series solutions of second order linear equations

1

CHA P T E R

5

Series Solutions of Second Order

Linear Equations

5.1

1. Apply the ratio test:

limn→∞

∣∣(x− 3)n+1∣∣

|(x− 3)n|= lim

n→∞|x− 3| = |x− 3| .

Hence the series converges absolutely for |x− 3| < 1 . The radius of convergenceis ρ = 1 . The series diverges for x = 2 and x = 4, since the n-th term does notapproach zero.

2. Use the ratio test:

limn→∞

∣∣(n+ 1)xn+1/2n+1∣∣

|nxn/2n|= limn→∞

n+ 1

n

1

2|x| = |x|

2.

Therefore the series converges absolutely for |x| < 2. For x = 2 and x = −2 the nth

term does not approach zero as n→∞ so the series diverge. Hence the radius ofconvergence is ρ = 2.

3. Applying the ratio test,

limn→∞

∣∣n!x2n+2∣∣

|(n+ 1)!x2n|= lim

n→∞

x2

n+ 1= 0 .

2 Chapter 5. Series Solutions of Second Order Linear Equations

The series converges absolutely for all values of x. Thus the radius of convergenceis ρ =∞ .


limn→∞

∣∣2n+1xn+1∣∣

|2nxn|= lim

n→∞2 |x| = 2 |x| .

Hence the series converges absolutely for 2 |x| < 1, or |x| < 1/2 . The radius ofconvergence is ρ = 1/2 . The series diverges for x = ±1/2 , since the n-th termdoes not approach zero.

5. Applying the ratio test,

limn→∞

∣∣n(x− x0)n+1∣∣

|(n+ 1)(x− x0)n|= lim

n→∞

n

n+ 1|(x− x0)| = |(x− x0)| .

Hence the series converges absolutely for |(x− x0)| < 1 . The radius of convergenceis ρ = 1 . At x = x0 + 1 , we obtain the harmonic series, which is divergent. At theother endpoint, x = x0 − 1 , we obtain

∞∑n=1

(−1)n

n,

which is conditionally convergent.

6. Apply the ratio test :

limn→∞

∣∣3n(n+ 1)2(x+ 2)n+1∣∣

|3n+1n2(x+ 2)n|= lim

n→∞

(n+ 1)2

3n2|(x+ 2)| = 1

3|(x+ 2)| .

Hence the series converges absolutely for 13 |x+ 2| < 1, or |x+ 2| < 3 . The radius

of convergence is ρ = 3 . At x = −5 and x = +1 , the series diverges, since the n-thterm does not approach zero.

7. For this problem f(x) = sinx, so f ′(x) = cosx, f ′′(x) = − sinx, f ′′′(x) = − cosx,f ′′′′(x) = sinx . . ., and thus f(0) = 0, f ′(0) = 1, f ′′(0) = 0, f ′′′(0) = −1 . . .. Theeven terms in the series vanish and the odd terms alternate in sign. We obtain thatsinx =

∑∞n=0(−1)nx2n+1/(2n+ 1)!. Also,

limn→∞

∣∣(−1)n+1x2n+3/(2n+ 3)!∣∣

|(−1)nx2n+1/(2n+ 1)!|= lim

n→∞x2

1

(2n+ 3)(2n+ 2)= 0,

so the series converges for all x and ρ =∞.

8. We have f(x) = ex, with f (n)(x) = ex, for n = 1, 2, . . . . Therefore f (n)(0) = 1.Hence the Taylor expansion about x0 = 0 is

ex =

∞∑n=0

xn

n!.

5.1 3

Applying the ratio test,

limn→∞

∣∣n!xn+1∣∣

|(n+ 1)!xn|= lim

n→∞

1

n+ 1|x| = 0.

The radius of convergence is ρ =∞ .

9. We have f(x) = x , with f ′(x) = 1 and f (n)(x) = 0 , for n = 2, . . . . Clearly,f(1) = 1 and f ′(1) = 1 , with all other derivatives equal to zero. Hence the Taylorexpansion about x0 = 1 is

x = 1 + (x− 1).

Since the series has only a finite number of terms, it converges absolutely for all x.

10. For this problem f(x) = x2. Hence f ′(x) = 2x, f ′′(x) = 2, and f (n)(x) = 0 for

n > 2. Then f(−1)=1, f ′(−1) = −2, f ′′(−1) = 2 and x2 =1−2(x+1)+2(x+1)2/2! = 1− 2(x+1) + (x+1)2. Since the series terminates after a finite number ofterms, it converges for all x. Thus ρ =∞.

11. For this problem f(x) = lnx. Hence f ′(x) = 1/x, f ′′(x) = −1/x2, f ′′(x) = 1 ·2/x3 . . ., and f (n)(x) = (−1)n+1(n− 1)!/xn. Then f(1) = 0, f ′(1) = 1, f ′′(1) = −1,f ′′′(1) = 1 · 2 . . ., f (n)(1) = (−1)n+1(n− 1)!. The Taylor series is

lnx = (x− 1)− (x− 1)2/2 + (x− 1)3/3− . . . =

∞∑n=1

(−1)n+1 (x− 1)n

n.

It follows from the ratio test that the series converges absolutely for |x− 1| < 1.Thus ρ = 1. The series diverges at x = 0 and converges at x = 2.

12. We have f(x) = 1/(1− x), f ′(x) = 1/(1− x)2, f ′′(x) = 2/(1− x)3, . . . withf (n)(x) = n!/(1− x)n+1, for n ≥ 1 . It follows that f (n)(0) = n!, for n ≥ 0 . Hencethe Taylor expansion about x0 = 0 is

1

1− x=

∞∑n=0

xn.


limn→∞

∣∣xn+1∣∣

|xn|= lim

n→∞|x| = |x| .

The series converges absolutely for |x| < 1 , but diverges at x = ± 1 .

13. We have f(x) = 1/(1− x), f ′(x) = 1/(1− x)2, f ′′(x) = 2/(1− x)3, . . . withf (n)(x) = n!/(1− x)n+1, for n ≥ 1 . It follows that f (n)(2) = (−1)n+1n! for n ≥ 0 .Hence the Taylor expansion about x0 = 2 is

1

1− x= −

∞∑n=0

(−1)n(x− 2)n.



limn→∞

∣∣(x− 2)n+1∣∣

| (x− 2)n|= lim

n→∞|x− 2| = |x− 2| .

The series converges absolutely for |x− 2| < 1 , but diverges at x = 1 and x = 3 .

14.(a,b,c) Applying the ratio test,

limn→∞

∣∣(n+ 1)xn+1∣∣

|nxn|= lim

n→∞

n+ 1

n|x| = |x| .

The series converges absolutely for |x| < 1 . Term-by-term differentiation results in

y ′ =

∞∑n=1

n2xn−1 = 1 + 4x+ 9x2 + 16x3 + . . .

y ′′ =

∞∑n=2

n2(n− 1)xn−2 = 4 + 18x+ 48x2 + 100x3 + . . .

Shifting the indices, we can also write

y ′ =

∞∑n=0

(n+ 1)2xn and y ′′ =

∞∑n=0

(n+ 2)2(n+ 1)xn.

15.(a,b,c) We have y=a0+a1x+a2x2+. . .+anx

n+. . ., so y′=a1 + 2a2x+3a3x2+

. . .+ (n+ 1)an+1xn + . . ., and then we get y′′ = 2a2 + 3 · 2a3x+ 4 · 3a4x2 + . . .+

(n+ 2)(n+ 1)an+2xn + . . .. If y′′ = y, we equate coefficients of like powers of x to

obtain 2a2 = a0, 3 · 2a3 = a1, 4 · 3a4 = a2, . . . (n+ 2)(n+ 1)an+2 = an. Thus a2 =a0/2, a3 = a1/6, a4 = a2/(4 · 3) = a0/4!, . . . an+2 = an/((n+2)(n+1)). These yieldthe desired results for n = 0, 1, 2, 3, . . .

16. Set m = n− 1 on the right hand side of the equation. Then n = m+ 1 and

when n = 1, m = 0. Thus the right hand side becomes∑∞m=0 am(x− 1)m+1, which

is the same as the left hand side when m is replaced by n.

17. Shifting the index in the second series, that is, setting n = k + 1 ,

∞∑k=0

akxk+1 =

∞∑n=1

an−1xn .

Hence∞∑k=0

ak+1xk +

∞∑k=0

akxk+1 =

∞∑k=0

ak+1xk +

∞∑k=1

ak−1xk

= a1 +

∞∑k=1

(ak+1 + ak−1)xk+1.

5.1 5

18. Shifting the index by 2 , that is, setting m = n− 2 ,

∞∑n=2

n(n− 1)anxn−2 =

∞∑m=0

(m+ 2)(m+ 1)am+2 xm

=

∞∑n=0

(n+ 2)(n+ 1)an+2 xn .

19. Multiplying each term of the first series by x yields

x

∞∑n=1

nanxn−1 =

∞∑n=1

nanxn =

∞∑n=0

nanxn,

where the last equality comes from nan = 0 for n = 0. Changing the index from kto n in the second series yields

∞∑n=0

nanxn +

∞∑n=0

anxn =

∞∑n=0

(n+ 1)anxn.

20.

∞∑m=2

m(m− 1)amxm−2 + x

∞∑k=1

kakxk−1 =

∞∑n=0

(n+ 2)(n+ 1)an+2xn +

∞∑k=1

kakxk =

∞∑n=0

[(n+ 2)(n+ 1)an+2 + nan]xn.

In the first case we have let n = m− 2 in the first summation and multiplied eachterm of the second summation by x. In the second case we have let n = k andnoted that for n = 0, nan = 0.

21. Clearly,

∞∑n=1

nan xn−1 + x

∞∑n=0

an xn =

∞∑n=1

nan xn−1 +

∞∑n=0

an xn+1.

Shifting the index in the first series, that is, setting k = n− 1 ,

∞∑n=1

nan xn−1 =

∞∑k=0

(k + 1)ak+1xk.

Shifting the index in the second series, that is, setting k = n+ 1 ,

∞∑n=0

an xn+1 =

∞∑k=1

ak−1xk.

Combining the series, and starting the summation at n = 1 ,

∞∑n=1

nan xn−1 + x

∞∑n=0

an xn = a1 +

∞∑n=1

[(n+ 1)an+1 + an−1]xn.


22. Shifting the index in the first series, that is, setting k = n− 2 ,

∞∑n=2

n(n− 1)an xn−2 =

∞∑k=0

(k + 2)(k + 1)ak+2 xk

Combining the series and starting the summation at n = 0,

∞∑n=2

n(n− 1)an xn−2 +

∞∑n=0

an xn =

∞∑n=0

[(n+ 2)(n+ 1)an+2 + an]xn.

23. If we shift the index of summation in the first sum by letting m = n− 1, wehave

∞∑n=1

nanxn−1 =

∞∑m=0

(m+ 1)am+1xm.

Substituting this into the given equation and letting m = n again, we obtain

∞∑n=0

(n+ 1)an+1xn + 2

∞∑n=0

anxn = 0, or

∞∑n=0

[(n+ 1)an+1 + 2an]xn = 0.

Hence an+1 = −2an/(n+ 1) for n = 0, 1, 2, 3, .... Thus a1 = −2a0, a2 = −2a1/2 =22a0/2, a3 = −2a2/3 = −23a0/2 · 3 = −23a0/3!... and an = (−1)n2na0/n!. Noticethat for n = 0 this formula reduces to a0, so we can write

∞∑n=0

anxn =

∞∑n=0

(−1)n2na0xn/n! = a0

∞∑n=0

(−2x)n/n! = a0e−2x.

5.2

1.(a,b,d) Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Then

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.

Substitution into the ODE results in∞∑n=0

(n+ 2)(n+ 1)an+2 xn −

∞∑n=0

anxn = 0

or∞∑n=0

[(n+ 2)(n+ 1)an+2 − an]xn = 0 .

Equating all the coefficients to zero,

(n+ 2)(n+ 1)an+2 − an = 0 , n = 0, 1, 2, . . . .

We obtain the recurrence relation

an+2 =an

(n+ 1)(n+ 2), n = 0, 1, 2, . . . .

5.2 7

The subscripts differ by two, so for k = 1, 2, . . .

a2k =a2k−2

(2k − 1)2k=

a2k−4(2k − 3)(2k − 2)(2k − 1)2k

= . . . =a0

(2k)!

and

a2k+1 =a2k−1

2k(2k + 1)=

a2k−3(2k − 2)(2k − 1)2k(2k + 1)

= . . . =a1

(2k + 1)!.

Hence

y = a0

∞∑k=0

x2k

(2k)!+ a1

∞∑k=0

x2k+1

(2k + 1)!.

The linearly independent solutions are

y1 = 1 +x2

2!+x4

4!+x6

6!+ . . . = cosh x

y2 = x+x3

3!+x5

5!+x7

7!+ . . . = sinh x .

(c) The Wronskian at 0 is 1.

2.(a) Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1 xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


[(n+ 2)(n+ 1)an+2 + 3(n+ 1)an+1] xn = 0 .

Setting the coefficients equal to zero, we have

(n+ 2)(n+ 1)an+2 + 3(n+ 1)an+1 = (n+ 1) [(n+ 2)an+2 + 3an+1] = 0

The recurrence relation can be written as

an+2 = − 3

n+ 2an+1 for n = 0, 1, 2, . . .

(b) If we set a0 = 1 and a1 = 0, we find that all other coefficients are 0 so y1 = 1is a solution to the differential equation. If we set a0 = 0 and a1 = 1, we find thata2 = −3/2, a3 = 3/2, and a4 = −9/8. Alternatively, we could note that if a0 = 0and a1 = 1, then an = (−3)n−1/n! for n = 1, 2, 3, . . . , so

y2 = x− 3

2!x2 +

32

3!x3 − 33

4!x4 + · · ·

is a solution to the differential equation.



(d) Combining y1 and y2 gives

K + x− 3

2!x2 +

32

3!x3 − 33

4!x4 + · · ·

as a general solution to the differential equation.

3.(a) y =∑∞n=0 anx

n; y′ =∑∞n=1 nanx

n−1 and since we must multiply y′ by x in

the D.E. we do not shift the index; and y′′ =∑∞n=2 n(n− 1)anx

n−2 =∑∞n=0(n+2)

(n+ 1)an+2xn. Substituting in the D.E., we obtain

∞∑n=0

(n+ 2)(n+ 1)an+2xn −

∞∑n=1

nanxn −

∞∑n=0

anxn = 0.

In order to have the starting point the same in all three summations, we let n = 0in the first and third terms to obtain the following:

(2 · 1a2 − a0)x0 +

∞∑n=1

[(n+ 2)(n+ 1)an+2 − (n+ 1)an]xn = 0.

Thus an+2 = an/(n+ 2) for n = 1, 2, 3, . . . Note that the recurrence relation is alsocorrect for n = 0.

(b) From the recurrence relation we have a2 = a0/2, a4 = a2/4 = a0/2 · 4, a6 =a4/6 = a0/2 · 4 · 6, so y1 = 1 + x2/2 + x4/2 · 4 + x6/2 · 4 · 6 + . . ., and a3 = a1/3,a5 = a3/5 = a1/3 · 5, a7 = a5/7 = a1/3 · 5 · 7, so y2 = x+ x3/3 + x5/3 · 5 + x7/3 ·5 · 7 + . . .

(c) W (y1, y2)(0) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 and thus y1, y2 form a fundamental set of solutions.

(d) From part (b) we see the even coefficients can be written as a2m = a0/2mm!. For

the odd coefficients notice that a3 = 2a1/(2 · 3) = 2a1/3!, that a5 = 2 · 4a1/(2 · 3 ·4 · 5) = 22 · 2a1/5!, and that a7 = 2 · 4 · 6a1/(2 · 3 · 4 · 5 · 6 · 7) = 23 · 3!a1/7!. Like-wise a9 = a7/9 = 23 · 3!a1/(7!)9 = 23 · 3!8a1/9! = 24 · 4!a1/9!. Continuing, we havea2m+1 = 2mm!a1/(2m+ 1)!. Thus

y = a0

∞∑m=0

x2m

2mm!+ a1

∞∑m=0

2mm!x2m+1

(2m+ 1)!.

4.(a)

y =

∞∑n=0

an(x− 1)n; y′ =

∞∑n=1

nan(x− 1)n−1 =

∞∑n=0

(n+ 1)an+1(x− 1)n,

and

y′′ =

∞∑n=2

n(n− 1)an(x− 1)n−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2(x− 1)n.

5.2 9

Substituting in the differential equation and setting x = 1 + (x− 1), we obtain

∞∑n=0

(n+ 2)(n+ 1)an+2(x− 1)n −∞∑n=0

(n+ 1)an+1(x− 1)n −∞∑n=1

nan(x− 1)n

−∞∑n=0

an(x− 1)n = 0,

where the third sum comes from

−(x− 1)y′ = −∞∑n=0

(n+ 1)an+1(x− 1)n+1 = −∞∑n=1

nan(x− 1)n.

Letting n = 0 in the first, second and fourth sums we obtain

(2 · 1 · a2 − 1 · a1 − a0)(x− 1)0+

+

∞∑n=1

[(n+ 2)(n+ 1)an+2 − (n+ 1)an+1 − (n+ 1)an] (x− 1)n = 0.

Setting the terms in the square brackets equal to zero and dividing by (n+ 1) givesus that (n+ 2)an+2 − an+1 − an = 0 for n = 1, 2, 3, . . . (which also holds for n = 0).This recurrence relation can be used to solve for a2 in terms of a0 and a1, then fora3 in terms of a1 and a2 and so on.

(b) In many cases it is easier to first take a0 = 0 and generate one solution and thentake a1 = 0 and generate a second solution. Thus, choosing a0 = 0, we find thata2 = a1/2, a3 = (a2 + a1)/3 = a1/2, a4 = (a3 + a2)/4 = a1/4, a5 = (a4 + a3)/5 =3a1/20, . . .. This yields the solution y2(x) = (x− 1) + (x− 1)2/2 + (x− 1)3/2 +(x− 1)4/4 + . . .. The second solution may be obtained by choosing a1 = 0. Thena2 = a0/2, a3 = (a2 + a1)/3 = a0/6, a4 = (a3 + a2)/4 = a0/6, a5 = (a4 + a3)/5 =a0/15, . . .. This yields the solution y1(x) = 1 + (x− 1)2/2 + (x− 1)3/6+(x−1)4/6 + (x− 1)5/15 + . . ..

(c) W (y1, y2)(1) =

∣∣∣∣ 1 00 1

∣∣∣∣ = 1 and thus y1 and y2 form a fundamental set of

solutions.

(d) A general term is not easily found in this case.


n + . . . . Then

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


(n+ 2)(n+ 1)an+2 xn + k2x2

∞∑n=0

anxn = 0 .

Rewriting the second summation,

∞∑n=0

(n+ 2)(n+ 1)an+2 xn +

∞∑n=2

k2an−2 xn = 0 ,


that is,

2a2 + 3 · 2 a3x+

∞∑n=2

[(n+ 2)(n+ 1)an+2 + k2an−2

]xn = 0 .

Setting the coefficients equal to zero, we have a2 = 0 , a3 = 0 , and

(n+ 2)(n+ 1)an+2 + k2an−2 = 0 , for n = 2, 3, 4, . . . .

The recurrence relation can be written as

an+2 = − k2an−2(n+ 2)(n+ 1)

, n = 2, 3, 4, . . . .

