Chapter 32: Modiﬁed Power Series Solutions and the...

Worked Solutions 295

Chapter 32: Modified Power Series Solutions and the Basic Method of Frobenius

32.2 a. Let y = y(x) = (x − 3)r . Then

(x − 3)2 y′′ − 2(x − 3)y′ + 2y = 0

→֒ (x − 3)2[

(x − 3)r]′′ − 2(x − 3)

[

(x − 3)r]′ + 2

[

(x − 3)r]

= 0

→֒ (x − 3)2[

r(r − 1)(x − 3)r−2]′′

− 2(x − 3)

[

r(x − 3)r−1]′

+ 2[

(x − 3)r]

= 0

→֒ (x − 3)r [r(r − 1) − 2r + 2] = 0 .

Dividing out the (x − 3)r and simplifying the rest yields

r2 − 3r + 2 = 0 ,

which factors to

(r − 2)(r − 1) = 0 ,

Hence, r = 2 or r = 1 , and a fundamental set of solutions is {y1, y2} with

y1(x) = (x − 3)2 and y1(x) = (x − 3)1 = x − 3 .

32.2 c. Letting y = y(x) = (x − 1)r ,

0 = (x − 1)2 y′′ − 5(x − 1)y′ + 9y

= (x − 1)2[

(x − 1)r]′′ − 5(x − 1)

[

(x − 1)r]′ + 9

[

(x − 1)r]

y

= (x − 1)2[

r(r − 1)(x − 1)r−2]′′

− 5(x − 1)[

r(x − 1)r−1]′

+ 9[

(x − 3)r]

= (x − 1)r [r(r − 1) − 5r + 9] .

Dividing out the (x − 1)r leaves us with

0 = r(r − 1) − 5r + 9 = r2 − 6r + 9 = (r − 3)3 .

Thus, r = 3 is the only solution to the indicial equation, and a fundamental pair of solutions

is given by

y1(x) = (x − 1)3 and y2(x) = (x − 1)3 ln |x − 1| .

32.2 e. Letting y = y(x) = (x − 5)r ,

0 = 3(x − 5)2 y′′ − 4(x − 5)y′ + 2y

= 3(x − 5)2[

(x − 5)r]′′ − 4(x − 5)

[

(x − 5)r]′ + 2

[

(x − 5)r]

= 3(x − 5)2[

r(r − 1)(x − 5)r−2]′′

− 4(x − 5)[

r(x − 5)r−1]′

+ 2[

(x − 5)r]

= (x − 5)r [3r(r − 1) − r4 + 2]︸︷︷︸

3r2−7r+2

.

296 Modified Power Series Solutions and the Basic Method of Frobenius

Dividing out the (x − 5)r leaves us with

0 = 3r2 − 7r + 2 .

Hence,

r = −[−7] ±√

(−7)2 − 4 · 3 · 2

2 · 3= 7 ± 5

6 r = 2 or r = 1

3.

So a fundamental pair of solutions is given by

y1(x) = (x − 5)2 and y2(x) = (x − 5)1/3 = 3

√x − 5 .

32.3 a. After multiplying through by x − 2 and by x + 2 , the differential equation becomes

x2(x − 2)(x + 2)︸︷︷︸

a(x)

y′′ + x(x + 2)︸︷︷︸

b(x)

y′ + 2(x − 2)︸︷︷︸

c(x)

y = 0 . (⋆)

Each coefficient is a polynomial; there are no factors common to all three, and the first is 0

if and only if x = 0 , x = 2 or x = −2 . So these values of x are the singular points for

the differential equation.

To determine whether each singular point is regular or not, we can either see if we can

rewrite the equation in quasi-Euler form about that point or use the test from theorem 32.2

on page 674. We’ll illustrate the use of both approaches.

For the singular point x0 = 0 , let us simply note that equation (⋆) is

(x − 0)2 (x − 2)(x + 2)︸︷︷︸

α(x)

y′′ + (x − 0) (x + 2)︸︷︷︸

β(x)

y′ + 2(x − 2)︸︷︷︸

γ (x)

y = 0 .

with

α(0) = (0 − 2)(0 + 2) 6= 0 .

This is a quasi-Euler form about x0 = 0 , and, so, x0 is a regular singular point.

