ , as in Example 6.1.3.

Example 6.1.4 (initial-value problem)

Solve by power series about $x_{0} = 0$ :

(x^{2} + 4) y^{''} + 6 x y^{'} + 6 y = 0, y (0) = 1, y^{'} (0) = - 1.

Where does the series converge?

Solution

Guaranteed radius. $a_{0} (x) = x^{2} + 4$ vanishes at $x = \pm 2 i$ . The distance from $0$ to $\pm 2 i$ is $2$ , so the series converges at least for $∣ x ∣ < 2$ .
Substitute and distribute. Splitting $(x^{2} + 4) y^{''} = x^{2} y^{''} + 4 y^{''}$ : $n = 2 \sum \infty n (n - 1) a_{n} x^{n} + n = 2 \sum \infty 4 n (n - 1) a_{n} x^{n - 2} + n = 1 \sum \infty 6 n a_{n} x^{n} + n = 0 \sum \infty 6 a_{n} x^{n} = 0.$
Line up. In the first sum the terms with $n = 0, 1$ would be zero (because of the factor $n (n - 1)$ ), so it may start at $n = 0$ ; likewise the third sum may start at $n = 0$ . Only the second sum needs a shift ( $k = n - 2$ ): $n = 2 \sum \infty 4 n (n - 1) a_{n} x^{n - 2} = n = 0 \sum \infty 4 (n + 2) (n + 1) a_{n + 2} x^{n} .$
Combine. $n = 0 \sum \infty [4 (n + 2) (n + 1) a_{n + 2} + (n (n - 1) + 6 n + 6) a_{n}] x^{n} = 0.$
Simplify. $n (n - 1) + 6 n + 6 = n^{2} + 5 n + 6 = (n + 2) (n + 3)$ . So each bracket is $(n + 2) [4 (n + 1) a_{n + 2} + (n + 3) a_{n}]$ , and the recurrence is $a_{n + 2} = - \frac{n + 3}{4 ( n + 1 )} a_{n}, n \geq 0.$
Initial conditions. $a_{0} = y (0) = 1$ and $a_{1} = y^{'} (0) = - 1$ .
Coefficients. Even: $a_{2} = - \frac{3}{4}$ , $a_{4} = - \frac{5}{12} \cdot (- \frac{3}{4}) = \frac{5}{16}$ , $a_{6} = - \frac{7}{20} \cdot \frac{5}{16} = - \frac{7}{64}$ , … Odd: $a_{3} = - \frac{4}{8} \cdot (- 1) = \frac{1}{2}$ , $a_{5} = - \frac{6}{16} \cdot \frac{1}{2} = - \frac{3}{16}$ , $a_{7} = - \frac{8}{24} \cdot (- \frac{3}{16}) = \frac{1}{16}$ , … In general $a_{2 k} = (2 k + 1) (- \frac{1}{4})^{k}, a_{2 k + 1} = - (k + 1) (- \frac{1}{4})^{k} .$
Sum the series. Put $u = - x^{2} /4$ . Differentiating the geometric series $\sum_{k \geq 0} u^{k} = \frac{1}{1 - u}$ gives $\sum_{k \geq 0} (k + 1) u^{k} = \frac{1}{( 1 - u ) ^{2}}$ , and since $2 k + 1 = 2 (k + 1) - 1$ , $y = k = 0 \sum \infty (2 k + 1) u^{k} - x k = 0 \sum \infty (k + 1) u^{k} = \frac{2}{( 1 - u ) ^{2}} - \frac{1}{1 - u} - \frac{x}{( 1 - u ) ^{2}} .$ With $\frac{1}{1 - u} = \frac{4}{x ^{2} + 4}$ this becomes $y = \frac{32 - 4 ( x ^{2} + 4 ) - 16 x}{( x ^{2} + 4 ) ^{2}} = \frac{4 ( 4 - 4 x - x ^{2} )}{( x ^{2} + 4 ) ^{2}} .$
Convergence. Both series in $u$ converge exactly when $∣ u ∣ < 1$ , that is $∣ x^{2} /4∣ < 1$ , or $∣ x ∣ < 2$ . (For $∣ u ∣ \geq 1$ their terms do not even tend to zero.) The guaranteed radius of Theorem 6.1.2 is attained.

Here is the surprise: the function $4 (4 - 4 x - x^{2}) / (x^{2} + 4)^{2}$ is smooth on the whole real line, but its power series about $0$ diverges for $∣ x ∣ > 2$ . The reason is invisible on the real axis: the function blows up at the complex points $\pm 2 i$ . Figure 6.1.4 shows the partial sums tearing away from the solution at $x = \pm 2$ .