The indices differ by four, so a4 , a8 , a12 , . . . are defined by

a4 = −k2a0

4 · 3, a8 = −k

2a48 · 7

, a12 = − k2a812 · 11

, . . . .

Similarly, a5 , a9 , a13 , . . . are defined by

a5 = −k2a1

5 · 4, a9 = −k

2a59 · 8

, a13 = − k2a913 · 12

, . . . .

The remaining coefficients are zero. Therefore the general solution is

y = a0

[1− k2

4 · 3x4 +

k4

8 · 7 · 4 · 3x8 − k6

12 · 11 · 8 · 7 · 4 · 3x12 + . . .

]+

+ a1

[x− k2

5 · 4x5 +

k4

9 · 8 · 5 · 4x9 − k6

13 · 12 · 9 · 8 · 4 · 4x13 + . . .

].

Note that for the even coefficients,

a4m = − k2a4m−4(4m− 1)4m

, m = 1, 2, 3, . . .

and for the odd coefficients,

a4m+1 = − k2a4m−34m(4m+ 1)

, m = 1, 2, 3, . . . .

Hence the linearly independent solutions are

y1(x) = 1 +

∞∑m=0

(−1)m+1(k2x4)m+1

3 · 4 · 7 · 8 . . . (4m+ 3)(4m+ 4)

y2(x) = x

[1 +

∞∑m=0

(−1)m+1(k2x4)m+1

4 · 5 · 8 · 9 . . . (4m+ 4)(4m+ 5)

].


6.

y =

∞∑n=0

anxn; y′ =

∞∑n=1

nanxn−1; y′′ =

∞∑n=2

n(n− 1)anxn−2.

Substituting in the differential equation and shifting the index in both summationsfor y′′ gives

5.2 11

∞∑n=0

(n+ 2)(n+ 1)an+2xn −

∞∑n=1

(n+ 1)nan+1xn +

∞∑n=0

anxn = (2 · 1 · a2 + a0)x0

+

∞∑n=1

[(n+ 2)(n+ 1)an+2 − (n+ 1)nan+1 + an]xn = 0.

Thus a2 = −a0/2 and an+2 = nan+1/(n+ 2)− an/(n+2)(n+1), n = 1, 2, . . .. Choo-sing a0 = 0 yields a2 = 0, a3 = −a1/6, a4 = 2a3/4 = −a1/12, a5 = 3a4/5− a3/20 =−a1/24, . . ., and hence y2(x) = a1(x− x3/6− x4/12− x5/24 + . . .). A second lin-early independent solution is obtained by choosing a1 = 0. Then a2 = −a0/2,a3 = a2/3 = −a0/6, a4 = 2a3/4− a2/12 = −a0/24, . . ., which gives y1(x) = a0(1−x2/2− x3/6− x4/24 + . . .).


n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


(n+ 2)(n+ 1)an+2 xn + x

∞∑n=0

(n+ 1)an+1xn + 2

∞∑n=0

anxn = 0 .

First write

x

∞∑n=0

(n+ 1)an+1xn =

∞∑n=1

nanxn.

We then obtain

2a2 + 2a0 +

∞∑n=1

[(n+ 2)(n+ 1)an+2 + nan + 2an]xn = 0 .

It follows that a2 = −a0 and an+2 = −an/(n+ 1) , n = 0, 1, 2, . . . . Note that theindices differ by two, so for k = 1, 2, . . .

a2k = − a2k−22k − 1

=a2k−4

(2k − 3)(2k − 1)= . . . =

(−1)ka01 · 3 · 5 . . . (2k − 1)

and

a2k+1 = −a2k−12k

=a2k−3

(2k − 2)2k= . . . =

(−1)ka12 · 4 · 6 . . . (2k)

.


y1(x) = 1− x2

1+

x4

1 · 3− x6

1 · 3 · 5+ . . . = 1 +

∞∑n=1

(−1)nx2n

1 · 3 · 5 . . . (2n− 1)

y2(x) = x− x3

2+

x5

2 · 4− x7

2 · 4 · 6+ . . . = x+

∞∑n=1

(−1)nx2n+1

2 · 4 · 6 . . . (2n).



8. If y =∑∞n=0 an(x− 1)n, then

xy = [1 + (x− 1)]y =

∞∑n=0

an(x− 1)n +

∞∑n=0

an(x− 1)n+1,

y′ =

∞∑n=1

nan(x− 1)n−1, and

xy′′ = [1 + (x− 1)]y′′ =

∞∑n=2

n(n− 1)an(x− 1)n−2 +

∞∑n=2

n(n− 1)an(x− 1)n−1.


n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.

Substitution into the ODE results in

(3− x2)

∞∑n=0

(n+ 2)(n+ 1)an+2 xn − 3x

∞∑n=0

(n+ 1)an+1xn −

∞∑n=0

anxn = 0 .

Before proceeding, write

x2∞∑n=0

(n+ 2)(n+ 1)an+2 xn =

∞∑n=2

n(n− 1)anxn

and

x

∞∑n=0

(n+ 1)an+1xn =

∞∑n=1

nanxn.

It follows that

6a2 − a0 + (−4a1 + 18a3)x+

+

∞∑n=2

[3(n+ 2)(n+ 1)an+2 − n(n− 1)an − 3nan − an]xn = 0.

We obtain a2 = a0/6 , 2a3 = a1/9 , and

3(n+ 2)an+2 = (n+ 1)an , n = 0, 1, 2, . . . .

The indices differ by two, so for k = 1, 2, . . .

a2k =(2k − 1)a2k−2

3(2k)=

(2k − 3)(2k − 1)a2k−432(2k − 2)(2k)

= . . . =3 · 5 . . . (2k − 1) a0

3k · 2 · 4 . . . (2k)

and

a2k+1 =(2k)a2k−13(2k + 1)

=(2k − 2)(2k)a2k−332(2k − 1)(2k + 1)

= . . . =2 · 4 · 6 . . . (2k) a1

3k · 3 · 5 . . . (2k + 1).

5.2 13


y1(x) = 1 +x2

6+x4

24+

5x6

432+ . . . = 1 +

∞∑n=1

3 · 5 . . . (2n− 1)x2n

3n · 2 · 4 . . . (2n)

y2(x) = x+2x3

9+

8x5

135+

16x7

945+ . . . = x+

∞∑n=1

2 · 4 · 6 . . . (2n)x2n+1

3n · 3 · 5 . . . (2n+ 1).



n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


2

∞∑n=0

(n+ 2)(n+ 1)an+2 xn + x

∞∑n=0

(n+ 1)an+1xn + 3

∞∑n=0

anxn = 0 .

First write

x

∞∑n=0

(n+ 1)an+1xn =

∞∑n=1

nanxn.

We then obtain

4a2 + 3a0 +

∞∑n=1

[2(n+ 2)(n+ 1)an+2 + nan + 3an]xn = 0 .

It follows that a2 = −3a0/4 and

2(n+ 2)(n+ 1)an+2 + (n+ 3)an = 0

for n = 0, 1, 2, . . . . The indices differ by two, so for k = 1, 2, . . .

a2k = − (2k + 1)a2k−22(2k − 1)(2k)

=(2k − 1)(2k + 1)a2k−4

22(2k − 3)(2k − 2)(2k − 1)(2k)= . . .

=(−1)k3 · 5 . . . (2k + 1)

2k (2k)!a0 .

and

a2k+1 = − (2k + 2)a2k−12(2k)(2k + 1)

=(2k)(2k + 2)a2k−3

22(2k − 2)(2k − 1)(2k)(2k + 1)= . . .

=(−1)k 4 · 6 . . . (2k)(2k + 2)

2k (2k + 1)!a1 .



y1(x) = 1− 3

4x2 +

5

32x4 − 7

384x6 + . . . =

∞∑n=0

(−1)n3 · 5 . . . (2n+ 1)

2n (2n)!x2n

y2(x) = x− 1

3x3 +

1

20x5 − 1

210x7 + . . . = x+

∞∑n=1

(−1)n4 · 6 . . . (2n+ 2)

2n (2n+ 1)!x2n+1.


11. We need to rewrite x+ 1 as 3 + (x− 2) in order to multiply x+ 1 times y′ asa power series about x0 = 2.

12.(a) From Problem 3, we have

y1(x) =

∞∑n=0

x2n

2n n!and y2(x) =

∞∑n=0

2n n!x2n+1

(2n+ 1)!.

Since a0 = y(0) and a1 = y ′(0) , we have y(x) = 2 y1(x) + y2(x) . That is,

y(x) = 2 + x+ x2 +1

3x3 +

1

4x4 +

1

15x5 +

1

24x6 + . . . .

The four- and five-term polynomial approximations are

p4 = 2 + x+ x2 + x3/3, and p5 = 2 + x+ x2 + x3/3 + x4/4 .

(b)

(c) The four-term approximation p4 appears to be reasonably accurate (within 10%)on the interval |x| < 0.7 .

13.(a) From Problem 7, the linearly independent solutions are

y1(x) = 1 +

∞∑n=1

(−1)nx2n

1 · 3 · 5 . . . (2n− 1)and y2(x) = x+

∞∑n=1

(−1)nx2n+1

2 · 4 · 6 . . . (2n).

5.2 15

Since a0 = y(0) and a1 = y ′(0) , we have y(x) = 4 y1(x)− y2(x) . That is,

y(x) = 4− x− 4x2 +1

2x3 +

4

3x4 − 1

8x5 − 4

15x6 + . . . .

The four- and five-term polynomial approximations are

p4 = 4− x− 4x2 +1

2x3, and p5 = 4− x− 4x2 +

1

2x3 +

4

3x4.

(b)

(c) The four-term approximation p4 appears to be reasonably accurate (within 10%)on the interval |x| < 0.5 .

15. Letting t = x− 1 yields (x− 1)2 = t2 and (x2 − 1) = t2 + 2t. Now let u(t) =y(t+ 1) and hence u′ = y′ and u′′ = y′′. Thus the differential equation transformsinto u′′(t) + t2u′(t) + (t2 + 2t)u(t) = 0. Assuming that u(t) =

∑∞n=0 ant

n, we haveu′(t) =

∑∞n=1 nant

n−1 and u′′(t) =∑∞n=2 n(n− 1)ant

n−2. Substituting in the dif-ferential equation and shifting indices yields

∞∑n=0

(n+ 2)(n+ 1)an+2tn +

∞∑n=2

(n− 1)an−1tn +

∞∑n=2

an−2tn +

∞∑n=1

2an−1tn = 0,

2 · 1 · a2t0 + (3 · 2 · a3 + 2 · a0)t1+

+

∞∑n=2

[(n+ 2)(n+ 1)an+2 + (n+ 1)an−1 + an−2] tn = 0.

It follows that a2 = 0, a3 = −a0/3, and an+2 = −an−1/(n+ 2)− an−2/(n+ 2)(n+1), n = 2, 3, 4, . . .. We obtain one solution by choosing a1 = 0. Then a4 = −a0/12,a5 = −a2/5− a1/20 = 0, a6 = −a3/6− a2/30 = a0/18, . . .. Thus one solution isu1(t) = a0(1− t3/3− t4/12 + t6/18 + . . .), so y1(x) = u1(x− 1) = 1− (x− 1)3/3−(x− 1)4/12 + (x− 1)6/18 + . . .. We obtain a second solution by choosing a0 = 0.Then a4 = −a1/4, a5 = −a2/5− a1/20 = −a1/20, a6 = −a3/6− a2/30 = 0, a7 =−a4/7− a3/42 = a1/28, . . .. Thus u2(t) = t− t4/4− t5/20 + t7/28 + . . ., or

y2(x) = u2(x− 1) = (x− 1)− (x− 1)4/4− (x− 1)5/20 + (x− 1)7/28 + . . . .


The Taylor series for x2 − 1 about x = 1 may be obtained by writing x = 1 + (x− 1)so x2 = 1 + 2(x− 1) + (x− 1)2 and x2 − 1 = 2(x− 1) + (x− 1)2. The differen-tial equation now appears as y′′ + (x− 1)2y′ +

[(x− 1)2 + 2(x− 1)

]y = 0, which

is identical to the transformed equation with t = x− 1.

16. The given recurrence relation is

(n+ 2)(n+ 1)an+2 + an = 0

and the proposed solution is given in Eq.(10) as

a2k+1 =(−1)k

(2k + 1)!a1, for k = 1, 2, 3, . . .

To prove the solution by induction, we first note that for k = 0,

(−1)0

(2(0) + 1)!a1 = a1

so Eq.(10) is satisfied for k = 0. Next we assume that Eq.(10) is true for k = m;that is, we assume that

a2m+1 =(−1)m

(2m+ 1)!a1

and compute a2(m+1)+1 = a2m+3:

a2m+3 =−1

(2m+ 3)(2m+ 2)a2m+1 (from the recurrence relation)

=−1

(2m+ 3)(2m+ 2)· (−1)m

(2m+ 1)!a1 (from the inductive hypothesis)

=(−1)m+1

(2m+ 3)!a1 =

(−1)m+1

(2(m+ 1) + 1)!a1

and Eq.(10) holds for k = m+ 1. Therefore Eq.(10) is true for k = 1, 2, 3, . . . .

17. Two linearly independent solutions of Airy’s equation (about x0 = 0) are

y1(x) = 1 +

∞∑n=1

x3n

2 · 3 . . . (3n− 1)(3n)

y2(x) = x+

∞∑n=1

x3n+1

3 · 4 . . . (3n)(3n+ 1).

Applying the ratio test to the terms of y1(x) ,

limn→∞

∣∣2 · 3 . . . (3n− 1)(3n)x3n+3∣∣

|2 · 3 . . . (3n+ 2)(3n+ 3)x3n|= limn→∞

1

(3n+ 1)(3n+ 2)(3n+ 3)|x|3 = 0 .

5.2 17

Similarly, applying the ratio test to the terms of y2(x) ,

limn→∞

∣∣3 · 4 . . . (3n)(3n+ 1)x3n+4∣∣

|3 · 4 . . . (3n+ 3)(3n+ 4)x3n+1|= limn→∞

1

(3n+ 2)(3n+ 3)(3n+ 4)|x|3 = 0 .

Hence both series converge absolutely for all x .

18. Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


(n+ 2)(n+ 1)an+2 xn − 2x

∞∑n=0

(n+ 1)an+1xn + λ

∞∑n=0

anxn = 0 .

First write

x

∞∑n=0

(n+ 1)an+1xn =

∞∑n=1

nanxn.

We then obtain

2a2 + λ a0 +

∞∑n=1

[(n+ 2)(n+ 1)an+2 − 2nan + λ an]xn = 0 .

Setting the coefficients equal to zero, it follows that

an+2 =(2n− λ)

(n+ 1)(n+ 2)an

for n = 0, 1, 2, . . . . Note that the indices differ by two, so for k = 1, 2, . . .

a2k =(4k − 4− λ)a2k−2

(2k − 1)2k=

(4k − 8− λ)(4k − 4− λ)a2k−4(2k − 3)(2k − 2)(2k − 1)2k

= (−1)kλ . . . (λ− 4k + 8)(λ− 4k + 4)

(2k)!a0 .

and

a2k+1 =(4k − 2− λ)a2k−1

2k(2k + 1)=

(4k − 6− λ)(4k − 2− λ)a2k−3(2k − 2)(2k − 1)2k(2k + 1)

= (−1)k(λ− 2) . . . (λ− 4k + 6)(λ− 4k + 2)

(2k + 1)!a1.

Hence the linearly independent solutions of the Hermite equation (about x0 = 0)are

y1(x) = 1− λ

2!x2 +

λ(λ− 4)

4!x4 − λ(λ− 4)(λ− 8)

6 !x6 + . . .

y2(x) = x− λ− 2

3 !x3 +

(λ− 2)(λ− 6)

5 !x5 − (λ− 2)(λ− 6)(λ− 10)

7 !x7 + . . . .


(b) Based on the recurrence relation

an+2 =(2n− λ)

(n+ 1)(n+ 2)an ,

the series solution will terminate as long as λ is a nonnegative even integer. Ifλ = 2m, then one or the other of the solutions in part (b) will contain at mostm/2 + 1 terms. In particular, we obtain the polynomial solutions corresponding toλ = 0, 2, 4, 6, 8, 10 :

λ = 0 y1(x) = 1

λ = 2 y2(x) = x

λ = 4 y1(x) = 1− 2x2

λ = 6 y2(x) = x− 2x3/3

λ = 8 y1(x) = 1− 4x2 + 4x4/3λ = 10 y2(x) = x− 4x3/3 + 4x5/15

(c) Observe that if λ = 2n , and a0 = a1 = 1 , then

a2k = (−1)k2n . . . (2n− 4k + 8)(2n− 4k + 4)

(2k)!

and

a2k+1 = (−1)k(2n− 2) . . . (2n− 4k + 6)(2n− 4k + 2)

(2k + 1)!.

for k = 1, 2, . . . [n/2]. It follows that the coefficient of xn, in y1 and y2 , is

an =

{(−1)k 4k k!

(2k)! for n = 2k

(−1)k 4k k!(2k+1)! for n = 2k + 1

Then by definition,

Hn(x) =

{(−1)k 2n (2k)!

4k k!y1(x) = (−1)k (2k)!

k! y1(x) for n = 2k

(−1)k 2n (2k+1)!4k k!

y2(x) = (−1)k 2 (2k+1)!k! y2(x) for n = 2k + 1

Therefore the first six Hermite polynomials are

H0(x) = 1

H1(x) = 2x

H2(x) = 4x2 − 2

H3(x) = 8x8 − 12x

H4(x) = 16x4 − 48x2 + 12

H5(x) = 32x5 − 160x3 + 120x

5.2 19

19.(a) Clearly, (sinx)′ = cosx =√

1− sin2 x (for −π/2 ≤ x ≤ π/2) and sin 0 = 0.

(b) y = a0 + a1x+ a2x2 + ... , y2 = a20 + 2a0a1x+

(2a0a2 + a21

)x2 + ... , y′ = a1 +

2a2x+ 3a3x2 + ... , and (y′)2 = a21 + 4a1a2x+

(6a1a3 + 4a22

)x2 + .... Substituting

these into (y′)2 = 1− y2 and collecting coefficients of like powers of x yields(a21 + a20 − 1

)+ (4a1a2 + 2a0a1)x+

(6a1a3 + 4a22 + 2a0a2 + a21

)x2 + ... = 0.

As in the earlier problems, each coefficient must be zero. The I.C. y(0) = 0 requiresthat a0 = 0, and thus a21 + a20− 1 = 0 gives a21 = 1. However, the D.E. indicatesthat y′ is always positive, so y′(0) = a1 > 0 implies a1 = 1. Then 4a1a2 + 2a0a1 = 0implies that a2 = 0; and 6a1a3 + 4a22 + 2a0a2 + a21 = 6a1a3 + a21 = 0 implies thata3 = −1/6. Thus y = x− x3/3! + ... , which are the first two terms of the Taylorseries or sinx.

20. The series solution is given by

y(x) = x− x3

2+

x5

2 · 4− x7

2 · 4 · 6+

x9

2 · 4 · 6 · 8− . . . .


y(x) = x− x3

12− x5

240− x7

2240− x9

16128− . . .

- 2 - 1 1 2

- 2

- 1

1

2



y(x) = 1− x4

12+

x8

672− x12

88704+ . . . .

23. Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Then

y ′ =

∞∑n=1

nanxn−1 =

∞∑n=0

(n+ 1)an+1xn

and

y ′′ =

∞∑n=2

n(n− 1)anxn−2 =

∞∑n=0

(n+ 2)(n+ 1)an+2 xn.