For the other two singular points, we will apply theorem 32.2, which involves the limits

limx→x0

(x − x0)b(x)

a(x)= lim

x→x0

(x − x0)x(x + 2)

x2(x − 2)(x + 2)= lim

x→x0

x − x0

x(x − 2)

and

limx→x0

(x − x0)2 c(x)

a(x)= lim

x→x0

(x − x0)2 2(x − 2)

x2(x − 2)(x + 2)= lim

x→x0

2(x − x0)2

x2(x + 2)

with x0 being the singular point in question.

For x0 = 2 , we have

limx→2

(x − 2)b(x)

a(x)= lim

x→2

x − 2

x(x − 2)= lim

x→2

1

x= 1

2

and

limx→2

(x − 2)2 c(x)

a(x)= lim

x→2

2(x − 2)2

x2(x + 2)= 0 .

Since both limits are finite, theorem 32.2 assures us that singular point x0 = 2 is regular.

For x0 = −2 , we have

limx→−2

(x − [−2])b(x)

a(x)= lim

x→−2

x + 2

x(x − 2)= 0


and

limx→−2

(x − [−2])2 c(x)

a(x)= lim

x→−2

2(x + 2)2

x2(x + 2)= lim

x→−2

2(x + 2)

x2= 0 .

Again, since both limits are finite, theorem 32.2 assures us that singular point x0 = −2 is

regular.

So all three singular points 0 , 2 and −2 are regular singular points. The closest to 0

other than 0 , itself, is either ±2 . Hence, the Frobenius radius about 0 is

R = |0 − [±2]| = 2 .

32.3 c. Since all three coefficients are polynomials and clearly without common factors, each

singular point x0 where the first coefficient is 0 :

0 = x3 − x4 = x3(1 − x) x0 = 0 or x0 = 1 .

To apply theorem 32.2, we need

limx→x0

(x − x0)b(x)

a(x)= lim

x→x0

(x − x0)3x − 1

x3 − x4= lim

x→x0

(x − x0)(3x − 1)

x3(1 − x)

and

limx→x0

(x − x0)2 c(x)

a(x)= lim

x→x0

(x − x0)2 827

x3 − x4= lim

x→x0

827(x − x0)2

x3(1 − x).

For x0 = 0 ,

limx→0

(x − 0)b(x)

a(x)= lim

x→0

(x − 0)(3x − 1)

x3(1 − x)= lim

x→0

(3x − 1)

x2(1 − x),

which is not finite. So, x0 is an irregular singular point.

For x0 = 1 ,

limx→1

(x − 1)b(x)

a(x)= lim

x→1

(x − 1)(3x − 1)

x3(1 − x)= lim

x→1

−(3x − 1)

x3= −2

and

limx→1

(x − 1)2 c(x)

a(x)= lim

x→1

827(x − 1)2

x3(1 − x)= lim

x→1

−827(x − 1)

x3= 0 .

Since both limits are finite, we know x0 = 1 is a regular singular point.

The closest singular point to 1 other than 1 , itself, is 0 . So the Frobenius radius about

x0 = 1 is

R = |1 − 0| = 1 .

32.3 e. By a number of arguments, it should be clear that x0 = 3 and x0 = 4 are the singular

points. For applying theorem 32.2, we have

limx→x0

(x − x0)b(x)

a(x)= lim

x→x0

(x − x0)1

(x − 3)2= lim

x→x0

x − x0

(x − 3)2

and

limx→x0

(x − x0)2 c(x)

a(x)= lim

x→x0

(x − x0)2 1

(x − 4)2= lim

x→x0

(x − x0)2

(x − 4)2.

For x0 = 3 ,

limx→3

(x − 3)b(x)

a(x)= lim

x→3

x − 3

(x − 3)2= lim

x→3

1

(x − 3),


which is not finite, telling us that x0 = 3 is an irregular singular point.

For x0 = 4 ,

limx→4

(x − 4)b(x)

a(x)= lim

x→4

x − 4

(x − 3)2= 0

and

limx→4

(x − 4)2 c(x)

a(x)= lim

x→4

(x − 4)2

(x − 4)2= lim

x→41 = 1 .

Both limits are finite, telling us that x0 = 4 is a regular singular point.

Since the closest singular point to 4 other than 4 , itself, is 3 , the Frobenius radius about

x0 = 4 is

R = |4 − 3| = 1 .

32.3 g. After multiplying through by x and by 1 + x2 , the differential equation becomes

A(x)y′′ + B(x)y′ + C(x)y = 0 (⋆)

where A , B and C are the polynomials

A(x) = x(

1 + x2) (

4x2 − 1)

,

B(x) =(

1 + x2)

(4x − 2)

and

C(x) = x(

1 − x2)

,

which, after using a little algebra, can be written in fully-factored form

A(x) = 4x(x − i)(x + i)(

x − 1

2

) (

x + 1

2

)

,

B(x) = 4(x − i)(x + i)(

x − 1

2

)

and

C(x) = x(x − 1)(x + 1) .