Figure 6.1.4. Example 6.1.4. Inside the shaded interval

∣ x ∣ < 2

the partial sums converge to

y = 4 (4 - 4 x - x^{2}) / (x^{2} + 4)^{2}

; outside it, adding terms makes things worse. The edge of convergence is set by the complex singular points

\pm 2 i

Checking the answer without series

Substituting $y = \frac{A ( 4 - x ^{2} ) + B x}{( x ^{2} + 4 ) ^{2}}$ into $(x^{2} + 4) y^{''} + 6 x y^{'} + 6 y$ gives $0$ for every choice of the constants $A$ and $B$ (a few lines of algebra, or one line of a computer algebra system). So these functions form the general solution, and $A = 4$ , $B = - 16$ matches the initial conditions. Closed forms like this are rare; the series method works whether or not one exists, and here it actually found it for us.

Expanding about a point other than zero

If the initial conditions are given at $x_{0} \neq = 0$ , the solution should be expanded in powers of $x - x_{0}$ ; this is also necessary when $x = 0$ is a singular point. The cleanest way is to substitute $t = x - x_{0}$ . Since $d t / d x = 1$ , derivatives with respect to $x$ and $t$ are the same, so only the coefficients change.

Example 6.1.5 (expansion about

x_{0} = 1

)

Find the first six terms of the series solution of

x y^{''} + 3 y^{'} - y = 0, y (1) = 1, y^{'} (1) = 0,

in powers of $x - 1$ , and state an interval on which it is guaranteed to converge.

Solution

Points. $a_{0} (x) = x$ vanishes only at $x = 0$ , so $x = 0$ is the only singular point and $x_{0} = 1$ is ordinary. The distance from $1$ to $0$ is $1$ : the series converges at least for $∣ x - 1∣ < 1$ , i.e. on $0 < x < 2$ . (A series in powers of $x$ would not work here, because $0$ is singular.)
Shift the variable. Let $t = x - 1$ , so $x = 1 + t$ . The problem becomes $(1 + t) y^{''} + 3 y^{'} - y = 0, y = 1, y^{'} = 0 at t = 0,$ where primes now mean $d / d t$ .
Substitute $y = \sum_{n = 0}^{\infty} a_{n} t^{n}$ : $n = 2 \sum \infty n (n - 1) a_{n} t^{n - 2} + n = 2 \sum \infty n (n - 1) a_{n} t^{n - 1} + n = 1 \sum \infty 3 n a_{n} t^{n - 1} - n = 0 \sum \infty a_{n} t^{n} = 0.$
Shift each sum to $t^{n}$ .
- First: $n \to n + 2$ gives $\sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} t^{n}$ .
- Second: $n \to n + 1$ gives $\sum_{n = 1}^{\infty} (n + 1) n a_{n + 1} t^{n}$ ; its $n = 0$ term is zero, so start it at $n = 0$ .
- Third: $n \to n + 1$ gives $\sum_{n = 0}^{\infty} 3 (n + 1) a_{n + 1} t^{n}$ .
Combine. Since $(n + 1) n + 3 (n + 1) = (n + 1) (n + 3)$ : $n = 0 \sum \infty [(n + 2) (n + 1) a_{n + 2} + (n + 1) (n + 3) a_{n + 1} - a_{n}] t^{n} = 0.$
Recurrence. This one links three consecutive coefficients: $a_{n + 2} = \frac{a _{n} - ( n + 1 ) ( n + 3 ) a _{n + 1}}{( n + 2 ) ( n + 1 )}, n \geq 0.$
Compute. With $a_{0} = 1$ , $a_{1} = 0$ : $n = 0 : n = 1 : n = 2 : n = 3 : a_{2} = \frac{1 - 3 \cdot 0}{2} = \frac{1}{2}, a_{3} = \frac{0 - 8 \cdot \frac{1}{2}}{6} = - \frac{2}{3}, a_{4} = \frac{\frac{1}{2} - 15 \cdot ( - \frac{2}{3} )}{12} = \frac{21/2}{12} = \frac{7}{8}, a_{5} = \frac{- \frac{2}{3} - 24 \cdot \frac{7}{8}}{20} = \frac{- 65/3}{20} = - \frac{13}{12} .$
Answer. $y = 1 + \frac{1}{2} (x - 1)^{2} - \frac{2}{3} (x - 1)^{3} + \frac{7}{8} (x - 1)^{4} - \frac{13}{12} (x - 1)^{5} + \dots,$ guaranteed to converge for $0 < x < 2$ .

With a three-term recurrence there is usually no neat formula for $a_{n}$ , and that is fine: the recurrence itself is an exact recipe that a computer (or the solver above, applied to $(1 + x) y^{''} + 3 y^{'} - y = 0$ ) can run as far as needed. The coefficients do not shrink quickly here, a hint that the radius is only $1$ : the singular point $x = 0$ is close. In Section 6.2 we will learn how to build solutions right at that singular point.