(1− x)

∞∑n=0

(n+ 2)(n+ 1)an+2 xn + x

∞∑n=0

(n+ 1)an+1xn − 2

∞∑n=0

anxn = 0 .

After appropriately shifting the indices, it follows that

2a2 − 2a0 +

∞∑n=1

[(n+ 2)(n+ 1)an+2 − (n+ 1)nan+1 + nan − 2 an]xn = 0.

We find that a2 = a0 and

(n+ 2)(n+ 1)an+2 − (n+ 1)nan+1 + (n− 2)an = 0

for n = 1, 2, . . . . Writing out the individual equations,

3 · 2 a3 − 2 · 1 a2 − a1 = 0

4 · 3 a4 − 3 · 2 a3 = 0

5 · 4 a5 − 4 · 3 a4 + a3 = 0

6 · 5 a6 − 5 · 4 a5 + 2 a4 = 0

...

5.3 21

Since a0 = 0 and a1 = 1 , the remaining coefficients satisfy the equations

3 · 2 a3 − 1 = 0

4 · 3 a4 − 3 · 2 a3 = 0

5 · 4 a5 − 4 · 3 a4 + a3 = 0

6 · 5 a6 − 5 · 4 a5 + 2 a4 = 0

...

That is, a3 = 1/6 , a4 = 1/12 , a5 = 1/24 , a6 = 1/45 , . . . . Hence the series solutionof the initial value problem is

y(x) = x+1

6x3 +

1

12x4 +

1

24x5 +

1

45x6 +

13

1008x7 + . . . .

5.3

1. The differential equation can be solved for y′′ to yield y′′ = −xy′ − y. If y = φ(x)is a solution, then φ′′(x) = −xφ′(x)− φ(x) and thus setting x = 0 we obtain thatφ′′(0) = −0− 1 = −1. Differentiating the equation for y′′ yields y′′′ = −xy′′ − 2y′

and hence setting y = φ(x) again yields φ′′′(0) = −0− 0 = 0. In a similar fash-ion y(4) = −xy′′′ − 3y′′ and thus φ(4)(0) = −0− 3(−1) = 3. The process can becontinued to calculate higher derivatives of φ(x).

2. Let y = φ(x) be a solution of the initial value problem. First write

y ′′ = −1 + x

x2y ′ − 3 ln x

x2y .

Differentiating twice,

y ′′′ =−1

x3[(x+ x2)y ′′ + (3x ln x− x− 2)y ′ + (3− 6 ln x)y

].

y(4) =−1

x4

[(x2 + x3)y ′′′ + (3x2 ln x− 2x2 − 4x)y ′′+

+ (6 + 8x− 12x ln x)y ′ + (18 ln x− 15)y].


Given that φ(1) = 2 and φ ′(1) = 0 , the first equation gives φ ′′(1) = 0 and the lasttwo equations give φ ′′′(1) = −6 and φ(4)(1) = 42 .

3. Let y = φ(x) be a solution of the initial value problem. First note that

y ′′ = −x2 y ′ − (sin x)y .

Differentiating twice,

y ′′′ = −x2 y ′′ − (2x+ sin x)y ′ − (cos x)y

y(4) = −x2 y ′′′ − (4x+ sin x)y ′′ − (2 + 2 cos x)y ′ + (sin x)y .

Given that φ(0) = a0 and φ ′(0) = a1, the first equation gives φ ′′(0) = 0 and thelast two equations give φ ′′′(0) = −a0 and φ(4)(0) = −4a1.

4. Clearly, p(x) = 4 and q(x) = 6x are analytic for all x . Hence the series solutionsconverge everywhere.

5. The zeros of P (x) = x2 − 2x− 3 are x = −1 and x = 3. For x0 = 4, x0 = −4,and x0 = 0 the distance to the nearest zero of P (x) is 1, 3, and 1, respectively. Thusa lower bound for the radius of convergence for series solutions in powers of (x− 4),(x+ 4), and x is ρ = 1, ρ = 3, and ρ = 1, respectively.

6. The zeros of P (x) = 1 + x3 are the three cube roots of −1 . They all lie on theunit circle in the complex plane. So for x0 = 0 , ρ = 1 . For x0 = 2 , the nearestroot is eiπ/3 = (1 + i

√3 )/2 , hence ρ =

√3 .

7.(Problem 1) Since p(x) = −1 and q(x) = 0, the radius of convergence about x0 = 0is ρ =∞.

(Problem 2) Since p(x) = 3 and q(x) = 0, the radius of convergence about x0 = 0is ρ =∞.

(Problem 3) Since p(x) = −x and q(x)=−1 the radius of convergence about x0 =0is ρ =∞.

(Problem 4) Since p(x) = −x and q(x)=−1 the radius of convergence about x0 =0is ρ =∞.

(Problem 5) Since p(x) = 0 and q(x) = kx2, the radius of convergence about x0 = 0is ρ =∞.

(Problem 6) Since p(x) = 0 and q(x) = 11−x , the radius of convergence about x0 = 0

is ρ = 1.

(Problem 7) Since p(x) = x and q(x) = 2, the radius of convergence about x0 = 0is ρ =∞.

(Problem 8) Since p(x) = − 1x and q(x) = 1, the radius of convergence about x0 = 1

is ρ = 1.

5.3 23

(Problem 9) Since p(x) = − 3x3−x2 and q(x) = − 1

3−x2 , the radius of convergence

about x0 = 0 is ρ =√

3.

(Problem 10) Since p(x) = x2 and q(x) = 3

2 , the radius of convergence about x0 = 1is ρ =∞.

(Problem 11) Since p(x) = x+12 and q(x) = − 3

2 , the radius of convergence aboutx0 = 0 is ρ =∞.

8.(a) If we assume that y =∑∞n=0 anx

n, then y′ =∑∞n=1 nanx

n−1 and also y′′ =∑∞n=2 n(n− 1)anx

n−2. Substituting in the D.E. gives

∞∑n=2

n(n− 1)anxn−2 −

∞∑n=2

n(n− 1)anxn −

∞∑n=1

nanxn + α2

∞∑n=0

anxn = 0.

Shifting indices of summation and collecting coefficients of like powers of x yieldsthe equation (

2 · 1 · a2 + α2a0)x0 +

[3 · 2 · a3 + (α2 − 1)a1

]x1+

+

∞∑n=2

[(n+ 2)(n+ 1)an+2 + (α2 − n2)an

]xn = 0.

Hence the recurrence relation is an+2 = (n2 − α2)an/(n+ 2)(n+ 1), n = 0, 1, 2, . . ..For the first solution we choose a1 = 0. We find that a2 = −α2a0/2 · 1, a3 =0, a4 = (22 − α2)a2/4 · 3 = −(22 − α2)α2a0/4!, . . ., and then by induction a2m =−[(2m− 2)2 − α2

]. . . (22 − α2)α2a0/(2m)!, and a2m+1 = 0, so

y1(x)=1− α2

2!x2− (22−α2)α2

4!x4−. . .− [(2m− 2)2−α2] . . . (22−α2)α2

(2m)!x2m−. . . ,

where we have set a0 = 1. For the second solution we take a0 = 0 and a1 = 1 inthe recurrence relation to obtain the desired solution.

(b) If α is an even integer 2k, then (2m− 2)2 − α2 = 4(m− 1)2 − 4k2. Thus whenm = k + 1 all terms in the series for y1(x) are zero after the x2k term. A similarargument shows that if α = 2k + 1, then all terms in y2(x) are zero after the x2k+1

term.

(c) Using the previous parts, we obtain that p0(x) = 1, p1(x) = x, p2(x) = 1− 2x2

and p3(x) = x− 4x3/3.

9. The Taylor series about x = 0 for sinx is sinx = x− x3/3! + x5/5!− . . .. Assu-

ming that y =∑∞n=2 anx

n, we find

y′′ + (sinx)y = 2a2 + 6a3x+ 12a4x2 + 20a5x

3 + 30a6x4 + 42a7x

5 + . . .+

+(x− x3/3! + x5/5!− . . .)(a0 + a1x+ a2x2 + a3x

3 + a4x4 + . . .) =

= 2a2 + (6a3 + a0)x+ (12a4 + a1)x2 + (20a5 + a2 − a0/6)x3+

+(30a6 + a3 − a1/6)x4 + (42a7 + a4 − a2/3! + a0/5!)x5 + . . . = 0.


Hence a2 = 0, a3 = −a0/6, a4 = −a1/12, a5 = a0/120, a6 = (a1 + a0)/180, a7 =−a0/7! + a1/504, . . .. We set a0 = 1 and a1 = 0 and obtain y1(x) = (1− x3/6 +x5/120 + x6/180 + . . .). Next we set a0 = 0 and a1 = 1 and obtain y2(x) = (x−x4/12 + x6/180 + x7/504 + . . .). Since p(x) = 1 and q(x) = sinx, both have ρ =∞;the solution in this case converges for all x, that is, ρ =∞.

10. The Taylor series expansion of ex, about x0 = 0 , is

ex =

∞∑n=0

xn

n !.

Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Substituting into the ODE,[ ∞∑n=0

xn

n !

][ ∞∑n=0

(n+ 2)(n+ 1)an+2 xn

]+ x

∞∑n=0

anxn = 0 .

First note that

x

∞∑n=0

anxn =

∞∑n=1

an−1xn = a0 x+ a1x

2 + a2x3 + . . .+ an−1x

n + . . . .

The coefficient of xn in the product of the two series is

cn = 2a21

n !+ 6a3

1

(n− 1) !+ 12a4

1

(n− 2) !+ . . .

. . .+ (n+ 1)nan+1 + (n+ 2)(n+ 1)an+2 .

Expanding the individual series, it follows that

2a2 + (2a2 + 6a3)x+ (a2 + 6a3 + 12a4)x2 + (a2 + 6a3 + 12a4 + 20a5)x3 + . . .

. . .+ a0 x+ a1x2 + a2x

3 + . . . = 0 .

Setting the coefficients equal to zero, we obtain the system 2a2 = 0, 2a2 + 6a3 +a0 = 0, a2 + 6a3 + 12a4 + a1 = 0 , a2 + 6a3 + 12a4 + 20a5 + a2 = 0 , . . . . Hence thegeneral solution is

y(x) = a0 + a1x− a0x3

6+ (a0 − a1)

x4

12+ (2a1 − a0)

x5

40+ (

4

3a0 − 2a1)

x6

120+ . . . .

We find that two linearly independent solutions (W (y1, y2)(0) = 1) are

y1(x) = 1− x3

6+x4

12− x5

40+ . . .

y2(x) = x− x4

12+x5

20− x6

60+ . . .

Since p(x) = 0 and q(x) = xe−x converge everywhere, ρ =∞ .

11. The Taylor series expansion of cos x , about x0 = 0 , is

cos x =

∞∑n=0

(−1)nx2n

(2n) !.

5.3 25

Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Substituting into the ODE,[ ∞∑n=0

(−1)nx2n

(2n) !

][ ∞∑n=0

(n+ 2)(n+ 1)an+2 xn

]+

∞∑n=1

nanxn − 2

∞∑n=0

anxn = 0 .

The coefficient of xn in the product of the two series is

cn = 2a2bn + 6a3bn−1 + 12a4bn−2 + . . .+ (n+ 1)nan+1b1 + (n+ 2)(n+ 1)an+2b0,

in which cos x = b0 + b1x+ b2x2 + . . .+ bnx

n + . . . . It follows that

2a2 − 2a0 +

∞∑n=1

cnxn +

∞∑n=1

(n− 2)anxn = 0 .

Expanding the product of the series, it follows that

2a2 − 2a0 + 6a3x+(−a2 + 12a4)x2 + (−3a3 + 20a5)x3 + . . .

. . .− a1 x+ a3x3 + 2a4x

4 + . . . = 0 .

Setting the coefficients equal to zero, a2 − a0 = 0 , 6a3 − a1 = 0 , −a2 + 12a4 = 0 ,−3a3 + 20a5 + a3 = 0 , . . . . Hence the general solution is

y(x) = a0 + a1x+ a0x2 + a1

x3

6+ a0

x4

12+ a1

x5

60+ a0

x6

120+ a1

x7

560+ . . . .

We find that two linearly independent solutions (W (y1, y2)(0) = 1) are

y1(x) = 1 + x2 +x4

12+

x6

120+ . . .

y2(x) = x+x3

6+x5

60+

x7

560+ . . .

The nearest zero of P (x) = cos x is at x = ±π/2 . Hence ρmin = π/2 .

12. If y1 = x and y2 = x2 are solutions, then substituting y2 into the ODE resultsin

2P (x) + 2xQ(x) + x2R(x) = 0 .

Setting x = 0 , we find that P (0) = 0 . Similarly, substituting y1 into the ODEresults in Q(0) = 0 . Therefore P (x)/Q(x) and R(x)/P (x) may not be analytic.If they were, Theorem 3.2.1 would guarantee that y1 and y2 were the only twosolutions. But note that an arbitrary value of y(0) cannot be a linear combinationof y1(0) and y2(0). Hence x0 = 0 must be a singular point.

13. Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Substituting into the ODE,

∞∑n=0

(n+ 1)an+1 xn −

∞∑n=0

anxn = 0 .

That is,∞∑n=0

[(n+ 1)an+1 − an]xn = 0 .


Setting the coefficients equal to zero, we obtain

an+1 =ann+ 1

for n = 0, 1, 2, . . . . It is easy to see that an = a0/(n !) . Therefore the generalsolution is

y(x) = a0

[1 + x+

x2

2 !+x3

3 !+ . . .

]= a0e

x.

The coefficient a0 = y(0), which can be arbitrary.

14. Let y = a0 + a1x+ a2x2 + . . .+ anx


∞∑n=0

(n+ 1)an+1 xn − x

∞∑n=0

anxn = 0 .

That is,∞∑n=0

(n+ 1)an+1 xn −

∞∑n=1

an−1xn = 0 .

Combining the series, we have

a1 +

∞∑n=1

[(n+ 1)an+1 − an−1] xn = 0 .

Setting the coefficient equal to zero, a1 =0 and an+1 = an−1/(n+1) for n = 1, 2, . . ..Note that the indices differ by two, so for k = 1, 2, . . .

a2k =a2k−2(2k)

=a2k−4

(2k − 2)(2k)= . . . =

a02 · 4 . . . (2k)

and

a2k+1 = 0 .

Hence the general solution is

y(x) = a0

[1 +

x2

2+

x4

22 2 !+

x6

23 3 !+ . . .+

x2n

2n n !+ . . .

]= a0e

x2/2.


15. Let y = a0 + a1x+ a2x2 + . . .+ anx


(1− x)

∞∑n=0

(n+ 1)an+1 xn −

∞∑n=0

anxn = 0 .

That is,∞∑n=0

(n+ 1)an+1 xn −

∞∑n=1

nan xn −

∞∑n=0

anxn = 0 .

Combining the series, we have

a1 − a0 +

∞∑n=1

[(n+ 1)an+1 − nan − an] xn = 0 .

5.3 27

Setting the coefficients equal to zero, a1 = a0 and an+1 = an for n = 0, 1, 2, . . . .Hence the general solution is

y(x) = a0[1 + x+ x2 + x3 + . . .+ xn + . . .

]= a0

1

1− x.


16. Substituting y =∑∞n=0 anx

n into the differential equation we obtain that∑∞n=1 nanx

n−1 −∑∞n=0 anx

n = x2. Shifting the indices then gives the equation∑∞n=0 [(n+ 1)an+1 − an]xn = x2. Equating coefficients of both sides then gives:

a1 − a0 = 0, 2a2 − a1 = 0, 3a3 − a2 = 1 and (n+ 1)an+1 = an for n = 3, 4, . . .. Thusa1 = a0, a2 = a1/2 = a0/2, a3 = 1/3 + a2/3 = 1/3 + a0/2 · 3, a4 = a3/4 = 1/3 · 4 +a0/2 · 3 · 4 = 2/4! + a0/4!, and in general an = an−1/n = 2/n! + a0/n!. Hence

y(x) = a0

(1 + x+

x2

2!+ . . .+

xn

n!. . .

)+ 2

(x3

3!+x4

4!+ . . .+

xn

n!+ . . .

).

Using the power series for ex, the first and second sums can be rewritten as a0ex +

2(ex − 1− x− x2/2), which is the same solution as we found using methods ofChap. 2.


n into the Legendre equation, shifting indices, andcollecting coefficients of like powers of x yields

[2 · 1 · a2 + α(α+ 1)a0]x0 + (3 · 2 · a3 − [2 · 1− α(α+ 1)]a1)x1+

+

∞∑n=2

((n+ 2)(n+ 1)an+2 − [n(n+ 1)− α(α+ 1)]an)xn = 0.

Thus a2 =−α(α+1)a0/2!, a3 =[2 · 1− α(α+ 1)]a1/3! = −(α− 1)(α+2)a1/3! andthe recurrence relation gives us (n+ 2)(n+ 1)an+2 = −[α(α+ 1)− n(n+ 1)]an =−(α− n)(α+ n+ 1)an, n = 2, 3, . . . . Setting a1 = 0, a0 = 1 yields a solution witha3 = a5 = a7 = . . . = 0 and a4 = α(α− 2)(α+ 1)(α+ 3)/4 . . ., and generally, a2m =(−1)m[α(α− 2) . . . (α− 2m+ 2)][(α+ 1) . . . (α+ 2m− 1)]/(2m)!. The second lin-early independent solution is obtained by setting a0 = 0 and a1 = 1. The coefficientsare then a2 = a4 = a6 = . . . = 0 and a3 = (α− 1)(α+ 2)/3!, and a5 = −(α− 3)(α+4)a3/5 · 4 = (α− 1)(α− 3)(α+ 2)(α+ 4)/5!.

18. If α = 0 , then y1(x) = 1 . If α = 2n , then a2m = 0 for m ≥ n+ 1 . As a result,

y1(x) = 1

+

n∑m=1

(−1)m2mn(n− 1) . . . (n−m+ 1)(2n+ 1)(2n+ 3) . . . (2n+ 2m− 1)

(2m)!x2m.

α = 0 1

α = 2 1− 3x2

α = 4 1− 10x2 + 353 x

4


If α = 2n+ 1 , then a2m+1 = 0 for m ≥ n+ 1 . As a result,

y2(x) = x

+

n∑m=1

(−1)m2mn(n− 1) . . . (n−m+ 1)(2n+ 3)(2n+ 5) . . . (2n+ 2m+ 1)

(2m+ 1)!x2m+1.

α = 1 x

α = 3 x− 53x

3

α = 5 x− 143 x

3 + 215 x

5

19.(a) Based on Problem 18,

α = 0 1 y1(1) = 1

α = 2 1− 3x2 y1(1) = −2

α = 4 1− 10x2 + 353 x

4 y1(1) = 83

Normalizing the polynomials, we obtain

P0(x) = 1

P2(x) = −1

2+

3

2x2

P4(x) =3

8− 15

4x2 +

35

8x4

α = 1 x y2(1) = 1

α = 3 x− 53x

3 y2(1) = − 23

α = 5 x− 143 x

3 + 215 x

5 y2(1) = 815

Similarly,

P1(x) = x

P3(x) = −3

2x+

5

2x3

P5(x) =15

8x− 35

4x3 +

63

8x5

(b)

5.3 29

(c) P0(x) has no roots. P1(x) has one root at x = 0 . The zeros of P2(x) are atx = ± 1/

√3 . The zeros of P3(x) are x = 0,±

√3/5 . The roots of P4(x) are given

by x2 = (15 + 2√

30)/35 , (15− 2√

30)/35 . The roots of P5(x) are given by x = 0and x2 = (35 + 2

√70)/63 , (35− 2

√70)/63 .