Since no factor is shared by all these polynomials, the singular points are the values of x for

which A(x) = 0 ; namely,

0 , i , − i ,1

2and − 1

2.

Rather than apply theorem 32.2, let us multiply equation (⋆) by x again, obtaining,

x A(x)y′′ + x B(x)y′ + xC(x)y = 0 ,

which is

x2α(x)y′′ + xβ(x)y′ + γ (x)y = 0 (⋆⋆)

with

x A(x) = x2(

1 + x2) (

4x2 − 1)

︸︷︷︸

α(x)

,

x B(x) = x(

1 + x2)

(4x − 2)︸︷︷︸

β(x)


and

xC(x) = x2(

1 − x2)

︸︷︷︸

γ (x)

.

Noting that

α(0) = · · · = −1 6= 0 ,

we see that equation (⋆⋆) is our differential equation in quasi-Euler form about x0 = 0 . So,

x0 = 0 is a regular singular point.

Similar computations/arguments, will show that the other singular points are also regular.

Alternatively, you can apply theorem 32.2 to verify that each singular point is regular.

Since the closest singular points to 0 other than 0, itself, are ± 1

2, the Frobenius radius

of convergence about 0 is

F =∣∣∣0 −

[

±1

2

]∣∣∣ = 1

2.

32.4 a. The equation is already in desired form.

Since x0 = 0 , we set

y(x) = (x − 0)r

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .

Differentiating this and plugging the results into the differential equation, we get

y′ = d

dx

∞∑

k=0

ak xk+r =∞∑

k=0

d

dx

[

ak xk+r]

=∞∑

k=0

ak(k + r)xk+r−1 ,

y′′ = d

dx

∞∑

k=0

ak(k + r)xk+r−1

=∞∑

k=0

d

dx

[

ak(k + r)xk+r−1]

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 = x2 y′′ − 2xy′ +(

x2 + 2)

y

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 − 2x

∞∑

k=0

ak(k + r)xk+r−1

+(

x2 + 2) ∞

∑

k=0

ak xk+r

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 − 2x

∞∑

k=0

ak(k + r)xk+r−1

+ x2∞∑

k=0

ak xk+r + 2

∞∑

k=0

ak xk+r


=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

(−2)ak(k + r)xk+r

+∞∑

k=0

ak xk+r+2 +∞∑

k=0

2ak xk+r .

Dividing out the xr and continuing:

0 =∞∑

k=0

ak(k + r)(k + r − 1)xk +∞∑

k=0

(−2)ak(k + r)xk +∞∑

k=0

ak xk+2 +∞∑

k=0

2ak xk

=∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

(−2)ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

ak xk+2

︸︷︷︸

n=k+2

+∞∑

k=0

2ak xk

︸︷︷︸

n=k

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

(−2)an(n + r)xn +∞∑

n=2

an−2xn +∞∑

n=0

2anxn

=[

a0(0 + r)(0 + r − 1)x0 + a1(1 + r)(1 + r − 1)x1 +∞∑

n=2

an(n + r)(n + r − 1)xn

]

+[

− 2a0(0 + r)x0 − 2a1(1 + r)x1 +∞∑

n=2

(−2)an(n + r)xn

]

+∞∑

n=2

an−2xn

+[

2a0x0 + 2a1x1 +∞∑

n=2

2anxn

]

= a0 [r(r − 1) − 2r + 2] x0 + a1 [(1 + r)r − 2(1 + r) + 2] x1

+∞∑

n=2

[

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

]

xn .

Simplifying slightly, we now have

0 = a0

[

r2 − 3r + 2]

x0 + a1

[

r2 − r]

x1

+∞∑

n=2

[

an

[

(n + r)2 − 3(n + r) + 2]

+ an−2

]

xn .(⋆)

Each term on the right side of equation (⋆) must be zero. In particular, the first term must

be zero no matter what the arbitrary constant a0 is. This yields the indicial equation

r2 − 3r + 2 = 0 ,

which factors to

(r − 2)(r − 1) = 0 .

Hence, the equation’s exponents are

r1 = 2 and r2 = 1 .