Coefficients that are not polynomials

Theorem 6.1.2 only needs $p$ and $q$ to be analytic. If, say, $q (x) = e^{x}$ , replace it by its Taylor series $1 + x + x^{2} /2 + \dots$ and multiply the series for $q$ and $y$ term by term (collect all products that give the same power). The recurrence is then longer, but the computation of the first few coefficients is still straightforward; Exercise 10 is of this type.

Exercises

Write as a single power series in $x$ : $x n = 2 \sum \infty n (n - 1) a_{n} x^{n - 2} - 2 n = 1 \sum \infty n a_{n} x^{n - 1} + n = 0 \sum \infty a_{n} x^{n + 2}$ .
Answer
$- 2 a_{1} - 2 a_{2} x + k = 2 \sum \infty [(k + 1) (k - 2) a_{k + 1} + a_{k - 2}] x^{k}$
For $(x^{2} - 6 x + 13) y^{''} + y^{'} + x y = 0$ , find the singular points and a lower bound for the radius of convergence of power series solutions about $x_{0} = 0$ and about $x_{0} = 3$ .
Answer
Singular points $3 \pm 2 i$ ; $R \geq 13$ about $0$ , $R \geq 2$ about $3$ .
For $x (x^{2} + 9) y^{''} + y^{'} + y = 0$ , find a lower bound for the radius of convergence about $x_{0} = 4$ and about $x_{0} = - 1$ .
Answer
Singular points $0, \pm 3 i$ . About $4$ : $R \geq min (4, 5) = 4$ . About $- 1$ : $R \geq min (1, 10) = 1$ .
Find the recurrence relation and two basic power series solutions about $0$ of $y^{''} - 2 x y^{'} + 4 y = 0$ . Show that one of them is a polynomial.
Answer
$a_{n + 2} = \frac{2 ( n - 2 )}{( n + 2 ) ( n + 1 )} a_{n}$ ; $y_{1} = 1 - 2 x^{2}$ (the series stops), $y_{2} = x - \frac{x ^{3}}{3} - \frac{x ^{5}}{30} - \frac{x ^{7}}{210} - \dots$ ; converge for all $x$ .
Solve $y^{''} + x^{2} y = 0$ by power series about $0$ (give the terms through $x^{9}$ ).
Answer
$a_{2} = a_{3} = 0$ , $a_{n + 2} = - \frac{a _{n - 2}}{( n + 2 ) ( n + 1 )}$ for $n \geq 2$ ; $y = a_{0} (1 - \frac{x ^{4}}{12} + \frac{x ^{8}}{672} - \dots) + a_{1} (x - \frac{x ^{5}}{20} + \frac{x ^{9}}{1440} - \dots)$ , valid for all $x$ .
Solve $y^{''} - 3 x y^{'} - 3 y = 0$ about $0$ and identify the even solution in closed form.
Answer
$a_{n + 2} = \frac{3}{n + 2} a_{n}$ ; $y = a_{0} e^{3 x^{2} /2} + a_{1} (x + x^{3} + \frac{3}{5} x^{5} + \frac{9}{35} x^{7} + \dots)$ .
Solve $(x^{2} - 4) y^{''} + 3 x y^{'} + y = 0$ about $0$ . What is the guaranteed radius of convergence?
Answer
$a_{n + 2} = \frac{( n + 1 )}{4 ( n + 2 )} a_{n}$ ; $y = a_{0} (1 + \frac{x ^{2}}{8} + \frac{3 x ^{4}}{128} + \dots) + a_{1} (x + \frac{x ^{3}}{6} + \frac{x ^{5}}{30} + \dots)$ , $R \geq 2$ . (In fact $y_{1} = 2/ 4 - x^{2}$ and $y_{2} = 4 arcsin (x /2) / 4 - x^{2}$ .)
Solve the initial-value problem $y^{''} + x y^{'} + 3 y = 0$ , $y (0) = 2$ , $y^{'} (0) = 0$ (terms through $x^{6}$ ).
Answer
$a_{n + 2} = - \frac{( n + 3 )}{( n + 2 ) ( n + 1 )} a_{n}$ ; $y = 2 - 3 x^{2} + \frac{5}{4} x^{4} - \frac{7}{24} x^{6} + \dots$
Solve $x y^{''} + y = 0$ , $y (1) = 0$ , $y^{'} (1) = 1$ , in powers of $x - 1$ (terms through $(x - 1)^{5}$ ). On which interval is convergence guaranteed?
Answer
$y = (x - 1) - \frac{( x - 1 ) ^{3}}{6} + \frac{( x - 1 ) ^{4}}{12} - \frac{( x - 1 ) ^{5}}{24} + \dots$ ; guaranteed on $0 < x < 2$ .
Find the terms through $x^{5}$ of the solution of $y^{''} + e^{x} y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ .
Answer
$y = 1 - \frac{x ^{2}}{2} - \frac{x ^{3}}{6} + 0 \cdot x^{4} + \frac{x ^{5}}{40} + \dots$