20. Using the chain rule, we have

dF (φ)

dφ=dF [φ(x)]

dx

dx

dφ= −f ′(x) sinφ(x) = −f ′(x)

√1− x2,

d2F (φ)

dφ2=

d

dx

[−f ′(x)

√1− x2

] dxdφ

= (1− x2)f ′′(x)− xf ′(x),

which when substituted into the D.E. yields the desired result.

21. We first compute the nth derivatives of (x2 − 1)n for n = 0, 1, 2, 3:

d0

dx0(x2 − 1)0 = 1

d

dx(x2 − 1) = 2x

d2

dx2(x2 − 1)2 = −4 + 12x2

d3

dx3(x2 − 1)3 = 120x3 − 72x

Thus we have that

1

200!

d0

dx0(x2 − 1)0 = 1 = P0(x)

1

211!

d

dx(x2 − 1) =

2x

2= x = P1(x)

1

222!

d2

dx2(x2 − 1)2 =

−4 + 12x2

8= −1

2+

3

2x2 = P2(x)

1

233!

d3

dx3(x2 − 1)3 =

−72x+ 120x3

48= −3

2x+

5

2x3 = P3(x)

22. Since [(1− x2)y′]′ = (1− x)2y′′ − 2xy′, the Legendre Equation, from Problem17, can be written as shown. Thus, carrying out the multiplications indicated yieldsthe two equations

Pm[(1− x2)P ′n

]′= −n(n+ 1)PnPm

Pn[(1− x2)P ′m

]′= −m(m+ 1)PnPm.

As long as n 6= m, the second equation can be subtracted from the first and theresult integrated from −1 to 1 to obtain∫ 1

−1

{Pm[(1− x2)P ′n

]′ − Pn [(1−x2)P ′m]′}

dx=[m(m+1)−n(n+1)]

∫ 1

−1PnPmdx.


The left side may be integrated by parts to yield[Pm(1− x2)P ′n − Pn(1− x2)P ′m

]1−1 +

∫ 1

−1

[P ′m(1− x2)P ′n − P ′n(1− x2)P ′m

]dx,

which is zero. Thus∫ 1

−1 Pn(x)Pm(x)dx = 0 for n 6= m.

23. Since the n+ 1 polynomials P0, P1, . . ., Pn are linearly independent, andthe degree of Pk is k , any polynomial f of degree n can be expressed as a linearcombination

f(x) =

n∑k=0

akPk(x) .

Multiplying both sides by Pm and integrating,∫ 1

−1f(x)Pm(x)dx =

n∑k=0

ak

∫ 1

−1Pk(x)Pm(x)dx .

Based on Problem 22, ∫ 1

−1Pk(x)Pm(x)dx =

2

2m+ 1δkm.

Hence ∫ 1

−1f(x)Pm(x)dx =

2

2m+ 1am .

5.4

1. Substitution of y = xr results in the quadratic equation F (r) = 0, where F (r) =r(r − 1) + 4r + 2 = r2 + 3r + 2. The roots are r = −2 , −1 . Hence the generalsolution, for x 6= 0 , is y = c1 x

−2 + c2 x−1.

2. This equation is of the form of an Euler equation with x replaced by x+ 1,so we seek solutions of the form y = (x+ 1)r for x+ 1 > 0. Substitution of y intothe D.E. yields F (r) = [r(r − 1) + 3r + 3/4](x+ 1)r = 0. Thus r2 + 2r + 3/4 = 0,which gives r = −3/2,−1/2. The general solution of the differential equation isthen y = c1|x+ 1|−1/2 + c2|x+ 1|−3/2, x 6= −1.

3. Substitution of y = xr results in the quadratic equation F (r) = 0 , where F (r) =r(r − 1)− 3r + 4 = r2 − 4r + 4. The root is r = 2 , with multiplicity two. Hencethe general solution, for x 6= 0 , is y = (c1 + c2 ln |x|)x2.

4. Substitution of y = xr results in the quadratic equation F (r) = 0 , where F (r) =r(r − 1)− r + 1 = r2 − 2r + 1. The root is r = 1 , with multiplicity two. Hence thegeneral solution, for x 6= 0 , is y = (c1 + c2 ln |x|)x.

5.4 31

5. Substitution of y = xr results in the quadratic equation F (r) = 0 , where F (r) =r2 + 5r − 1. The roots are r = −(5 ±

√29)/2 . Hence the general solution, for

x 6= 0 , is y = c1 |x|−(5+√29)/2

+ c2 |x|−(5−√29)/2

.

6. Substitution of y = xr results in the quadratic equation F (r) = 0 , where F (r) =r2 − 3r + 3. The roots are complex, with r = (3 ± i

√3 )/2 . Hence the general

solution, for x 6= 0, is

y = c1 |x|3/2 cos(

√3

2ln |x|) + c2 |x|3/2 sin(

√3

2ln |x|).

7. Again let y = xr to obtain F (r) = r(r − 1)− 5r + 9 = 0, or (r − 3)2 = 0. Thus

the roots are x = 3, 3 and y = c1x3 + c2x

3 ln |x|, x 6= 0, is the solution of the dif-ferential equation.

8. Substitution of y = (x− 2)r results in the quadratic equation F (r) = 0 , whereF (r) = r2 + 4r + 8. The roots are complex, with r = −2 ± 2i . Hence the generalsolution, for x 6= 2, is y=c1 (x− 2)−2 cos(2 ln |x− 2|)+c2(x− 2)−2 sin(2 ln |x−2|).

9. In this case F (r) = 2r(r − 1) + r − 3 = 2r2 − r − 3 = (2r − 3)(r + 1) = 0, so y =

c1x3/2 + c2x

−1 (since x0 = 1, we do not need |x|) and y′ = (3/2)c1x1/2 − c2x−2.

Setting x = 1 in y and y′, we obtain c1 + c2 = 1 and (3/2)c1 − c2 = 4, which yieldc1 = 2 and c2 = −1. Hence y = 2x3/2 − x−1. As x→ 0+, we have y → −∞ due tothe second term.

10. Substitution of y = xr results in the quadratic equation F (r) = 0 , whereF (r) = 4r2 + 4r + 17. The roots are complex, with r = −1/2 ± 2i . Hence thegeneral solution, for x > 0, is y = c1 x

−1/2 cos(2 ln x) + c2 x−1/2 sin(2 ln x). Invok-

ing the initial conditions, we obtain the system of equations

c1 = 2, −1

2c1 + 2c2 = −3.

Hence the solution of the initial value problem is

y(x) = 2x−1/2 cos(2 ln x)− x−1/2 sin(2 ln x).


As x → 0+, the solution decreases without bound.

11. Substitution of y = xr results in the quadratic equation F (r) = 0 , whereF (r) = r2 − 4r + 4. The root is r = 2 , with multiplicity two. Hence the generalsolution, for x < 0 , is y = (c1 + c2 ln |x|)x2. Invoking the initial conditions, weobtain the system of equations

c1 = 2, −2c1 − c2 = 3.


y(x) = (2− 7 ln |x|)x2.

We find that y(x) → 0 as x → 0−.

12. Since the coefficients of y, y′, and y′′ have no common factors, and since P (x)vanishes only at x = 0, we conclude that x = 0 is a singular point. Writing thedifferential equation in the form y′′ + p(x)y′ + q(x)y = 0, we get p(x) = (1− x)/xand q(x) = 1. Thus for the singular point we have limx→0 xp(x)=limx→0 1−x=1,and limx→0 x

2q(x) = 0, thus x = 0 is a regular singular point.

13. We see that P (x) = 0 when x = 0 and 1 . Since the three coefficients have nofactors in common, both of these points are singular points. Near x = 0,

limx→0

x p(x) = limx→0

x2x

x2(1− x)2= 2 .

limx→0

x2q(x) = limx→0

x24

x2(1− x)2= 4 .

5.4 33

The singular point x = 0 is regular. Considering x = 1,

limx→1

(x− 1)p(x) = limx→1

(x− 1)2x

x2(1− x)2.

The latter limit does not exist. Hence x = 1 is an irregular singular point.

14. P (x) = 0 when x = 0 and 1 . Since the three coefficients have no commonfactors, both of these points are singular points. Near x = 0,

limx→0

x p(x) = limx→0

xx− 2

x2(1− x).

The limit does not exist, and so x = 0 is an irregular singular point. Consideringx = 1,

limx→1

(x− 1)p(x) = limx→1

(x− 1)x− 2

x2(1− x)= 1 .

limx→1

(x− 1)2q(x) = limx→1

(x− 1)2−3x

x2(1− x)= 0 .

Hence x = 1 is a regular singular point.

15. P (x) = 0 when x = 0 and ± 1 . Since the three coefficients have no commonfactors, both of these points are singular points. Near x = 0,

limx→0

x p(x) = limx→0

x2

x3(1− x2).

The limit does not exist, and so x = 0 is an irregular singular point. Near x = −1,

limx→−1

(x+ 1)p(x) = limx→−1

(x+ 1)2

x3(1− x2)= −1 .

limx→−1

(x+ 1)2q(x) = limx→−1

(x+ 1)22

x3(1− x2)= 0 .

Hence x = −1 is a regular singular point. At x = 1 ,

limx→1

(x− 1)p(x) = limx→1

(x− 1)2

x3(1− x2)= −1 .

limx→1

(x− 1)2q(x) = limx→1

(x− 1)22

x3(1− x2)= 0 .


16. Writing the differential equation in the form y′′ + p(x)y′ + q(x)y = 0, we find

p(x) = x/(1− x)(1 + x)2 and q(x) = (1 + x)/(1− x2)2. Therefore x = ±1 are sin-gular points. Since limx→1(x− 1)p(x) and limx→1(x− 1)2q(x) both exist, we con-clude that x = 1 is a regular singular point. Finally, since limx→−1(x+ 1)p(x) doesnot exist, we conclude that x = −1 is an irregular singular point.


17. The only singular point is at x = 0 . We find that

limx→0

x p(x) = limx→0

xx

x2= 1 .

limx→0

x2q(x) = limx→0

x2x2 − ν2

x2= −ν2.


18. Dividing the ODE by (x+ 2)2(x− 1), we find that

p(x) =3

(x+ 2)2and q(x) =

−2

(x+ 2)(x− 1).

The singular points are at x = −2 and 1 . For x = −2,

limx→−2

(x+ 2)p(x) = limx→−2

(x+ 2)3

(x+ 2)2.

The limit does not exist. Hence x = −2 is an irregular singular point. For x = 1,

limx→1

(x− 1)p(x) = limx→1

(x− 1)3

(x+ 2)2= 0 .

limx→1

(x− 1)2q(x) = limx→1

(x− 1)2−2

(x+ 2)(x− 1)= 0 .


19. P (x) = 0 when x = 0 and 3 . Since the three coefficients have no commonfactors, both of these points are singular points. Near x = 0,

limx→0

x p(x) = limx→0

xx+ 1

x(3− x)=

1

3.

limx→0

x2q(x) = limx→0

x2−2

x(3− x)= 0 .

Hence x = 0 is a regular singular point. For x = 3,

limx→3

(x− 3)p(x) = limx→3

(x− 3)x+ 1

x(3− x)= −4

3.

limx→3

(x− 3)2q(x) = limx→3

(x− 3)2−2

x(3− x)= 0 .


20. Writing the differential equation in the form y′′ + p(x)y′ + q(x)y = 0, we seethat p(x) = ex/x and q(x) = (3 cosx)/x. Thus x = 0 is a singular point. Sincexp(x) = ex is analytic at x = 0 and x2q(x) = 3x cosx is analytic at x = 0, the pointx = 0 is a regular singular point.

21. Note that p(x)=ln |x| and q(x)=3x . Evidently, p(x) is not analytic at x0 = 0.Furthermore, the function x p(x) = x ln |x| does not have a Taylor series aboutx0 = 0. Hence x = 0 is an irregular singular point.

5.4 35

22. Writing the differential equation in the form y′′ + p(x)y′ + q(x)y = 0, we seethat p(x) = x/ sinx and q(x) = 4/ sinx. Since limx→0 q(x) does not exist, the pointx0 = 0 is a singular point and since neither limx→±nπ p(x) nor limx→±nπ q(x) exist,either, the points x0 = ±nπ are also singular points. To determine whether thesingular points are regular or irregular we must use Eq. (31) and the result #7 ofmultiplication and division of power series from Section 5.1. For x0 = 0, we have

xp(x) =x2

sinx=

x2

x− x3

6 + . . .= x

[1 +

x2

6+ . . .

]= x+

x3

6+ . . . ,

which converges about x0 = 0 and thus xp(x) is analytic at x0 = 0. x2q(x), bysimilar steps, is also analytic at x0 = 0 and thus x0 = 0 is a regular singular point.For x0 = nπ ,we have

(x− nπ)p(x) =(x− nπ)x

sinx=

(x− nπ)[(x− nπ) + nπ]

±(x− nπ)∓ (x−nπ)36 ± . . .

= ±[(x− nπ) + nπ]

[1 +

(x− nπ)2

6+ . . .

],

which converges about x0 = nπ and thus (x− nπ)p(x) is analytic at x = nπ. Sim-ilarly (x+ nπ)p(x) and (x± nπ)2q(x) are analytic and thus x0 = ±nπ are regularsingular points.

23. The singular points are located at x = ±nπ , n = 0, 1, . . . . Dividing the ODEby x sin x , we find that x p(x) = 3 csc x and x2q(x) = x2 csc x . Evidently, x p(x)is not even defined at x = 0. Hence x = 0 is an irregular singular point. On theother hand, the Taylor series of x csc x, about x = 0, is

x csc x = 1 + x2/6 + 7x4360 + . . . .

Noting that csc(x ∓ nπ) = (−1)n csc x ,

(x ∓ nπ)p(x) = 3(−1)n(x ∓ nπ) csc(x ∓ nπ)/x

= 3(−1)n(x ∓ nπ) csc(x ∓ nπ)

[1

(x ∓ nπ) ± nπ

].

It is apparent that (x ∓ nπ)p(x) is analytic at x = ±nπ . Similarly,

(x ∓ nπ)2q(x) = (x ∓ nπ)2 csc x = (−1)n(x ∓ nπ)2 csc(x ∓ nπ),

which is also analytic at x = ±nπ . Hence all other singular points are regular.

24. Substituting y = xr, we find that r(r − 1) + αr + 5/2 = 0 or r2 + (α− 1)r +

5/2 = 0. Thus r1, r2 =[−(α− 1)±

√(α− 1)2 − 10

]/2. In order for solutions to

approach zero as x→ 0, it is necessary that the real parts of r1 and r2 be positive.Suppose that α > 1, then

√(α− 1)2 − 10 is either imaginary or real and less than

α− 1; hence the real parts of r1 and r2 will be negative. Suppose that α = 1,then r1, r2 = ±i

√10 and the solutions are oscillatory. Suppose that α < 1, then√

(α− 1)2 − 10 is either imaginary or real and less than 1− α; hence the real partsof r1 and r2 will be positive. Thus if α < 1, the solutions of the differential equationwill approach zero as x→ 0.


25. Substitution of y = xr results in the quadratic equation r2 − r + β = 0 . Theroots are

r =1 ±

√1− 4β

2.

If β > 1/4 , the roots are complex, with r1,2 = (1 ± i√

4β − 1 )/2 . Hence thegeneral solution, for x 6= 0 , is

y = c1 |x|1/2 cos(1

2

√4β − 1 ln |x|) + c2 |x|1/2 sin(

1

2

√4β − 1 ln |x|).

Since the trigonometric factors are bounded, y(x) → 0 as x → 0 . If β = 1/4 , theroots are equal, and

y = c1 |x|1/2 + c2 |x|1/2 ln |x| .

Since limx→ 0

√|x| ln |x| = 0 , y(x) → 0 as x → 0 . If β < 1/4 , the roots are real,

with r1,2 = (1 ±√

1− 4β )/2 . Hence the general solution, for x 6= 0 , is

y = c1 |x|1/2+√1−4β/2

+ c2 |x|1/2−√1−4β/2

.

Evidently, solutions approach zero as long as 1/2−√

1− 4β/2 > 0 . That is,

0 < β < 1/4 .

Hence all solutions approach zero for β > 0 .

26. Substitution of y = xr results in the quadratic equation r2 − r − 2 = 0 . Theroots are r = −1 , 2 . Hence the general solution, for x 6= 0 , is y = c1x

−1 + c2 x2.

Invoking the initial conditions, we obtain the system of equations

c1 + c2 = 1, −c1 + 2c2 = γ


y(x) =2− γ

3x−1 +

1 + γ

3x2.

The solution is bounded, as x → 0 , if γ = 2 .

27. In all cases the roots of F (r) = 0 are given by Eq. (6) and the forms of thesolution are given in Eqs. (25), (26), and (27).

(a) The real part of the root must be positive so, from Eq. (6), α < 1. Also β > 0,since the

√(α− 1)2 − 4β term must be less than |α− 1|.

(b) Similarly to part (a), if α < 1, then here we need β ≥ 0 (a single zero eigenvalueis allowed in this case) or if α = 1, then we need β > 0.

(c) The real part of the root has to be negative, so α > 1, and β > 0 (a negative βvalue would give us a positive root, β = 0 would give us a zero root).

(d) The real part of the root must be negative, so α > 1, with β ≥ 0 (for β = 0 oneroot is zero, which gives a bounded solution as x→∞). If α = 1, then the rootsare ±

√−4β, so β > 0 will yield oscillatory solutions as x→∞, which are bounded.

(e) According to (b) and (d) this happens when α = 1 and β > 0.

5.4 37

28. Assume that y = v(x)xr1 . Then we obtain that y′ = v(x)r1xr1−1 + v′(x)xr1 and

y′′ = v(x)r1(r2 − 1)xr1−2 + 2v′(x)r1xr1−1 + v′′(x)xr1 . Substituting in the D.E. and

collecting terms yields xr1+2v′′+(α+2r1)xr1+1v′+[r1(r1−1)+αr1+β]xr1v=0. Nowwe make use of the fact that r1 is a double root of f(r) = r(r − 1) + αr + β. Thismeans that f(r1) = 0 and f ′(r1) = 2r1 − 1 + α = 0. Hence the D.E. for v reducesto xr1+2v′′ + xr1+1v′. Since x > 0, we may divide by xr1+1 to obtain xv′′ + v′ = 0.Thus v(x) = lnx and a second solution is y = xr1 lnx.

29. To show thatW [xλ cos(µ lnx), xλ sin(µ lnx)](x) = µx2λ−1, we proceed by direct

computation. Let y1 = xλ cos(µ lnx) and y2 = xλ sin(µ lnx). Then

y′1 = λxλ−1 cos(µ lnx) + xλ(− sin(µ lnx) · µ

x

)= xλ−1 (λ cos(µ lnx)− µ sin(µ lnx))

y′2 = λxλ−1 sin(µ lnx) + xλ(

cos(µ lnx) · µx

)= xλ−1 (λ sin(µ lnx) + µ cos(µ lnx))

and the Wronskian will be

W [y1, y2](x) =

∣∣∣∣ y1 y2y′1 y′2

∣∣∣∣=

∣∣∣∣ xλ cos(µ lnx) xλ sin(µ lnx)xλ−1 (λ cos(µ lnx)− µ sin(µ lnx)) xλ−1 (λ sin(µ lnx) + µ cos(µ lnx))

∣∣∣∣= xλ cos(µ lnx)xλ−1 (λ sin(µ lnx) + µ cos(µ lnx))

− xλ sin(µ lnx)xλ−1 (λ cos(µ lnx)− µ sin(µ lnx))

= x2λ−1(λ cos(µ lnx) sin(µ lnx) + µ cos2(µ lnx)

−λ sinµ lnx cos(µ lnx) + µ sin2(µ lnx))

= µx2λ−1(cos2(µ lnx) + sin2(µ lnx)

)= µx2λ−1


n into the differential equation yields

2

∞∑n=2

n(n− 1)anxn−1 + 3

∞∑n=1

nanxn−1 +

∞∑n=0

anxn+1 = 0.