Plugging the larger exponent, r = r1 = 2 , into equation (⋆) (and simplifying the terms

in the series a little more):

0 = a0

[

22 − 3 · 2 + 2]

x0 + a1

[

22 − 2]

x1

+∞∑

n=2

[

an

[

(n + 2)2 − 3(n + 2) + 2︸︷︷︸

(

(n+2)−2)(

(n+2)−1)

]

+ an−2

]

xn

= a0[0]x0 + 2a1x1 +∞∑

n=2

[

ann(n + 1) + an−2

]

xn .

Since each term must be zero, we have

2a1 = 0 a1 = 0

and, for n ≥ 2 ,

ann(n + 1) + an−2 = 0 an = −1

n(n + 1)an−2 .

So the recursion formula when r = 2 is

ak = −1

k(k + 1)ak−2 for k ≥ 2 .

Applying the above:

a2 = −1

2(2 + 1)a2−2 = −1

3 · 2a0 ,

a3 = −1

3(3 + 1)a3−2 = −1

4 · 3a1 = −1

4 · 3· 0 = 0 ,

a4 = −1

4(4 + 1)a4−2 = −1

5 · 4a2 = −1

5 · 4· −1

3 · 2a0 = (−1)2

5!a0 ,

a5 = −1

5(5 + 1)a5−2 = −1

6 · 5a3 = −1

6 · 5· 0 = 0 ,

a6 = −1

6(6 + 1)a6−2 = −1

7 · 6a4 = −1

7 · 6· (−1)2

5!a0 = (−1)3

7!a0 ,

...

In general, ak = 0 if k is odd, and

a2m = (−1)m

(2m + 1)!a0 for m = 1, 2, 3, . . . .

And since

a2·0 = (−1)0

(2 · 0 + 1)!a0 = 1

1!a0 = a0 ,

the above formula for a2m also holds for m = 0 . So, going back to the original series

formula for y , we have

y(x) = xr1

∞∑

k=0

ak xk

= x2[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 · · ·]


= x2[

a0 + 0x − 1

3!a0x2 + 0x3 + 1

5!a0x4 + 0x5 − 1

7!a0x6 + · · ·

]

= a0x2[

1 − 1

3!x2 + 1

5!x4 − 1

7!x6 + · · ·

]

That is, y(x) = ay1(x) where a is an arbitrary constant, and

y1(x) = x2[

1 − 1

3!x2 + 1

5!x4 − 1

7!x6 + · · ·

]

= x2∞∑

m=0

(−1)m

(2m + 1)!x2m .

Note that, in fact,

y1(x) = x2∞∑

m=0

(−1)m

(2m + 1)!x2m = x

∞∑

m=0

(−1)m

(2m + 1)!x2m+1 = x sin(x) .

Next, we plug the smaller exponent, r = r2 = 1 into equation (⋆) (and further simplify

the terms):

0 = a0

[

r2 − 3r + 2]

x0 + a1

[

r2 − r]

x1

+∞∑

n=2

[

an

[

(n + r)2 − 3(n + r) + 2]

+ an−2

]

xn

= a0

[

12 − 3 · 1 + 2]

x0 + a1

[

12 − 1]

x1

+∞∑

n=2

[

an

[

(n + 1)2 − 3(n + 1) + 2︸︷︷︸

(

(n+1)−2)(

(n+1)−1)

]

+ an−2

]

xn

= a0[0]x0 + a1[0]x1 +∞∑

n=2

[

an(n − 1)n + an−2

]

xn .

Thus,

0a0 = 0 , 0a1 = 0

and, for n ≥ 2 ,

an(n − 1)n + an−2 = 0 an = −1

n(n − 1)an−2 .

The first two equations tells us that both a0 and a1 can be arbitrary, but, following the advice

given in step 8 on page 685 we will keep a0 as arbitrary and set

a1 = 0 .

From above, we also have the recursion formula

ak = −1

k(k − 1)ak−2 for k = 2, 3, 4, 5, . . . .

Using the above,

a2 = −1

2(2 − 1)a2−2 = −1

2 · 1a0 ,

a3 = −1

3(3 − 1)a3−2 = −1

3 · 2a1 = −1

3 · 2· 0 = 0 ,


a4 = −1

4(4 − 1)a4−2 = −1

4 · 3a2 = −1

4 · 3· −1

2 · 1a0 = (−1)2

4!a0 ,

a5 = −1

5(4 − 1)a5−2 = −1

5 · 4a3 = −1

5 · 4· 0 = 0 ,

a6 = −1

6(6 − 1)a6−2 = −1

6 · 5a4 = −1

6 · 5· (−1)2

4!a0 = (−1)3

6!a0 ,

a7 = −1

7(6 − 1)a7−2 = −1

7 · 6a5 = −1

7 · 6· 0 = 0 ,

a8 = −1

8(8 − 1)a8−2 = −1

8 · 7a6 = −1

8 · 7· (−1)4

6!a0 = (−1)4

8!a0 ,

...