The last sum becomes∑∞n=2 an−2x

n−1 (let m = n+ 2 and then replace m by n),the first term of the middle sum is 3a1, and thus we have

3a1 +

∞∑n=2

{[2n(n− 1) + 3n]an + an−2}xn−1 = 0.

Hence a1 = 0 and an = −an−2/n(2n+ 1), which is the desired recurrence relation.Thus all even coefficients are found in terms of a0 and all odd coefficients are zero,thereby yielding only one solution of the desired form. The result is

y = a0(1− x2

2 · 5+

x4

2 · 4 · 5 · 9− . . .).


31. x = 0 is the only singular point. Dividing the ODE by 2x2, we have p(x) =3/(2x) and q(x) = −x−2(1 + x)/2. It follows that

limx→0

x p(x) = limx→0

x3

2x=

3

2,

limx→0

x2q(x) = limx→0

x2−(1 + x)

2x2= −1

2,

so x = 0 is a regular singular point. Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . ..Substitution into the ODE results in

2x2∞∑n=0

(n+ 2)(n+ 1)an+2 xn + 3x

∞∑n=0

(n+ 1)an+1xn − (1 + x)

∞∑n=0

anxn = 0 .

That is,

2

∞∑n=2

n(n− 1)an xn + 3

∞∑n=1

nanxn −

∞∑n=0

anxn −

∞∑n=1

an−1xn = 0 .

It follows that

−a0 + (2a1 − a0)x+

∞∑n=2

[2n(n− 1)an + 3nan − an − an−1]xn = 0 .

Equating the coefficients to zero, we find that a0 = 0 , 2a1 − a0 = 0 , and

(2n− 1)(n+ 1)an = an−1 , n = 2, 3, . . . .

We conclude that all the an are equal to zero. Hence y(x) = 0 is the only solutionthat can be obtained.

32. If ξ = 1/x, thendy

dx=dy

dξ

dξ

dx= − 1

x2dy

dξ= −ξ2 dy

dξ,

d2y

dx2=

d

dξ

(−ξ2 dy

dξ

)dξ

dx=

(−2ξ

dy

dξ− ξ2 d

2y

dξ2

)(− 1

x2

)=

ξ4d2y

dξ2+ 2ξ3

dy

dξ.

Substituting in the differential equation, we have

P (1/ξ)

[ξ4d2y

dξ2+ 2ξ3

dy

dξ

]+Q(1/ξ)

[−ξ2 dy

dξ

]+R(1/ξ)y = 0,

or

ξ4P (1/ξ)d2y

dξ2+[2ξ3P (1/ξ)− ξ2Q(1/ξ)

] dydξ

+R(1/ξ)y = 0.

The result follows from the theory of singular points at ξ = 0.

33. Based on Problem 32, the change of variable, x = 1/ξ , transforms the ODEinto the form

ξ4d2y

dξ2+ 2ξ3

dy

dξ+ y = 0 .

5.4 39

Evidently, ξ = 0 is a singular point. Now p(ξ) = 2/ξ and q(ξ) = 1/ξ4. Since thevalue of limξ→0 ξ

2q(ξ) does not exist, ξ = 0 (x =∞ ) is an irregular singular point.

34. Since p(x) = x2, Q(x) = x, and R(x) = −4, we have

f(ξ) =[2P (1/ξ)/ξ −Q(1/ξ)/ξ2

]/P (1/ξ) = 2/ξ − 1/ξ = 1/ξ

and g(ξ) = R(1/ξ)/ξ4P (1/ξ) = −4/ξ2. Thus the point at infinity is a singular point.Since both ξf(ξ) and ξ2g(ξ) are analytic at ξ = 0, the point at infinity is a regularsingular point.

35. Under the transformation x = 1/ξ , the ODE becomes

ξ4(1− 1

ξ2)d2y

dξ2+

[2ξ3(1− 1

ξ2) + 2ξ2

1

ξ

]dy

dξ+ α(α+ 1)y = 0 ,

that is,

(ξ4 − ξ2)d2y

dξ2+ 2ξ3

dy

dξ+ α(α+ 1)y = 0 .

Therefore ξ = 0 is a singular point. Note that

p(ξ) =2ξ

ξ2 − 1and q(ξ) =

α(α+ 1)

ξ2(ξ2 − 1).

It follows that

limξ→0

ξ p(ξ) = limξ→0

ξ2ξ

ξ2 − 1= 0 ,

limξ→0

ξ2q(ξ) = limξ→0

ξ2α(α+ 1)

ξ2(ξ2 − 1)= −α(α+ 1) .

Hence ξ = 0 (x =∞) is a regular singular point.


ξ4d2y

dξ2+

[2ξ3 + 2ξ2

1

ξ

]dy

dξ+ λ y = 0 ,

that is,

ξ4d2y

dξ2+ 2(ξ3 + ξ)

dy

dξ+ λ y = 0 .


p(ξ) =2(ξ2 + 1)

ξ3and q(ξ) =

λ

ξ4.

It immediately follows that the limit limξ→0 ξ p(ξ) does not exist. Hence ξ = 0(x =∞) is an irregular singular point.


ξ4d2y

dξ2+ 2ξ3

dy

dξ− 1

ξy = 0 .



p(ξ) =2

ξand q(ξ) =

−1

ξ5.

We find that

limξ→0


ξ2

ξ= 2 ,

but

limξ→0


ξ2(−1)

ξ5.

The latter limit does not exist. Hence ξ = 0 (x =∞) is an irregular singular point.

5.5

1.(a) P (x) = 0 when x = 0 . Since the three coefficients have no common factors,x = 0 is a singular point. Near x = 0,

limx→0

x p(x) = limx→0

x1

2x=

1

2.

limx→0

x2q(x) = limx→0

x21

2= 0 .


(b) Let

y = xr(a0 + a1x+ a2x2 + . . .+ anx

n + . . .) =

∞∑n=0

an xr+n.

Then

y ′ =

∞∑n=0

(r + n)anxr+n−1

and

y ′′ =

∞∑n=0

(r + n)(r + n− 1)anxr+n−2.


2

∞∑n=0

(r + n)(r + n− 1)an xr+n−1 +

∞∑n=0

(r + n)anxr+n−1 +

∞∑n=0

anxr+n+1 = 0 .

That is,

2

∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)anxr+n +

∞∑n=2

an−2 xr+n = 0 .

5.5 41

It follows that

a0 [2r(r − 1) + r]xr + a1 [2(r + 1)r + r + 1]xr+1

+

∞∑n=2

[2(r + n)(r + n− 1)an + (r + n)an + an−2]xr+n = 0 .

Assuming that a0 6= 0 , we obtain the indicial equation 2r2 − r = 0, with rootsr1 = 1/2 and r2 = 0 . It immediately follows that a1 = 0 . Setting the remainingcoefficients equal to zero, we have

an =−an−2

(r + n) [2(r + n)− 1], n = 2, 3, . . . .

(c) For r = 1/2 , the recurrence relation becomes

an =−an−2

n(1 + 2n), n = 2, 3, . . . .

Since a1 = 0 , the odd coefficients are zero. Furthermore, for k = 1, 2, . . . ,

a2k =−a2k−2

2k(1 + 4k)=

a2k−4(2k − 2)(2k)(4k − 3)(4k + 1)

=(−1)ka0

2k k! 5 · 9 · 13 . . . (4k + 1).

(d) For r = 0 , the recurrence relation becomes

an =−an−2

n(2n− 1), n = 2, 3, . . . .

Since a1 = 0 , the odd coefficients are zero, and for k = 1, 2, . . . ,

a2k =−a2k−2

2k(4k − 1)=

a2k−4(2k − 2)(2k)(4k − 5)(4k − 1)

=(−1)ka0

2k k! 3 · 7 · 11 . . . (4k − 1).

The two linearly independent solutions are

y1(x) =√x

[1 +

∞∑k=1

(−1)k x2k

2k k! 5 · 9 · 13 . . . (4k + 1)

]

y2(x) = 1 +

∞∑k=1

(−1)k x2k

2k k! 3 · 7 · 11 . . . (4k − 1).

2.(a) If the D.E. is put in the standard form y′′ + p(x)y + q(x)y = 0, then p(x) =

x−1 and q(x) = 1− 1/9x2. Thus x = 0 is a singular point. Since xp(x)→ 1 andx2q(x)→ −1/9 as x→ 0, it follows that x = 0 is a regular singular point.

(b) In determining a series solution of the D.E. it is more convenient to leave theequation in the form given rather then divide by x2, the coefficient of y′′. If wesubstitute y =

∑∞n=0 anx

n+r, we have

∞∑n=0

(n+ r)(n+ r − 1)anxn+r +

∞∑n=0

(n+ r)anxn+r +

(x2 − 1

9

) ∞∑n=0

anxn+r = 0.


Note that

x2∞∑n=0

anxn+r =

∞∑n=0

anxn+r+2 =

∞∑n=2

an−2xn+r.

Thus we have[r(r − 1) + r − 1

9

]a0x

r +

[(r + 1)r + a(r + 1)− 1

9

]a1x

r+1+

+

∞∑n=2

{[(n+ r)(n+ r − 1) + (n+ r)− 1

9

]an + an−2

}xn+r = 0.

From the first term, the indicial equation is r2 − 1/9 = 0 with roots r1 = 1/3 andr2 = −1/3. For either value of r it is necessary to take a1 = 0 in order that thecoefficient of xr+1 be zero. The recurrence relation is an = −an−2/[(n+ r)2 − 1/9].

(c) For r = 1/3 we have

an =−an−2

(n+ 1/3)2 − (1/3)

2 = − an−2(n+ 2/3)n

, n = 2, 3, 4 . . . .

Since a1 = 0, it follows from the recurrence relation that a3 = a5 = a7 = . . . = 0.For the even coefficients it is convenient to let n = 2m,m = 1, 2, 3, . . .. Then a2m =−a2m−2/22m (m+ 1/3). The first few coefficients are given by

a2 =(−1)a0

22 (1 + 1/3) 1, a4 =

(−1)a222 (2 + 1/3) 2

=a0

24 (1 + 1/3) (2 + 1/3) 2!

a6 =(−1)a4

22 (3 + 1/3) 3=

(−1)a026 (1 + 1/3) (2 + 1/3) (3 + 1/3) 3!

,

and the coefficient of x2m for m = 1, 2, . . . is

a2m =(−1)ma0

22mm! (1 + 1/3) (2 + 1/3) . . . (m+ 1/3).

Thus one solution (on setting a0 = 1) is

y1(x) = x1/3

[1 +

∞∑m=1

(−1)m

m! (1 + 1/3) (2 + 1/3) . . . (m+ 1/3)

(x2

)2m]

(d) Since r2 = −1/3 6= r1 and r1 − r2 = 2/3 is not an integer, we can calculatea second series solution corresponding to r = −1/3. The recurrence relation isn(n− 2/3)an = −an−2, which yields the desired solution following the steps in part(c). Note that a1 = 0, as in the first solution, and thus all the odd coefficients arezero.

3.(a) Note that x p(x) = 0 and x2q(x) = x , which are both analytic at x = 0 .

(b) Set y = xr(a0 + a1x+ a2x2 + . . .+ anx

n + . . .). Substitution into the ODEresults in

∞∑n=0

(r + n)(r + n− 1)an xr+n−1 +

∞∑n=0

anxr+n = 0 ,

5.5 43

and after multiplying both sides of the equation by x ,

∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=1

an−1xr+n = 0 .

It follows that

a0 [r(r − 1)]xr +

∞∑n=1

[(r + n)(r + n− 1)an + an−1]xr+n = 0 .

Setting the coefficients equal to zero, the indicial equation is r(r − 1) = 0 . Theroots are r1 = 1 and r2 = 0 . Here r1 − r2 = 1 . The recurrence relation is

an =−an−1

(r + n)(r + n− 1), n = 1, 2, . . . .

(c) For r = 1 ,

an =−an−1n(n+ 1)

, n = 1, 2, . . . .

Hence for n ≥ 1 ,

an =−an−1n(n+ 1)

=an−2

(n− 1)n2(n+ 1)= . . . =

(−1)na0n!(n+ 1)!

.

Therefore one solution is

y1(x) = x

∞∑n=0

(−1)n xn

n!(n+ 1)!.

4.(a) Putting the D.E. in the form y′′ + p(x)y′ + q(x)y = 0, we see that p(x) =1/x and q(x) = −1/x. Thus x = 0 is a singular point, and since xp(x)→ 1 andx2q(x)→ 0, as x→ 0, x = 0 is a regular singular point.

(b) Substituting y =∑∞n=0 anx

n+r in xy′′ + y′ − y = 0 and shifting indices, weobtain∞∑

n=−1an+1(r + n+ 1)(r + n)xn+r +

∞∑n=−1

an+1(r + n+ 1)xn+r −∞∑n=0

anxn+r = 0,

or

[r(r − 1) + r]a0x−1+r +

∞∑n=0

[(r + n+ 1)2an+1 − an

]xn+r = 0.

From the first coefficient we find r2 = 0 is the indicial equation, and from thecoefficient of xn+r we find the recurrence relation is an+1 = an/(n+ 1 + r)2.

(c) Setting r = 0 in the reccurence relation, we find (n+ 1)2an+1 = an, n = 0, 1,2, . . . . The coefficients are a1 = a0, a2 = a1/2

2 = a0/22, a3 = a2/3

2 = a0/32 · 22,

a4 = a3/42 = a0/4

2 · 32 · 22, . . . and an = a0/(n!)2. Thus one solution (on settinga0 = 1) is y =

∑∞n=0 x

n/(n!)2.


(d) Since the indicial equation has only one root, we only have one solution of theform y = xr

∑∞n=0 anx

n+r.

5.(a) Note that x p(x) = 1 and x2q(x) = x− 2, which are both analytic at x = 0 .



∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)an xr+n

+∞∑n=0

anxr+n+1 − 2

∞∑n=0

anxr+n = 0 .

After adjusting the indices in the second-to-last series, we obtain

a0 [r(r − 1) + r − 2]xr

+

∞∑n=1

[(r + n)(r + n− 1)an + (r + n)an − 2 an + an−1]xr+n = 0.

Assuming a0 6= 0 , the indicial equation is r2 − 2 = 0 , with roots r = ±√

2 . Set-ting the remaining coefficients equal to zero, the recurrence relation is

an =−an−1

(r + n)2 − 2, n = 1, 2, . . . .

Note that (r + n)2 − 2 = (r + n+√

2 )(r + n−√

2 ).

(c) For r =√

2 ,

an =−an−1

n(n+ 2√

2 ), n = 1, 2, . . . .

It follows that

an =(−1)na0

n!(1 + 2√

2 )(2 + 2√

2 ) . . . (n+ 2√

2 ), n = 1, 2, . . . .

(d) For r = −√

2 ,

an =−an−1

n(n− 2√

2 ), n = 1, 2, . . . ,

and therefore

an =(−1)na0

n!(1− 2√

2 )(2− 2√

2 ) . . . (n− 2√

2 ), n = 1, 2, . . . .

The two linearly independent solutions are

y1(x) = x√2

[1 +

∞∑n=1

(−1)n xn

n!(1 + 2√

2 )(2 + 2√

2 ) . . . (n+ 2√

2 )

]

y2(x) = x−√2

[1 +

∞∑n=1

(−1)n xn

n!(1− 2√

2 )(2− 2√

2 ) . . . (n− 2√

2 )

].

5.5 45

6.(a) Here x p(x) = 1− x and x2q(x) = −x , which are both analytic at x = 0 .



∞∑n=0

(r + n)(r + n− 1)an xr+n−1 +

∞∑n=0

(r + n)an xr+n−1

−∞∑n=0

(r + n)an xr+n −

∞∑n=0

anxr+n = 0 .

After multiplying both sides by x ,

∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)an xr+n

−∞∑n=0

(r + n)an xr+n+1 −

∞∑n=0

anxr+n+1 = 0.

After adjusting the indices in the last two series, we obtain

a0 [r(r − 1) + r]xr

+

∞∑n=1

[(r + n)(r + n− 1)an + (r + n)an − (r + n)an−1]xr+n = 0 .

Assuming a0 6= 0 , the indicial equation is r2 = 0 , with roots r1 = r2 = 0 . Settingthe remaining coefficients equal to zero, the recurrence relation is

an =an−1r + n

, n = 1, 2, . . . .

(c) With r = 0 ,

an =an−1n

, n = 1, 2, . . . .

Hence one solution is

y1(x) = 1 +x

1!+x2

2!+ . . .+

xn

n!+ . . . = ex.

7. (a) If we make the change of variable t = x− 1 and let y = u(t), then the Legen-

dre equation transforms to (t2 + 2t)u′′(t) + 2(t+ 1)u′(t)− α(α+ 1)u(t) = 0. Sincex = 1 is a regular singular point of the original equation, we know that t = 0 is aregular singular point of the transformed equation. Substituting u =

∑∞n=0 ant

n+r

in the transformed equation and shifting indices, we obtain

∞∑n=0

(n+ r)(n+ r − 1)antn+r + 2

∞∑n=−1

(n+ r + 1)(n+ r)an+1tn+r+

+2

∞∑n=0

(n+ r)antn+r + 2

∞∑n=−1

(n+ r + 1)an+1tn+r − α(α+ 1)

∞∑n=0

antn+r = 0,


or

[2r(r − 1) + 2r]a0tr−1+

+

∞∑n=0

{2(n+ r + 1)2an+1 + [(n+ r)(n+ r + 1)− α(α+ 1)]an}tn+r = 0.

The indicial equation is 2r2 = 0, so r = 0 is a double root. Thus there will be onlyone series solution of the form y =

∑∞n=0 ant

n+r.

(b) The recurrence relation is

2(n+ 1)2an+1 = [α(α+ 1)− n(n+ 1)]an, n = 0, 1, 2, . . . .

We have

a1 = [α(α+ 1)]a0/2 · 12, a2 = [α(α+ 1)][α(α+ 1)− 1 · 2]a0/22 · 22 · 12,

a3 = [α(α+ 1)][α(α+ 1)− 1 · 2][α(α+ 1)− 2 · 3]a0/23 · 32 · 22 · 12, . . . , and

an = [α(α+ 1)][α(α+ 1)− 1 · 2] . . . [α(α+ 1)− (n− 1) · n]a0/2n(n!)2.

Reverting to the variable x, it follows that one solution of the Legendre equationin powers of x− 1 is

y1(x) =

∞∑n=0

[α(α+ 1)][α(α+ 1)− 1 · 2] . . . [α(α+ 1)− (n− 1) · n](x− 1)n/2n(n!)2

where we have set a0 = 1, which is equivalent to the answer in the text if a (−1) isfactored out of each square bracket.

8.(a) Dividing through by the leading coefficient, the ODE can be written as

y ′′ − x

1− x2y ′ +

α2

1− x2y = 0 .