In general, ak = 0 if k is odd, and

a2m = (−1)m

(2m)!a0 for m = 0, 1, 2, 3, . . . .

Thus,

y(x) = xr1

∞∑

k=0

ak xk

= x1[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 · · ·]

= x[

a0 + 0x − 1

2!a0x2 + 0x3 + 1

4!a0x4 + 0x5 − 1

6!a0x6 + · · ·

]

= a0x[

1 − 1

2!x2 + 1

4!x4 − 1

6!x6 + · · ·

]

That is, y(x) = ay2(x) where a is an arbitrary constant, and

y1(x) = x[

1 − 1

2!x2 + 1

4!x4 − 1

6!x6 + · · ·

]

= x

∞∑

m=0

(−1)m

(2m)!x2m .

Note that, in fact,

y2(x) = x

∞∑

m=0

(−1)m

(2m)!x2m = x cos(x) .

Finally, the general solution to this differential equation is given by

y(x) = c1 y1(x) + c2 y2(x)

where y1 and y2 are as described above.

32.4 c. The equation is already in desired form.


y(x) = (x − 0)r

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2


and

0 = x2 y′′ + xy′ + (4x − 4)y

= x2 y′′ + xy′ + 4xy − 4y

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 + x

∞∑

k=0

ak(k + r)xk+r−1

+ 4x

∞∑

k=0

ak xk+r − 4

∞∑

k=0

ak xk+r

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

ak(k + r)xk+r

+∞∑

k=0

4ak xk+r+1 +∞∑

k=0

(−4)ak xk+r .


0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

4ak xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

(−4)ak xk

︸︷︷︸

n=k

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

an(n + r)xn

+∞∑

n=1

4an−1xn +∞∑

n=0

(−4)anxn

=[

a0(0 + r)(0 + r − 1)x0 +∞∑

n=1

an(n + r)(n + r − 1)xn

]

+[

a0(0 + r)x0 +∞∑

n=2

an(n + r)xn

]

+∞∑

n=1

4an−1xn

+[

− 4a0x0 +∞∑

n=1

(−4)anxn

]

= a0 [r(r − 1) + r − 4] x0

+∞∑

n=1

[

an [(n + r)(n + r − 1) + (n + r) − 4] + 4an−1

]

xn .


Simplifying slightly, we now have

0 = a0

[

r2 − 4]

x0 +∞∑

n=1

[

an

[

(n + r)2 − 4]

+ 4an−1

]

xn . (⋆)

Each term on the right side of equation (⋆) must be zero. In particular, the first term must

be zero no matter what the arbitrary constant a0 is. This yields the indicial equation

r2 − 4 = 0 ,

which factors to

(r − 2)(r + 2) = 0 .

Hence, the equation’s exponents are

r1 = 2 and r2 = −2 .

Plugging the larger exponent, r = r1 = 2 , into equation (⋆) (and simplifying the terms

in the series a little more):

0 = a0

[

22 − 4]

x0 +∞∑

n=1

[

an

[

(n + 2)2 − 4]

+ 4an−1

]

xn

= a0[0]x0 +∞∑

n=1

[

an(n + 4)n + 4an−1

]

xn .

Since each term must be zero, we have, for n ≥ 1 ,

an(n + 4)n + 4an−1 = 0 an = 4

(n + 4)nan−1 .


ak = −4

(k + 4)kak−1 for k ≥ 1 .

Applying the above:

a1 = −4

(1 + 4)1a1−1 = −4

5 · 1a0 ,

a2 = −4

(2 + 4)2a2−1 = −4

6 · 2a1 = −4

6 · 2· −4

5 · 1a0 = (−4)2

(6 · 5)(2 · 1)a0 ,

a3 = −4

(3 + 4)3a3−1 = −4

7 · 3a2 = −4

7 · 3· (−4)2

(6 · 5)(2 · 1)a0 = (−4)3

(7 · 6 · 5)(3 · 2 · 1)a0

= (−4)34 · 3 · 2 · 1

(7 · 6 · 5 · 4 · 3 · 2 · 1)(3 · 2 · 1)a0 = (−4)34!

7! 3!a0 ,

a4 = −4

(4 + 4)4a4−1 = −4

(4 + 4)4a3 = −4

8 · 4· (−4)34!

7! 3!)a0 = (−4)44!

8! 4!a0 ,

...

In general, you can verify that

ak = (−4)k4!(k + 4)! k!

a0 for k = 0, 1, 2, 3, . . . .


So,

y(x) = xr1

∞∑

k=0

ak xk

= x2[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 · · ·]

= x2

[

a0 + −4

5 · 1a0x + (−4)2

(6 · 5)(2 · 1)a0x2 + (−4)3

(7 · 6 · 5)(3 · 2 · 1)a0x3 + · · ·

]

= a0x2

[

1 − 4

5x + 42

(6 · 5)(2 · 1)x2 − (−4)3

(7 · 6 · 5)(3 · 2 · 1)x3 + · · ·

]

.

Equivalently, y(x) = ay1(x) where a is an arbitrary constant, and

y1(x) = x2

[

1 − 4

5x + 42

(6 · 5)(2 · 1)x2 − (−4)3

(7 · 6 · 5)(3 · 2 · 1)x3 + · · ·

]

= · · · = x2∞∑

k=0

(−4)k4!(k + 4)!(k!)

xk .

Next, we plug the smaller exponent, r = r2 = −2 into equation (⋆):

0 = a0

[

r2 − 4]

x0 +∞∑

n=1

[

an

[

(n + r)2 − 4]

+ 4an−1

]

xn

= a0

[

(−2)2 − 4]

x0 +∞∑

n=1

[

an

[

(n − 2)2 − 4]

+ 4an−1

]

xn

= a00x0 +∞∑

n=1

[

ann(n − 4) + 4an−2

]

xn .

Thus, for n ≥ 1 ,

ann(n − 4) + 4an−2 = 0 an = −4

n(n − 4)an−1 .

This gives our recursion formula

ak = −4

k(k − 4)ak−1 for k = 1, 2, 3, 4, 5, . . . .

(The fact that we get zero in our denominator when k = 4 should be a warning that we might

not get another series solution.)

Using the above,

a1 = −4

1(1 − 4)ak−1 = 4

3a0 ,

a2 = −4

2(2 − 4)ak−1 = 1 · a1 = 4

3a0 ,

a3 = −4

3(3 − 4)ak−1 = 4

3a2 = 4

3· 4

3a0 = 42

32a0 ,

a4 = −4

4(4 − 4)a4−1 = −4

0a3 = −4

0· 42

32a0 = “ ∞ ” .


Since a4 “blows up” when a0 6= 0 , there cannot be a series solution of the form

y(x) = xr2

∞∑

k=0

ak xk with a0 6= 0 .

32.4 e. First, multiply the differential equation by 1 − x to get it into the desired form,

(

x2 − x3)

︸︷︷︸

x2(1−x)

y′′ +(

x2 − x)

︸︷︷︸

x(1−x)

y′ + y = 0 .

Then, set

y(x) = xr

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 =(

x2 − x3)

y′′ +(

x2 − x)

y′ + y

= x2 y′′ − x3 y′′ + x2 y′ − xy′ + y

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 − x3∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

+ x2∞∑

k=0

ak(k + r)xk+r−1 − x

∞∑

k=0

ak(k + r)xk+r−1 +∞∑

k=0

ak xk+r

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

(−1)ak(k + r)(k + r − 1)xk+r+1

+∞∑

k=0

ak(k + r)xk+r+1 +∞∑

k=0

(−1)ak(k + r)xk+r +∞∑

k=0

ak xk+r .


0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

(−1)ak(k + r)(k + r − 1)xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

ak(k + r)xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

(−1)ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

ak xk

︸︷︷︸

n=k


=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=1

(−1)an−1(n − 1 + r)(n − 2 + r)xn

+∞∑

n=1

an−1(n − 1 + r)xn +∞∑

n=0

(−1)an(n + r)xn +∞∑

n=0

anxn+r

=[

a0(0 + r)(0 + r − 1)x0 +∞∑

n=1

an(n + r)(n + r − 1)xn

]

+∞∑

n=1

(−1)an−1(n − 1 + r)(n − 2 + r)xn +∞∑

n=1

an−1(n − 1 + r)xn

+[

− a0(0 + r)x0 +∞∑

n=1

(−1)an(n + r)xn

]

+[

a0x0 +∞∑

n=1

anxn+r

]

= a0 [r(r − 1) − r + 1] x0

+∞∑

n=1

[

an [(n + r)(n + r − 1) − (n + r) + 1]

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

xn ,

which, after a bit of algebra, simplifies to

0 = a0

[

r2 − 2r + 1]

x0

+∞∑

n=1

[

an(n + r − 1)2 − an−1(n + r − 1)(n + r − 3)]

xn .(⋆)

From the first term, we get the indicial equation

0 = r2 − 2r + 1︸︷︷︸

(r−1)2

.