For x = 1 ,

p0 = limx→1

(x− 1)p(x) = limx→1

x

x+ 1=

1

2.

q0 = limx→1

(x− 1)2q(x) = limx→1

α2(1− x)

x+ 1= 0 .

For x = −1 ,

p0 = limx→−1

(x+ 1)p(x) = limx→−1

x

x− 1=

1

2.

q0 = limx→−1

(x+ 1)2q(x) = limx→−1

α2(x+ 1)

(1− x)= 0 .

Hence x = −1 and x = 1 are regular singular points. As shown in Example 1, theindicial equation is given by

r(r − 1) + p0r + q0 = 0 .

In this case, both sets of roots are r1 = 1/2 and r2 = 0 .

5.5 47

(b) Let t = x− 1 , and u(t) = y(t+ 1). Under this change of variable, the differen-tial equation becomes

(t2 + 2t)u ′′ + (t+ 1)u ′ − α2u = 0 .

Based on part (a), t = 0 is a regular singular point. Set u =∑∞n=0 an t

r+n. Sub-stitution into the ODE results in∞∑n=0

(r + n)(r + n− 1)an tr+n + 2

∞∑n=0

(r + n)(r + n− 1)an tr+n−1

+

∞∑n=0

(r + n)an tr+n +

∞∑n=0

(r + n)an tr+n−1 − α2

∞∑n=0

antr+n = 0 .

Upon inspection, we can also write

∞∑n=0

(r + n)2an tr+n + 2

∞∑n=0

(r + n)(r + n− 1

2)an t

r+n−1 − α2∞∑n=0

antr+n = 0.

After adjusting the indices in the second series, it follows that

a0

[2r(r − 1

2)

]tr−1

+

∞∑n=0

[(r + n)2an + 2(r + n+ 1)(r + n+

1

2)an+1 − α2an

]tr+n = 0.

Assuming that a0 6= 0 , the indicial equation is 2r2 − r = 0 , with roots r = 0 , 1/2 .The recurrence relation is

(r + n)2an + 2(r + n+ 1)(r + n+1

2)an+1 − α2an = 0 , n = 0, 1, 2, . . . .

With r1 = 1/2 , we find that for n ≥ 1 ,

an =4α2 − (2n− 1)2

4n(2n+ 1)an−1 = (−1)n

[1− 4α2

] [9− 4α2

]. . .[(2n− 1)2 − 4α2

]2n(2n+ 1)!

a0 .

With r2 = 0 , we find that for n ≥ 1 ,

an =α2 − (n− 1)2

n(2n− 1)an−1 = (−1)n

α(−α)[1− α2

] [4− α2

]. . .[(n− 1)2 − α2

]n! · 3 · 5 . . . (2n− 1)

a0 .

The two linearly independent solutions of the Chebyshev equation are

y1(x)= |x−1|1/2(

1+

∞∑n=1

(−1)n(1− 4α2)(9− 4α2) . . . ((2n− 1)2 − 4α2)

2n(2n+ 1)!(x−1)n

)

y2(x) = 1 +

∞∑n=1

(−1)nα(−α)(1− α2)(4− α2) . . . ((n− 1)2 − α2)

n! · 3 · 5 . . . (2n− 1)(x− 1)n .

9.(a) Here xp(x) = 1− x and x2q(x) = λx, which are both analytic at x = 0. Infact,

p0 = limx→0

x p(x) = 1 and q0 = limx→0

x2q(x) = 0 .


(b) The indicial equation is r(r − 1) + r = 0 , with roots r1,2 = 0 .

(c) Set

y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . .


n(n− 1)an xn−1 +

∞∑n=1

nan xn−1 −

∞∑n=0

nan xn + λ

∞∑n=0

anxn = 0 .

That is,

∞∑n=1

n(n+ 1)an+1 xn +

∞∑n=0

(n+ 1)an+1 xn −

∞∑n=1

nan xn + λ

∞∑n=0

anxn = 0 .

It follows that

a1 + λ a0 +

∞∑n=1

[(n+ 1)2an+1 − (n− λ)an

]xn = 0 .

Setting the coefficients equal to zero, we find that a1 = −λa0 , and

an =(n− 1− λ)

n2an−1 , n = 2, 3, . . . .

That is, for n ≥ 2 ,

an =(n− 1− λ)

n2an−1 = . . . =

(−λ)(1− λ) . . . (n− 1− λ)

(n!)2a0 .

Therefore one solution of the Laguerre equation is

y1(x) = 1 +

∞∑n=1

(−λ)(1− λ) . . . (n− 1− λ)

(n!)2xn.

Note that if λ = m , a positive integer, then an = 0 for n ≥ m+ 1 . In that case,the solution is a polynomial

y1(x) = 1 +

m∑n=1

(−λ)(1− λ) . . . (n− 1− λ)

(n!)2xn .

10. (a) The standard form is y′′ + p(x)y′ + q(x)y = 0, with p(x)=1/x and q(x)=1.

Thus x = 0 is a singular point; and since xp(x)→ 1 and x2q(x)→ 0 as x→ 0, x = 0is a regular singular point.


n+r into x2y′′ + xy′ + x2y = 0 and shifting indicesappropriately, we obtain

∞∑n=0

(n+ r)(n+ r − 1)anxn+r +

∞∑n=0

(n+ r)anxn+r +

∞∑n=2

an−2xn+r = 0,

or

[r(r − 1) + r]a0xr + [(1 + r)r + 1 + r]a1x

r+1 +

∞∑n=2

[(n+ r)2an + an−2

]xn+r = 0.

5.5 49

The indicial equation is, r2 = 0 so r = 0 is a double root. It is necessary to takea1 = 0 in order that the coefficient of xr+1 be zero.

(c) The recurrence relation is n2an = −an−2, n = 2, 3, . . .. Since a1 = 0 it followsthat a3 = a5 = a7 = . . . = 0. For the even coefficients we let n = 2m, m = 1, 2, . . ..Then a2m = −a2m−2/22m2 so a2 = −a0/22 · 12, a4 = a0/2

2 · 22 · 12 · 22, . . ., so weget a2m = (−1)ma0/2

2m(m!)2. Thus one solution of the Bessel equation of orderzero is J0(x) = 1 +

∑∞m=1(−1)mx2m/22m(m!)2 where we have set a0 = 1.

(d) Using the ratio test, it can be shown that the series converges for all x. Alsonote that J0(x)→ 1 as x→ 0.

11. In order to determine the form of the integral for x near zero, we must studythe integrand for x small. Using the above series for J0, we have

1

x[J0(x)]2=

1

x[1− x2/2 + . . .]2=

1

x[1− x2 + . . .]=

1

x[1 + x2 + . . .]

for x small. Thus

y2(x) = J0(x)

∫dx

x[J0(x)]2= J0(x)

∫ [1

x+ x+ . . .

]dx = J0(x)

[lnx+

x2

x+ . . .

],

and it is clear that y2(x) will contain a logarithmic term.

12.(a) Putting the D.E. in the standard form y′′ + p(x)y′ + q(x)y = 0, we see that

p(x) = 1/x and q(x) = (x2 − 1)/x2. Thus x = 0 is a singular point and sincexp(x)→ 1 and x2q(x)→ −1 as x→ 0, x = 0 is a regular singular point.


n+r into x2y′′ + xy′ + (x2 − 1)y = 0, shifting indicesappropriately and collecting coefficients of common powers of x, we obtain

[r(r − 1) + r − 1]a0xr + [(1 + r)r + 1 + r − 1]a1x

r+1

+

∞∑n=2

{[(n+ r)2 − 1

]an + an−2

}xn+r = 0.

The indicial equation is r2 − 1 = 0 so the roots are r1 = 1 and r2 = −1.

(c) For either value of r it is necessary to take a1 = 0 in order that the coef-ficient of xr+1 be zero. The recurrence relation is [(n+ r)2 − 1]an = −an−2, n =2, 3, 4, . . .. For r = 1 we have an = −an−2/[n(n+ 2)], n = 2, 3, 4, . . .. Since a1 = 0 itfollows that a3 = a5 = a7 = . . . = 0. Let n = 2m. Then a2m = −a2m−2/22m(m+1), m = 1, 2, . . ., so a2 = −a0/22 · 1 · 2, a4 = −a2/22 · 1 · 2 · 3 = a0/2

2 · 22 · 1 · 2 · 2 ·3, . . ., and a2m = (−1)ma0/2

2mm!(m+ 1)!. Thus one solution (set a0 = 1/2) of theBessel equation of order 1 is

J1(x) =x

2

∞∑n=0

(−1)nx2n

(n+ 1)!n!22n.

(d) The ratio test shows that the series converges for all x. Also note that J1(x)→ 0as x→ 0.


(e) For r = −1 the recurrence relation is [(n− 1)2 − 1]an = −an−2, n = 2, 3, . . ., sofor n = 2 the coefficient of a2 is zero and we cannot calculate a2. Consequently, itis not possible to find a series solution of the form x−1

∑∞n=0 bnx

n.

5.6

1.(a) The differential equation has the form P (x)y′′ +Q(x)y′ +R(x)y = 0 withP (x) = x, Q(x) = 2x, and R(x) = 6ex. From this we find p(x) = Q(x)/P (x) = 2and q(x) = R(x)/P (x) = 6ex/x, and thus x = 0 is a singular point. Since xp(x) =2x and x2q(x) = 6xex are analytic at x = 0, we conclude that x = 0 is a regularsingular point.

(b) We have xp(x)→ 0 = p0 and x2q(x)→ 0 = q0 as x→ 0 and thus Eq. (7), theindicial equation, is F (r) = r(r − 1) = 0 which has roots r1 = 1 and r2 = 0. Theseare the exponents of the singularity at x = 0.

2.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = −2− x and x2q(x) = 2 + x2.It follows that

p0 = limx→0

(−2− x) = −2

q0 = limx→0

(2 + x2) = 2

and therefore x = 0 is a regular singular point.

(b) The indicial equation is given by r(r − 1)− 2r + 2 = 0, that is, r2 − 3r + 2 = 0 ,with roots r1 = 2 and r2 = 1 .

3. The coefficients P (x) , Q(x) , and R(x) are analytic for all x ∈ R . Hence thereare no singular points.

4.(a) P (x) = 0 for x = 0 and x = −2. We note that p(x) = x−1(x+ 2)−1/2, andq(x) = −(x+ 2)−1/2. For the singularity at x = 0,

p0 = limx→0

1

2(x+ 2)=

1

4

q0 = limx→0

−x2

2(x+ 2)= 0

and therefore x = 0 is a regular singular point.For the singularity at x = −2,

p0 = limx→−2

(x+ 2)p(x) = limx→−2

1

2x= −1

4

q0 = limx→−2

(x+ 2)2q(x) = limx→−2

−(x+ 2)

2= 0

and therefore x = −2 is a regular singular point.

5.6 51

(b) For x = 0: the indicial equation is given by r(r − 1) + r/4 = 0, that is, r2 −3r/4 = 0, with roots r1 = 3/4 and r2 = 0 .

For x = −2: the indicial equation is given by r(r − 1)− r/4 = 0, that is, r2 −5r/4 = 0 , with roots r1 = 5/4 and r2 = 0 .

5.(a) P (x)=0 only for x=0. Furthermore, xp(x)=1/2+sinx/ 2x and x2q(x)=1.It follows that

p0 = limx→0

xp(x) = 1

q0 = limx→0

x2q(x) = 1


(b) The indicial equation is given by

r(r − 1) + r + 1 = 0 ,

that is, r2 + 1 = 0 , with complex conjugate roots r = ± i .

6.(a) For this differential equation,

p(x) =−(1 + x)

x2(1− x)and q(x) =

2

x(1− x),

and thus x = 0, 1 are singular points. Since xp(x) is not analytic at x = 0, x = 0 isnot a regular singular point. Looking at

(x− 1)p(x) =1 + x

x2and (x− 1)2q(x) =

2(1− x)

x,

we see that x = 1 is a regular singular point.

(b) As in Example (1),

p0 = limx→1

(x− 1)p(x) = 2 and q0 = limx→1

(x− 1)2q(x) = 0.

Thus the indicial equation is F (r) = r2 + r and r1 = 0 and r2 = −1.

7.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x(x− 2)−2(x+ 2)−1,and q(x) = 3(x− 2)−1(x+ 2)−1. For the singularity at x = 2,

limx→2

(x− 2)p(x) = limx→2

2x

x2 − 4,

which is undefined. Therefore x = 2 is an irregular singular point. For the singu-larity at x = −2,

p0 = limx→−2

(x+ 2)p(x) = limx→−2

2x

(x− 2)2= −1

4

q0 = limx→−2

(x+ 2)2q(x) = limx→−2

3(x+ 2)

x− 2= 0



(b) The indicial equation is given by r(r − 1)− r/4 = 0, that is, r2 − 5r/4 = 0 ,with roots r1 = 5/4 and r2 = 0 .

8.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x/(4− x2), and q(x) =3/(4− x2). For the singularity at x = 2,

p0 = limx→2

(x− 2)p(x) = limx→2

−2x

x+ 2= −1

q0 = limx→2

(x− 2)2q(x) = limx→2

3(2− x)

x+ 2= 0

and therefore x = 2 is a regular singular point.For the singularity at x = −2 ,

p0 = limx→−2

(x+ 2)p(x) = limx→−2

2x

2− x= −1

q0 = limx→−2

(x+ 2)2q(x) = limx→−2

3(x+ 2)

2− x= 0


(b) For x=2: the indicial equation is given by r(r−1)−r=0, that is, r2 − 2r=0 ,with roots r1 = 2 and r2 = 0 .

For x=−2: the indicial equation is given by r(r−1)−r=0, that is, r2−2r=0 ,with roots r1 = 2 and r2 = 0 .

9.(a) Note that p(x) = 1/x and q(x) = −1/x. Furthermore, xp(x) = 1 and x2q(x) =−x. It follows that

p0 = limx→0

(1) = 1 and q0 = limx→0

(−x) = 0,


(b) The indicial equation is given by r(r − 1) + r = 0, that is, r2 = 0 , with rootsr1 = r2 = 0 . This implies we can use Eq.(18) for the second solution.

(c) Let y = a0 + a1x+ a2x2 + . . .+ anx

n + . . . . Substitution into the ODE resultsin

∞∑n=0

(n+ 2)(n+ 1)an+2 xn+1 +

∞∑n=0

(n+ 1)an+1xn −

∞∑n=0

anxn = 0 .

After adjusting the indices in the first series, we obtain

a1 − a0 +

∞∑n=1

[n(n+ 1)an+1 + (n+ 1)an+1 − an]xn = 0 .

Setting the coefficients equal to zero, it follows that for n ≥ 0 ,

an+1 =an

(n+ 1)2.

So for n ≥ 1 ,

an =an−1n2

=an−2

n2(n− 1)2= . . . =

1

(n!)2a0 .

5.6 53

With a0 = 1 , one solution is

y1(x) = 1 + x+1

4x2 +

1

36x3 + . . .+

1

(n!)2xn + . . . .

For a second solution, set y2(x) = y1(x) ln x + b1x+ b2x2 + . . .+ bnx

n + . . . . Sub-stituting into the ODE, we obtain

L [y1(x)] · ln x+ 2 y ′1(x) + L

[ ∞∑n=1

bn xn

]= 0 .

Since L [y1(x)] = 0 , it follows that

L

[ ∞∑n=1

bn xn

]= −2 y ′1(x) .

More specifically,

b1+

∞∑n=1

[n(n+1)bn+1+(n+1)bn+1 − bn]xn = −2− x− 1

6x2 − 1

72x3 − 1

1440x4 − . . .

Equating the coefficients, we obtain the system of equations b1 =−2, 4b2 − b1 =−1,9b3 − b2 = −1/6, 16b4 − b3 = −1/72, . . .. Solving these equations for the coefficients,b1 = −2, b2 = −3/4, b3 = −11/108, b4 = −25/3456 , . . . . Therefore a second solu-tion is

y2(x) = y1(x) ln x+

[−2x− 3

4x2 − 11

108x3 − 25

3456x4 − . . .

].

10.(a) Here x p(x) = 2x and x2q(x) = 6xex . Both of these functions are analyticat x = 0 , therefore x = 0 is a regular singular point. Note that p0 = q0 = 0 .

(b) The indicial equation is given by r(r − 1) = 0, that is, r2 − r = 0 , with rootsr1 = 1 and r2 = 0 .

(c) In order to find the solution corresponding to r1 = 1 , set y = x∑∞n=0 anx

n.Upon substitution into the ODE, we have

∞∑n=0

(n+ 2)(n+ 1)an+1 xn+1 + 2

∞∑n=0

(n+ 1)anxn+1 + 6 ex

∞∑n=0

anxn+1 = 0 .

After adjusting the indices in the first two series, and expanding the exponentialfunction,

∞∑n=1

n(n+ 1)an xn + 2

∞∑n=1

nan−1xn + 6 a0x+ (6a0 + 6a1)x2

+(6a2 + 6a1 + 3a0)x3 + (6a3 + 6a2 + 3a1 + a0)x4 + . . . = 0 .


Equating the coefficients, we obtain the system of equations

2a1 + 2a0 + 6a0 = 0

6a2 + 4a1 + 6a0 + 6a1 = 0

12a3 + 6a2 + 6a2 + 6a1 + 3a0 = 0

20a4 + 8a3 + 6a3 + 6a2 + 3a1 + a0 = 0

...

Setting a0 = 1 , solution of the system results in a1 = −4, a2 = 17/3, a3 = −47/12,a4 = 191/120 , . . . . Therefore one solution is

y1(x) = x− 4x2 +17

3x3 − 47

12x4 + . . . .

The exponents differ by an integer. So for a second solution, set

y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnx

n + . . . .

Substituting into the ODE, we obtain

aL [y1(x)] · ln x+ 2a y ′1(x) + 2a y1(x)− ay1(x)

x+ L

[1 +

∞∑n=1

cn xn

]= 0 .


L

[1 +

∞∑n=1

cn xn

]= −2a y ′1(x)− 2a y1(x) + a

y1(x)

x.

More specifically,

∞∑n=1

n(n+ 1)cn+1xn + 2

∞∑n=1

n cnxn + 6 + (6 + 6c1)x

+(6c2 + 6c1 + 3)x2 + . . . = −a+ 10ax− 61

3ax2 +

193

12ax3 + . . . .


6 = −a2c2 + 8c1 + 6 = 10a

6c3 + 10c2 + 6c1 + 3 = −61

3a

12c4 + 12c3 + 6c2 + 3c1 + 1 =193

12a

...

Solving these equations for the coefficients, a = −6 . In order to solve the remainingequations, set c1 = 0 . Then c2 = −33, c3 = 449/6 , c4 = −1595/24 , . . . . Thereforea second solution is

y2(x) = −6 y1(x) ln x+

[1− 33x2 +

449

6x3 − 1595

24x4 + . . .

].

5.6 55

11.(a) After multiplying both sides of the ODE by x , we find that x p(x) = 0 andx2q(x) = x . Both of these functions are analytic at x = 0 , hence x = 0 is a regularsingular point.

(b) Furthermore, p0 = q0 = 0 . So the indicial equation is r(r − 1) = 0 , with rootsr1 = 1 and r2 = 0 .

(c) In order to find the solution corresponding to r1 = 1 , set y = x∑∞n=0 anx

n.Upon substitution into the ODE, we have

∞∑n=1

n(n+ 1)an xn +

∞∑n=0

anxn+1 = 0 .

That is,∞∑n=1

[n(n+ 1)an + an−1] xn = 0 .