So r1 = r2 = r = 1 .

Letting r = 1 , equation (⋆) becomes

0 = a0

[

12 − 2 · 1 + 1]

x0 +∞∑

n=1

[

an(n + 1 − 1)2 − an−1(n + 1 − 1)(n + 1 − 3)]

xn

= a0[0]x0 +∞∑

n=1

[

ann2 − an−1n(n − 2)]

xn .

So, for n ≥ 1 ,

ann2 − an−1n(n − 2) = 0 an = n − 2

nan−1 .

This is our recursion formula. Using it:

a1 = 1 − 2

1a1−1 = −a0 ,

a2 = 2 − 2

2a2−1 = 0a1 = 0 ,


a3 = 3 − 2

3a3−1 = 1

3a2 = 1

30 = 0 ,

a4 = 4 − 2

4a4−1 = 2

4a3 = 2

40 = 0 ,

...

Clearly, ak = 0 for k ≥ 2 . So

y(x) = xr1

∞∑

k=0

ak xk

= x[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 · · ·]

= x[

a0 − a0x + 0x2 + · · ·]

= a0 y1(x)

where

y1(x) = x[1 − x] = x − x2 .

Since there is only one value of r , no other series solution can be found by the basic

Frobenius method.

32.4 g. First, multiply the differential equation by x2 to get it into the form

x2 y′′ + xy′ +[

x2 − 1]

y = 0 .

Then, set

y(x) = xr

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 = x2 y′′ + xy′ +[

x2 − 1]

y

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 + x

∞∑

k=0

ak(k + r)xk+r−1

+[

x2 − 1] ∞∑

k=0

ak xk+r

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

ak(k + r)xk+r

+∞∑

k=0

ak xk+r+2 +∞∑

k=0

(−1)ak xk+r .



0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

ak xk+2

︸︷︷︸

n=k+2

+∞∑

k=0

(−1)ak xk

︸︷︷︸

n=k

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

an(n + r)xn +∞∑

n=0

(−1)anxn

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

an(n + r)xn +∞∑

n=2

an−2xn +∞∑

n=0

(−1)anxn

=[

a0(0 + r)(0 + r − 1)x0 + a1(1 + r)(1 + r − 1)x1 +∞∑

n=2

an(n + r)(n + r − 1)xn

]

+[

a0(0 + r)x0 + a1(1 + r)x1 +∞∑

n=2

an(n + r)xn

]

+∞∑

n=2

an−2xn

+[

− a0x0 − a1x1 +∞∑

n=2

(−1)anxn

]

= a0 [r(r − 1) + r − 1] x0 + a1 [(1 + r)r + (1 + r) − 1] x1

+∞∑

n=2

[

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

]

xn ,

which simplifies to

0 = a0

[

r2 − 1]

x0 + a1

[

r2 + 2r]

x1 +∞∑

n=2

[

an

[

(n + r)2 − 1]

+ an−2

]

xn . (⋆)


0 = r2 − 1 ,

which has solutions r = 1 and r = −1 . So the equation’s exponents are

r1 = 1 and r2 = −1 .

Plugging the larger exponent, r = r1 = 1 , into equation (⋆):

0 = a0 [0] x0 + a1 [1 + 2] x1 +∞∑

n=2

[

an

[

(n + 1)2 − 1]

+ an−2

]

xn

= 3a1x1 +∞∑

n=2

[

n(n + 2)an + an−2

]

xn .

Since each term must be zero, we have a1 = 0 and, for n ≥ 2 ,

n(n + 2)an + an−2 = 0 an = −1

n(n + 2)an−2 .



ak = −1

k(k + 2)ak−2 for k ≥ 2 .