Setting the coefficients equal to zero, we find that for n ≥ 1 ,

an =−an−1n(n+ 1)

.

It follows that

an =−an−1n(n+ 1)

=an−2

(n− 1)n2(n+ 1)= . . . =

(−1)na0(n!)2(n+ 1)

.


y1(x) = x− 1

2x2 +

1

12x3 − 1

144x4 +

1

2880x5 + . . . .

The exponents differ by an integer. So for a second solution, set

y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnx

n + . . . .

Substituting into the ODE, we obtain

aL [y1(x)] · ln x+ 2a y ′1(x)− ay1(x)

x+ L

[1 +

∞∑n=1

cn xn

]= 0 .


L

[1 +

∞∑n=1

cn xn

]= −2a y ′1(x) + a

y1(x)

x.

Now

L

[1 +

∞∑n=1

cn xn

]= 1 + (2c2 + c1)x+ (6c3 + c2)x2 + (12c4 + c3)x3

+(20c5 + c4)x4 + (30c6 + c5)x5 + . . . .

Substituting for y1(x) , the right hand side of the ODE is

−a+3

2ax− 5

12ax2 +

7

144ax3 − 1

320ax4 + . . . .



1 = −a

2c2 + c1 =3

2a

6c3 + c2 = − 5

12a

12c4 + c3 =7

144a

...

Evidently, a = −1 . In order to solve the second equation, set c1 = 0 . We then findthat c2 = −3/4 , c3 = 7/36 , c4 = −35/1728 , . . . . Therefore a second solution is

y2(x) = −y1(x) ln x+

[1− 3

4x2 +

7

36x3 − 35

1728x4 + . . .

].

12.(a) We have

p(x) =sinx

x2and q(x) =

cosx

x2,

so that x = 0 is a singular point. Note that xp(x) = sinx/x→ 1 = p0 as x→ 0 andx2q(x) = − cosx→ −1 = q0 as x→ 0. In order to assert that x = 0 is a regularsingular point we must demonstrate that xp(x) and x2q(x), with xp(x) = 1 at x = 0and x2q(x) = −1 at x = 0, have convergent power series (are analytic) about x = 0.We know that cosx is analytic, so we need only consider sinx/x. Now

sinx =

∞∑n=0

(−1)nx2n+1/(2n+ 1)!

for −∞ < x <∞, so

sinx/x =

∞∑n=0

(−1)nx2n/(2n+ 1)!

and hence is analytic. Thus we conclude that x = 0 is a regular singular point.

(b) From part (a) it follows that the indicial equation is r(r − 1) + r − 1 = r2−1=0and the roots are r1 = 1, r2 = −1.

(c) To find the first few terms of the solution corresponding to r1 = 1, assume that

y(x) = x

∞∑n=0

anxn = x(a0 + a1x+ a2x

2 + . . .) = a0x+ a1x2 + a2x

3 + . . . .

Substituting this series for y in the differential equation and expanding sinx andcosx about x = 0 yields

x2(2a1 + 6a2x+ 12a3x2 + 20a4x

3 + . . .)+

+(x− x3/3! + x5/5!− . . .)(a0 + 2a1x+ 3a2x2 + 4a3x

3 + 5a4x4 + . . .)

−(1− x2/2! + x4/4!− . . .)(a0x+ a1x2 + a2x

3 + a3x4 + a4x

5 + . . .) = 0.

5.6 57

Collecting terms, we have

(a0 − a0)x+ (2a1 + 2a1 − a1)x2 + (6a2 + 3a2 − a0/6− a2 + a0/2)x3+

+(12a3 + 4a3 − 2a1/6− a3 + a1/2)x4+

+(20a4 + 5a4 − 3a2/6 + a0/120− a4 + a2/2− a0/24)x5 + . . . = 0.

Simplifying yields 3a1x2 + (8a2 + a0/3)x3 + (15a3 + a1/6)x4 + (24a4 − a0/30)x5 +

. . . = 0. Thus, a1 = 0, a2 = −a0/4!, a3 = 0, a4 = a0/6!, . . .. Hence y1(x) = x−x3/4! + x5/6! + . . . where we have set a0 = 1. For the second solution we use avariation of Equation (24) similar to Equation (18):

y2(x) = ay1(x) lnx+ x−1

(1 +

∞∑n=1

cnxn

)=

ay1(x) lnx+1

x+ c1 + c2x+ c3x

2 + c4x3 + . . . ,

so we obtain that y′2 = ay′1 lnx+ ay1x−1 − x−2 + c2 + 2c3x+ 3c4x

2 + . . ., and y′′2 =ay′′1 lnx+ 2ay′1x

−1 − ay1x−2 + 2x−3 + 2c3 + 3c4x+ . . .. When these are substi-tuted in the given differential equation the terms including lnx will appear asa[x2y′′1 + sinxy′1 − cosxy1], which is zero since y1 is a solution. For the remainder ofterms, use y1 = x− x3/24 + x5/720 and the cosx and sinx series as shown earlier toobtain −c1 + (2/3 + 2a)x+ (3c3 + c1/2)x2+(4/45 + c2/3 + 8c4)x3+. . .=0. Theseyield c1 = 0, a = −1/3, c3 = 0, and c4 = −c2/24− 1/90. We may take c2 = 0, sincethis term will simply generate y1(x) over again. Thus

y2(x) = −1

3y1(x) lnx+ x−1 − 1

90x3.

If a computer algebra system is used, then additional terms in each series may beobtained without much additional effort. The next terms, in each case, are shownhere:

y1(x) = x− x3

24+

x5

720− 43x7

1451520+ . . . and

y2(x) = −1

3y1(x) lnx+

1

x

[1− x4

90+

41x6

120960− . . .

].

13.(a) We first write the D.E. in the standard form as given for Theorem 5.6.1except that we are expanding in powers of (x− 1) rather than powers of x:

(x− 1)2y′′ + (x− 1)[(x− 1)/2 lnx]y′ + [(x− 1)2/ lnx]y = 0.

Since ln 1 = 0, x = 1 is a singular point. To show it is a regular singular point ofthis D.E. we must show that (x− 1)/ lnx is analytic at x = 1; it will then followthat (x− 1)2/ lnx = (x− 1)[(x− 1)/ lnx] is also analytic at x = 1. If we expandlnx in a Taylor series about x = 1, we find that

lnx = (x− 1)− 1

2(x− 1)2 +

1

3(x− 1)3 − . . . .


Thus

(x− 1)/ lnx =

[1− 1

2(x− 1) +

1

3(x− 1)2 − . . .

]−1= 1 +

1

2(x− 1) + . . .

has a power series expansion about x = 1, and hence is analytic.

(b) We can use the above result to obtain the indicial equation at x = 1. We have

(x− 1)2y′′ + (x− 1)

[1

2+

1

4(x− 1) + . . .

]y′ +

[(x− 1) +

1

2(x− 1)2 + . . .

]y = 0.

Thus p0 = 1/2, q0 = 0, and the indicial equation is r(r − 1) + r/2 = 0. Hence r =1/2 and r = 0.

(c) In order to find the first three nonzero terms in a series solution correspondingto r = 1/2, it is better to keep the differential equation in its original form and tosubstitute the above power series for lnx:[

(x− 1)− 1

2(x− 1)2 +

1

3(x− 1)3 − 1

4(x− 1)4 + . . .

]y′′ +

1

2y′ + y = 0.

Next we substitute y = a0(x− 1)1/2 + a1(x− 1)3/2 + a2(x− 1)5/2 + . . . and collectcoefficients of like powers of (x− 1), which are then set equal to zero. This requiressome algebra before we find that 6a1/4 + 9a0/8 = 0 and 5a2 + 5a1/8− a0/12 = 0.These equations yield a1 = −3a0/4 and a2 = 53a0/480. With a0 = 1 we obtain thesolution

y1(x) = (x− 1)1/2 − 3

4(x− 1)3/2 +

53

480(x− 1)5/2 + . . . .

(d) Since the radius of convergence of the Taylor series for (x− 1)/ lnx is 1, wewould expect ρ = 1.

14.(a) After dividing by the leading coefficient, we find that

p0 = limx→0

x p(x) = limx→0

γ − (1 + α+ β)x

1− x= γ .

q0 = limx→0

x2q(x) = limx→0

−αβ x1− x

= 0 .

Hence x = 0 is a regular singular point. The indicial equation is r(r − 1) + γ r = 0 ,with roots r1 = 1− γ and r2 = 0 .

(b) For x = 1,

p0 = limx→1

(x− 1)p(x) = limx→1

−γ + (1 + α+ β)x

x= 1− γ + α+ β .

q0 = limx→1

(x− 1)2q(x) = limx→1

αβ(x− 1)

x= 0 .

5.6 59

Hence x = 1 is a regular singular point. The indicial equation is

r2 − (γ − α− β) r = 0 ,

with roots r1 = γ − α− β and r2 = 0 .

(c) Given that r1 − r2 is not a positive integer, we can set y =∑∞n=0 anx

n. Sub-stitution into the ODE results in

x(1−x)

∞∑n=2

n(n−1)anxn−2 + [γ − (1+α+ β)x]

∞∑n=1

nanxn−1 − αβ

∞∑n=0

anxn = 0.

That is,

∞∑n=1

n(n+ 1)an+1xn −

∞∑n=2

n(n− 1)anxn + γ

∞∑n=0

(n+ 1)an+1xn

−(1 + α+ β)

∞∑n=1

nanxn − αβ

∞∑n=0

anxn = 0.

Combining the series, we obtain

γ a1 − αβ a0 + [(2 + 2γ)a2 − (1 + α+ β + αβ)a1]x+

∞∑n=2

An xn = 0 ,

in which

An = (n+ 1)(n+ γ)an+1 − [n(n− 1) + (1 + α+ β)n+ αβ] an .

Note that n(n− 1) + (1 + α+ β)n+ αβ = (n+ α)(n+ β) . Setting the coefficientsequal to zero, we have γ a1 − αβ a0 = 0 , and

an+1 =(n+ α)(n+ β)

(n+ 1)(n+ γ)an

for n ≥ 1 . Hence one solution is

y1(x)=1+αβ

γ · 1!x+

α(α+1)β(β+1)

γ(γ+1) · 2!x2+

α(α+1)(α+2)β(β+1)(β + 2)

γ(γ + 1)(γ + 2) · 3!x3 + . . . .

Since the nearest other singularity is at x = 1 , the radius of convergence of y1(x)will be at least ρ = 1 .

(d) Given that r1 − r2 is not a positive integer, we can set y = x1−γ∑∞n=0 bnx

n.Then substitution into the ODE results in

x(1− x)

∞∑n=0

(n+ 1− γ)(n− γ)anxn−γ−1

+ [γ − (1 + α+ β)x]

∞∑n=0

(n+ 1− γ)anxn−γ − αβ

∞∑n=0

anxn+1−γ = 0.


That is,

∞∑n=0

(n+ 1− γ)(n− γ)anxn−γ −

∞∑n=0

(n+ 1− γ)(n− γ)anxn+1−γ

+γ

∞∑n=0

(n+ 1− γ)anxn−γ − (1 + α+ β)

∞∑n=0

(n+ 1− γ)anxn+1−γ

− αβ∞∑n=0

anxn+1−γ = 0.

After adjusting the indices,

∞∑n=0

(n+ 1− γ)(n− γ)anxn−γ −

∞∑n=1

(n− γ)(n− 1− γ)an−1xn−γ

+γ

∞∑n=0

(n+1−γ)anxn−γ−(1+α+β)

∞∑n=1

(n−γ)an−1xn−γ−αβ

∞∑n=1

an−1xn−γ =0.

Combining the series, we obtain

∞∑n=1

Bn xn−γ = 0 ,

in which

Bn = n(n+ 1− γ)bn − [(n− γ)(n− γ + α+ β) + αβ] bn−1.

Note that (n− γ)(n− γ + α+ β) + αβ = (n+ α− γ)(n+ β − γ). Setting Bn = 0,it follows that for n ≥ 1 ,

bn =(n+ α− γ)(n+ β − γ)

n(n+ 1− γ)bn−1.

Therefore a second solution is

y2(x) = x1−γ[1 +

(1 + α− γ)(1 + β − γ)

(2− γ)1!x

+(1 + α− γ)(2 + α− γ)(1 + β − γ)(2 + β − γ)

(2− γ)(3− γ)2 !x2 + . . .

].

(e) Under the transformation x = 1/ξ , the ODE becomes

ξ41

ξ(1− 1

ξ)d2y

dξ2+

{2ξ3

1

ξ(1− 1

ξ)− ξ2

[γ − (1 + α+ β)

1

ξ

]}dy

dξ− αβ y = 0 .

That is,

(ξ3 − ξ2)d2y

dξ2+[2ξ2 − γ ξ2 + (−1 + α+ β)ξ

] dydξ− αβ y = 0 .

5.6 61


p(ξ) =(2− γ) ξ + (−1 + α+ β)

ξ2 − ξand q(ξ) =

−αβξ3 − ξ2

.

It follows that

p0 = limξ→0


(2− γ) ξ + (−1 + α+ β)

ξ − 1= 1− α− β ,

q0 = limξ→0


−αβξ − 1

= αβ .

Hence ξ = 0 (x =∞) is a regular singular point. The indicial equation is

r(r − 1) + (1− α− β)r + αβ = 0 ,

or r2 − (α+ β)r + αβ = 0 . Evidently, the roots are r = α and r = β .

15.(a) If we write the D.E. in the standard form as given in Theorem 5.6.1, we obtain

x2y′′ + x[α/x]y′ + [β/x]y = 0 where xp(x) = α/x and x2g(x) = β/x. Neither ofthese terms are analytic at x = 0, so x = 0 is an irregular singular point.

(b) Substituting y = xr∑∞n=0 anx

n in x3y′′ + αxy′ + βy = 0 gives

∞∑n=0

(n+ r)(n+ r − 1)anxn+r+1 + α

∞∑n=0

(n+ r)anxn+r + β

∞∑n=0

anxn+r = 0.

Shifting the index in the first series and collecting coefficients of common powersof x, we obtain

(αr + β)a0xr +

∞∑n=1

{(n+ r − 1)(n+ r − 2)an−1 + [α(n+ r) + β]an}xn+r = 0.

Thus the indicial equation is αr + β = 0 with the single root r = −β/α.

(c) From part (b), the recurrence relation is

an = − (n+ r − 1)(n+ r − 2)an−1α(n+ r) + β

= −

(n− β

α − 1)(

n− βα − 2

)an−1

αn,

n = 1, 2, . . . for r=−β/α. For β/α=−1, an = −n(n− 1)an−1/αn, so that a1 =0 ·a0 = 0. Since all other an are multiples of a1, and hence are zero, y(x) = x isthe solution. Similarly for β/α = 0, an = −(n− 1)(n− 2)an−1/αn and again forn = 1, a1 = 0 and y(x) = 1 is the solution. Continuing in this fashion, we see thatthe series solution will terminate for β/α any positive integer as well as 0 and −1.For other values of β/α, we have∣∣∣∣ anan−1

∣∣∣∣ =

(n− β

α − 1)(

n− βα − 2

)αn

,

which approaches ∞ as n→∞ and thus the ratio test yields a zero radius ofconvergence.


16.(a) Note that

p(x) =α

xsand q(x) =

β

xt.

It follows that

limx→0

x p(x) = limx→0

αx1−s and limx→0

x2q(x) = limx→0

β x2−t.

Hence if s > 1 or t > 2 , one or both of the limits does not exist. Therefore x = 0is an irregular singular point.


n+r in the differential equation gives

∞∑n=0

(n+ r)(n+ r − 1)anxn+r + α

∞∑n=0

(n+ r)anxn+r+1−s + β

∞∑n=0

anxn+r+2−t = 0.

If s = 2 and t = 2 the first term in each of the three series is r(r − 1)a0xr, αra0x

r−1,and βa0x

r, respectively. Thus the indicial equation is F (r) = αra0 = 0, whichrequires r = 0. Hence there is at most one solution of the assumed form.

(c) Let y = a0xr + a1x

r+1 + . . .+ anxr+n + . . . . Write the ODE as

x3y ′′ + αx2y ′ + β y = 0 .

Substitution of the assumed solution results in∞∑n=0

(n+ r)(n+ r − 1)anxn+r+1 + α

∞∑n=0

(n+ r)anxn+r+1 + β

∞∑n=0

anxn+r = 0.

Adjusting the indices, we obtain

∞∑n=1

(n− 1 + r)(n+ r − 2)an−1xn+r + α

∞∑n=1

(n− 1 + r)an−1xn+r+

+β

∞∑n=0

anxn+r = 0.

Combining the series,

β a0xr +

∞∑n=1

An xn+r = 0 ,

in which An = β an + (n− 1 + r)(n+ r + α− 2)an−1. Setting the coefficients equalto zero, we have a0 = 0 . But for n ≥ 1 ,

an =(n− 1 + r)(n+ r + α− 2)

βan−1.

Therefore, regardless of the value of r, it follows that an = 0 , for n = 1, 2, . . . .

(d) In order for the indicial equation to be quadratic in r it is necessary that thefirst term in the first series contribute to the indicial equation. This means thatthe first term in the second and the third series cannot have powers less than xr.The first terms are r(r − 1)a0x

r, αra0xr+1−s and βa0x

r+2−t, respectively. Thus ifs ≤ 1 and t ≤ 2, the quadratic term will appear in the indicial equation.

5.7 63

5.7

1. It is clear that x = 0 is a singular point. The differential equation is in the stan-

dard form given in Theorem 5.6.1 with xp(x) = 2 and x2q(x) = x. Both are analyticat x = 0, so x = 0 is a regular singular point. Substituting y =

∑∞n=0 anx

n+r inthe differential equation, shifting indices appropriately, and collecting coefficientsof like powers of x yields

[r(r − 1) + 2r]a0xr +

∞∑n=1

[(r + n)(r + n+ 1)an + an−1]xr+n = 0.

The indicial equation is F (r) = r(r + 1) = 0 with roots r1 = 0, r2 = −1. Treatingan as a function of r, we see that an(r) = −an−1(r)/F (r + n), n = 1, 2, . . . if F (r +n) 6= 0. Thus a1(r) = −a0/F (r + 1), a2(r) = a0/F (r + 1)F (r + 2), . . ., and an(r) =(−1)na0/F (r + 1)F (r + 2) . . . F (r + n), provided F (r + n) 6= 0 for n = 1, 2, . . .. Forthe case r1 = 0, we have an(0) = (−1)na0/F (1)F (2) . . . F (n) = (−1)na0/n!(n+ 1)!,so one solution is

y1(x) =

∞∑n=0

(−1)nxn/n!(n+ 1)!

where we have set a0 = 1. If we try to use the above recurrence relation for thecase r2 = −1, we find that an(−1) = −an−1/n(n− 1), which is undefined for n = 1.Thus we must follow the procedure described at the end of Section 5.6 to calculatea second solution of the form given in Equation (24). Specifically, we use Equations(19) and (20) of Section 5.6 to calculate a and cn(r2), where r2 = −1. Since r1 −r2 = 1 = N , we have aN (r) = a1(r) = −1/F (r + 1), with a0 = 1. Hence

a = limr→−1

[(r + 1)(−1)/F (r + 1)] = limr→−1

[−(r + 1)/(r + 1)(r + 2)] = −1.

Next,

cn(−1) =d

dr[(r + 1)an(r)]|r=−1 = (−1)n

d

dr

[r + 1

F (r + 1) . . . F (r + n)

]|r=−1

where we again have set a0 = 1. Observe that

(r+1)/F (r+1) . . . F (r+n) = 1/[(r+2)2(r+3)2 . . . (r+n)2(r+n+1)] = 1/Gn(r).

Hence cn(−1)=(−1)n+1G′n(−1)/G2n(−1). ThenGn(−1)=12 · 22 · 32 . . . (n− 1)2n =

(n− 1)!n! and G′n(−1)/Gn(−1)=2[1/1+1/2+1/3+. . .+ 1/(n− 1)] + 1/n = Hn +Hn−1. Thus cn(−1) = (−1)n+1(Hn +Hn−1)/(n− 1)!n!. From Equation (24) ofSection 5.6 we obtain the second solution

y2(x) = −y1(x) lnx+ x−1

[1−

∞∑n=1

(−1)n(Hn +Hn−1)xn/n!(n− 1)!

].


2. It is clear that x = 0 is a singular point. The D.E. is in the standard form given

in Theorem 5.6.1 with xp(x) = 3 and x2q(x) = 1 + x. Both are analytic at x = 0,so x = 0 is a regular singular point. Substituting

y =

∞∑n=0

anxn+r

in the D.E., shifting indices appropriately, and collecting coefficients of like powersof x yields

[r(r − 1) + 3r + 1]a0xr +

∞∑n=1

{[(r + n)(r + n+ 2) + 1]an + an−1}xn+r = 0.

The indicial equation is F (r) = r2 + 2r + 1 = (r + 1)2 = 0 with the double rootr1 = r2 = −1. Treating an as a function of r, we can see immediately that an(r) =−an−1(r)/F (r + n), n = 1, 2, . . .. Thus we obtain that a1(r) = −a0/F (r + 1), thena2(r) = a0/F (r + 1)F (r + 2), . . . , an(r) = (−1)na0/F (r + 1)F (r + 2) . . . F (r + n).Setting r = −1, we find that an(−1) = (−1)na0/(n!)2, n = 1, 2, . . .. Hence onesolution is

y1(x) = x−1∞∑n=0

(−1)nxn/(n!)2

where we have set a0 = 1. To find a second solution we follow the procedure des-cribed in Section 5.6 for the case when the roots of the indicial equation are equal.Specifically, the second solution will have the form given in Eq. (17) of that sec-tion. We must calculate a′n(−1). Let us denote Gn(r) = F (r + 1) . . . F (r + n) =(r + 2)2(r + 3)2 . . . (r + n+ 1)2 and take a0 = 1, then a′n(−1) = (−1)n/[1/Gn(r)]′

evaluated at r = −1. Hence a′n(−1) = (−1)n+1G′n(−1)/G2n(−1). But Gn(−1) =

(n!)2 and G′n(−1)/Gn(−1) = 2[1/1 + 1/2 + 1/3 + . . .+ 1/n] = 2Hn. Thus a secondsolution is

y2(x) = y1(x) lnx− 2x−1∞∑n=1

(−1)nHnxn/(n!)2.

3. Here x p(x) = 1 and x2q(x) = 2x , which are both analytic everywhere. We sety = xr(a0 + a1x+ a2x

2 + . . .+ anxn + . . .). Substitution into the ODE results in

∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)an xr+n + 2

∞∑n=0

anxr+n+1 = 0 .

After adjusting the indices in the last series, we obtain

a0 [r(r − 1) + r]xr +

∞∑n=1

[(r + n)(r + n− 1)an + (r + n)an + 2 an−1]xr+n = 0.

Assuming a0 6= 0 , the indicial equation is r2 = 0 , with double root r = 0 . Settingthe remaining coefficients equal to zero, we have for n ≥ 1 ,

an(r) = − 2

(n+ r)2an−1(r) .

5.7 65

It follows that

an(r) =(−1)n 2n

[(n+ r)(n+ r − 1) . . . (1 + r)]2 a0 , n ≥ 1 .

Since r = 0 , one solution is given by

y1(x) =

∞∑n=0

(−1)n 2n

(n!)2xn .

For a second linearly independent solution, we follow the discussion in Section 5.6.First note that

a ′n(r)

an(r)= −2

[1

n+ r+

1

n+ r − 1+ . . .+

1

1 + r

].

Setting r = 0 ,

a ′n(0) = −2Hn an(0) = −2Hn(−1)n 2n

(n!)2.

Therefore,

y2(x) = y1(x) ln x− 2

∞∑n=0

(−1)n 2nHn

(n!)2xn .

4. Since x = 0 is a regular singular point, substitute

y =

∞∑n=0

anxn+r

in the differential equation, shift indices appropriately, and collect coefficients oflike powers of x to obtain

[r2 − 9/4]a0xr + [(r + 1)2 − 9/4]a1x

r+1+

+

∞∑n=2

{[(r + n)2 − 9/4]an + an−2

}xn+r = 0.

The indicial equation is F (r) = r2 − 9/4 = 0 with roots r1 = 3/2, r2 = −3/2. Treat-ing an as a function of r, we see that an(r) = −an−2(r)/F (r + n), n = 2, 3, . . . ifF (r + n) 6= 0. For the case r1 = 3/2, F (r1 + 1), which is the coefficient of xr1+1,is 6= 0 so we must set a1 = 0. It follows that a3 = a5 = . . . = 0. For the even coef-ficients, set n = 2m so a2m(3/2) = a2m−2(3/2)/F (3/2 + 2m) = −a2m−2/22m(m+3/2), m = 1, 2, . . .. Thus a2(3/2) = −a0/22 · 1(1 + 3/2), a4(3/2) = a0/2

4 · 2!(1 +3/2)(2 + 3/2), . . ., and a2m(3/2) = (−1)m/22mm!(1 + 3/2) . . . (m+ 3/2). Hence onesolution is

y1(x) = x3/2

[1 +

∞∑m=1

(−1)m

m!(1 + 3/2)(2 + 3/2) . . . (m+ 3/2)

(x2

)2m],

where we have set a0 = 1. For this problem, the roots r1 and r2 of the indicialequation differ by an integer: r1 − r2 = 3. Hence we can anticipate that there maybe difficulty in calculating a second solution corresponding to r = r2. This diffi-culty will occur in calculating a3(r) = −a1(r)/F (r + 3) since when r = r2 = −3/2we have F (r2 + 3) = F (r1) = 0. However, in this problem we are fortunate because


a1 = 0 and it will not be necessary to use the theory described at the end of Sec-tion 5.6. Notice for r = r2 − 3/2 that the coefficient of xr2+1 is [(r2 + 1)2 − 9/4]a1,which does not vanish unless a1 = 0. Thus the recurrence relation for the oddcoefficients yields a5 = −a3/F (7/2), a7 = −a5/F (11/2) = a3/F (11/2)F (7/2), andso forth. Substituting these terms into the assumed form, we see that a multipleof y1(x) has been obtained and thus we may take a3 = 0 without loss of general-ity. Hence a3 = a5 = a7 = . . . = 0. The even coefficients are given by a2m(−3/2) =−a2m−2(−3/2)/F (2m− 3/2), m = 1, 2, . . .. Thus a2(−3/2)=−a0/22 · 1 · (1−3/2),a4(−3/2) = a0/2

4 · 2!(1− 3/2)(2− 3/2), . . ., and a2m(−3/2)=(−1)ma0/22mm!(1−

3/2)(2− 3/2) . . . (m− 3/2). Thus a second solution is

y2(x) = x−3/2

[1 +

∞∑m=1

(−1)m

m!(1− 3/2)(2− 3/2) . . . (m− 3/2)

(x2

)2m].

5. Let y(x) = v(x)/√x . Then y ′ = x−1/2 v ′ − x−3/2 v/2 and y ′′ = x−1/2 v ′′ −

x−3/2 v ′ + 3x−5/2 v/4 . Substitution into the ODE results in[x3/2 v ′′ − x1/2 v ′ + 3x−1/2 v/4

]+[x1/2 v ′ − x−1/2 v/2

]+ (x2 − 1

4)x−1/2 v = 0 .

Simplifying, we find thatv ′′ + v = 0 ,

with general solution v(x) = c1 cos x+ c2 sin x . Hence

y(x) = c1x−1/2 cos x+ c2 x

−1/2 sin x .


limm→∞

|(−1)m+1x2m+2/22m+2[(m+ 1)!]2||(−1)mx2m/22m(m!)2|

= |x2| limm→∞

1

22(m+ 1)2= 0

for every x. Thus the series for J0(x) converges absolutely for all x.

7. The absolute value of the ratio of consecutive terms is∣∣∣∣a2m+2 x2m+2

a2m x2m

∣∣∣∣ =|x|2m+2

22m(m+ 1)!m!

|x|2m 22m+2(m+ 2)!(m+ 1)!=

|x|2

4(m+ 2)(m+ 1).


limm→∞

∣∣∣∣a2m+2 x2m+2

a2m x2m

∣∣∣∣ = limm→∞

|x|2

4(m+ 2)(m+ 1)= 0 .

Hence the series for J1(x) converges absolutely for all values of x . Furthermore,since the series for J0(x) also converges absolutely for all x, term-by-term differen-tiation results in

J ′0(x) =

∞∑m=1

(−1)mx2m−1

22m−1m!(m− 1)!=

∞∑m=0

(−1)m+1 x2m+1

22m+1(m+ 1)!m!=

= −x2

∞∑m=0

(−1)m x2m

22m(m+ 1)!m!.

Therefore, J ′0(x) = −J1(x) .

5.7 67

8.(a) Note that x p(x) = 1 and x2q(x) = x2 − ν2, which are both analytic at x = 0 .Thus x = 0 is a regular singular point. Furthermore, p0 = 1 and q0 = −ν2. Hencethe indicial equation is r2 − ν2 = 0 , with roots r1 = ν and r2 = −ν .


n + . . .). Substitution into the ODEresults in∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)an xr+n

+

∞∑n=0

anxr+n+2 − ν2

∞∑n=0

anxr+n = 0 .


a0[r(r − 1) + r − ν2

]xr + a1

[(r + 1)r + (r + 1)− ν2

]+

∞∑n=2

[(r + n)(r + n− 1)an + (r + n)an − ν2an + an−2

]xr+n = 0.

Setting the coefficients equal to zero, we find that a1 = 0 , and

an =−1

(r + n)2 − ν2an−2 ,

for n ≥ 2 . It follows that a3 = a5 = . . . = a2m+1 = . . . = 0 . Furthermore, withr = ν ,

an =−1

n(n+ 2ν)an−2 .

So for m = 1, 2, . . . ,

a2m =−1

2m(2m+ 2ν)a2m−2 =

(−1)m

22mm!(1 + ν)(2 + ν) . . . (m− 1 + ν)(m+ ν)a0 .


y1(x) = xν

[1 +

∞∑m=1

(−1)m

m!(1 + ν)(2 + ν) . . . (m− 1 + ν)(m+ ν)(x

2)2m

].

(c) Assuming that r1 − r2 = 2ν is not an integer, simply setting r = −ν in theabove results in a second linearly independent solution

y2(x) = x−ν

[1 +

∞∑m=1

(−1)m

m!(1− ν)(2− ν) . . . (m− 1− ν)(m− ν)(x

2)2m

].

(d) The absolute value of the ratio of consecutive terms in y1(x) is∣∣∣∣a2m+2 x2m+2

a2m x2m

∣∣∣∣ =|x|2m+2

22mm!(1 + ν) . . . (m+ ν)

|x|2m 22m+2(m+ 1)!(1 + ν) . . . (m+ 1 + ν)

=|x|2

4(m+ 1)(m+ 1 + ν).



limm→∞

∣∣∣∣a2m+2 x2m+2

a2m x2m

∣∣∣∣ = limm→∞

|x|2

4(m+ 1)(m+ 1 + ν)= 0 .

Hence the series for y1(x) converges absolutely for all values of x . The same canbe shown for y2(x) . Note also, that if ν is a positive integer, then the coefficientsin the series for y2(x) are undefined.

9.(a) It suffices to calculate L [J0(x) ln x]. Indeed,

[J0(x) ln x]′

= J ′0(x) ln x+J0(x)

x

and

[J0(x) ln x]′′

= J ′′0 (x) ln x+ 2J ′0(x)

x− J0(x)

x2.

Hence

L [J0(x) ln x] = x2J ′′0 (x) ln x+ 2xJ ′0(x)− J0(x)

+ xJ ′0(x) ln x+ J0(x) + x2J0(x) ln x .

Since x2J ′′0 (x) + xJ ′0(x) + x2J0(x) = 0 ,

L [J0(x) ln x] = 2xJ ′0(x) .

(b) Given that L [y2(x)] = 0 , after adjusting the indices in part (a), we have

b1x+ 22b2 x2 +

∞∑n=3

(n2bn + bn−2)xn = −2xJ ′0(x) .

Using the series representation of J ′0(x) in Problem 8 ,

b1x+ 22b2 x2 +

∞∑n=3

(n2bn + bn−2)xn = −2

∞∑n=1

(−1)n(2n)x2n

22n(n!)2.

(c) Equating the coefficients on both sides of the equation, we find that

b1 = b3 = . . . = b2m+1 = . . . = 0 .

Also, with n = 1, 22b2 = 1/(1!)2, that is, b2 = 1/[22(1!)2

]. Furthermore, for m ≥ 2,

(2m)2b2m + b2m−2 = −2(−1)m(2m)

22m(m!)2.

More explicitly,

b4 = − 1

22 42(1 +

1

2)

b6 =1

22 42 62(1 +

1

2+

1

3)

...

5.7 69

It can be shown, in general, that

b2m = (−1)m+1 Hm

22m(m!)2.

10. Bessel’s equation of order one is

x2 y ′′ + x y ′ + (x2 − 1)y = 0 .

Based on Problem 9, the roots of the indicial equation are r1 = 1 and r2 = −1 .Set y = xr(a0 + a1x+ a2x

2 + . . .+ anxn + . . .). Substitution into the ODE results

in∞∑n=0

(r + n)(r + n− 1)an xr+n +

∞∑n=0

(r + n)an xr+n

+

∞∑n=0

anxr+n+2 −

∞∑n=0

anxr+n = 0 .


a0 [r(r − 1) + r − 1]xr + a1 [(r + 1)r + (r + 1)− 1]

+

∞∑n=2

[(r + n)(r + n− 1)an + (r + n)an − an + an−2]xr+n = 0.

Setting the coefficients equal to zero, we find that a1 = 0 , and

an(r) =−1

(r + n)2 − 1an−2(r) =

−1

(n+ r + 1)(n+ r − 1)an−2(r),

for n ≥ 2 . It follows that a3 = a5 = . . . = a2m+1 = . . . = 0. Solving the recurrencerelation,

a2m(r) =(−1)m

(2m+ r + 1)(2m+ r − 1)2 . . . (r + 3)2(r + 1)a0.

With r = r1 = 1,

a2m(1) =(−1)m

22m(m+ 1)!m!a0.

For a second linearly independent solution, we follow the discussion in Section 5.6.Since r1 − r2 = N = 2 , we find that

a2(r) = − 1

(r + 3)(r + 1),

with a0 = 1 . Hence the leading coefficient in the solution is

a = limr→−1

(r + 1) a2(r) = −1

2.

Further,

(r + 1) a2m(r) =(−1)m

(2m+ r + 1) [(2m+ r − 1) . . . (3 + r)]2 .


Let An(r) = (r + 1) an(r) . It follows that

A ′2m(r)

A2m(r)= − 1

2m+ r + 1− 2

[1

2m+ r − 1+ . . .+

1

3 + r

].

Setting r = r2 = −1 , we calculate

c2m(−1) = −1

2(Hm +Hm−1)A2m(−1)

= −1

2(Hm +Hm−1)

(−1)m

2m [(2m− 2) . . . 2]2 = −1

2(Hm +Hm−1)

(−1)m

22m−1m!(m− 1)!.

Note that a2m+1(r) = 0 implies that A2m+1(r) = 0 , so

c2m+1(−1) =

[d

drA2m+1(r)

]r=r2

= 0 .

Therefore,

y2(x) = −1

2

[x

∞∑m=0

(−1)m

(m+ 1)!m!(x

2)2m

]ln x

+1

x

[1−

∞∑m=1

(−1)m(Hm +Hm−1)

m!(m− 1)!(x

2)2m

].

Based on the definition of J1(x),

y2(x) = −J1(x) ln x+1

x

[1−

∞∑m=1

(−1)m(Hm +Hm−1)

m!(m− 1)!(x

2)2m

].

11. Consider a solution of the form y(x) =√x f(αxβ). Then

y ′ =df

dξ· αβ x

β

√x

+f(ξ)

2√x

in which ξ = αxβ . Hence

y ′′ =d2f

dξ2· α

2β2 x2β

x√x

+df

dξ· αβ

2 xβ

x√x− f(ξ)

4x√x

,

and

x2 y ′′ = α2β2 x2β√xd2f

dξ2+ αβ2 xβ

√xdf

dξ− 1

4

√x f(ξ) .


α2β2 x2βd2f

dξ2+ αβ2 xβ

df

dξ− 1

4f(ξ) + (α2β2 x2β +

1

4− ν2β2)f(ξ) = 0 .

Simplifying, and setting ξ = αxβ , we find that

ξ2d2f

dξ2+ ξ

df

dξ+ (ξ2 − ν2)f(ξ) = 0 , (∗)

5.7 71

which is a Bessel equation of order ν . Therefore, the general solution of the givenODE is

y(x) =√x[c1 f1(αxβ) + c2 f2(αxβ)

],

in which f1(ξ) and f2(ξ) are the linearly independent solutions of (∗).

12. To compare y′′ − xy = 0 with the differential equation of Problem 12, we must

multiply by x2 to get x2y′′−x3y = 0. Thus 2β=3, α2β2 = −1 and 1/4− ν2β2 =0.Hence β = 3/2, α = 2i/3, and ν2 = 1/9, which yields the desired result.

13. First we verify that J0(λjx) satisfies the D.E. We know that J0(t) is a solutionof the Bessel equation of order zero:

t2J ′′0 (t) + tJ ′0(t) + t2J0(t) = 0 or

J ′′0 (t) + t−1J ′0(t) + J0(t) = 0.

Let t = λjx. Then

d

dxJ0(λjx) =

d

dtJ0(t)

dt

dx= λjJ

′0(t),

d2

dx2J0(λjx) = λj

d

dt[J ′0(t)]

dt

dx= λ2jJ

′′0 (t).

Substituting y = J0(λjx) in the given D.E. and making use of these results, we have

λ2jJ′′0 (t) + (λj/t)λjJ

′0(t) + λ2jJ0(t) = λ2j

[J ′′0 (t) + t−1J ′0(t) + J0(t)

]= 0.

Thus y = J0(λjx) is a solution of the given D.E. For the second part of the problemwe follow the hint. First, rewrite the D.E. by multiplying by x to yield xy′′ + y′ +λ2jxy = 0, which can be written as (xy′)′ = −λ2jxy. Now let yi(x) = J0(λix) and

yj(x) = J0(λjx) and we have, respectively, (xy′i)′ = −λ2ixyi and (xy′j)

′ = −λ2jxyj .Now, multiply the first equation by yj , the second by yi, integrate each from 0 to1, and subtract the second from the first:∫ 1

0

[yj (xy′i)

′ − yi(xy′j)′]

dx = −(λ2i − λ2j

) ∫ 1

0

xyiyjdx.

If we integrate each term on the left side once by parts and note that yi = yj = 0and x = 1, we find that the left side of this equation is identically zero. Hence theright side is identically zero and for λi 6= λj this gives the desired result.

series solutions of second order linear equations

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