Applying the above:

a1 = 0 ,

a2 = −1

2(2 + 2)a2−2 = −1

2 · 4a0 ,

a3 = −1

3(3 + 2)a3−2 = −1

3 · 5a1 = −1

3 · 50 = 0 ,

a4 = −1

4(4 + 2)a4−2 = −1

4 · 6a2 = −1

4 · 6· −1

2 · 4a0 = (−1)2

(4 · 2)(6 · 4)a0

= (−1)2

([2 · 2] · [2 · 1])([2 · 3] · [2 · 2])a0 = (−1)2

24(2 · 1)(3 · 2)a0 = (−1)2

242! 3!a0 ,

a5 = −1

5(5 + 2)a5−2 = −1

5 · 7a3 = −1

5 · 70 = 0 ,

a6 = −1

6(6 + 2)a6−2 = −1

6 · 8a4 = −1

6 · 8· (−1)2

(4 · 2)(6 · 4)a0 = (−1)3

(6 · 4 · 2)(8 · 6 · 4)a0

= (−1)3

(

[2 · 3] · [2 · 2] · [2 · 1])(

[2 · 4] · [2 · 3] · [2 · 2])a0 = (−1)3

263! 4!a0 ,

...

For odd values of k , we clearly have ak = 0 . For even values of k , you can verify that

ak = a2m = (−1)m

22mm! (m + 1)!a0 .

Thus,

y(x) = xr1

∞∑

k=0

ak xk

= x1[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 · · ·]

= x

[

a0 + 0x + −1

2 · 4a0x2 + 0x3 + (−1)2

(4 · 2)(6 · 4)a0x4 + 0x3

+ (−1)3

(6 · 4 · 2)(8 · 6 · 4)a0x6 + · · ·

]

= a0 y1(x)

where

y1(x) = x

[

1 + −1

2 · 4x2 + (−1)2

(4 · 2)(6 · 4)x4 + (−1)3

(6 · 4 · 2)(8 · 6 · 4)x6 + · · ·

]

= · · · = x

∞∑

m=0

(−1)m

22mm!(m + 1)!x2m .


Next, we plug the smaller exponent, r = r2 = −1 into equation (⋆):

0 = a0

[

(−1)2 − 1]

x0 + a1

[

(−1)2 + 2(−1)]

x1

+∞∑

n=2

[

an

[

(n − 1)2 − 1]

+ an−2

]

xn

= a0 [0] x0 + a1 [−1] x1 +∞∑

n=2

[

an [n(n − 2)] + an−2

]

xn .

Thus, a1 = 0 and, for n ≥ 2 ,

an [n(n − 2)] + an−2 = 0 an = −1

n(n − 2)an−2 .

This give our recursion formula

ak = −1

k(k − 2)ak−2 for k = 2, 3, 4, 5, . . . .

(The fact that we get zero in our denominator when k = 4 should be a warning that we might

not get another series solution.)

Applying the above:

a1 = 0 ,

but

a2 = −1

2(2 − 2)a2−2 = −1

0a0 = “ ∞ ” .

Since a2 “blows up” when a0 6= 0 , there cannot be a series solution of the form

y(x) = xr2

∞∑

k=0

ak xk with a0 6= 0 .

32.4 i. The equation is already in desired form.


y(x) = (x − 0)r∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 = x2 y′′ −(

5x + 2x2)

y′ + (9 + 4x) y

= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 −(

5x + 2x2) ∞

∑

k=0

ak(k + r)xk+r−1

+ (9 + 4x)

∞∑

k=0

ak xk+r


=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

(−5)ak(k + r)xk+r

+∞∑

k=0

(−2)ak(k + r)xk+r+1 +∞∑

k=0

9ak xk+r +∞∑

k=0

4ak xk+r+1 .

Dividing out xr and continuing:

0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

(−5)ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

(−2)ak(k + r)xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

9ak xk

︸︷︷︸

n=k

+∞∑

k=0

4ak xk+1

︸︷︷︸

n=k+1

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

(−5)an(n + r)xn

+∞∑

n=1

(−2)an−1(n − 1 + r)xn +∞∑

n=0

9anxn +∞∑

n=1

4an−1xn

=

[

a0(0 + r)(0 + r − 1)x0 +∞∑

n=1

an(n + r)(n + r − 1)xn

]

+

[

−5a0(0 + r)x0 +∞∑

n=1

(−5)an(n + r)xn

]

+∞∑

n=1

(−2)an−1(n − 1 + r)xn +[

9a0x0 +∞∑

n=1

9anxn

]

+∞∑

n=1

4an−1xn

= a0 [r(r − 1) − 5r + 9] x0

+∞∑

n=1

[

an [(n + r)(n + r − 1) − 5(n + r) + 9] + an−1 [−2(n − 1 + r) + 4]]

xn ,

which, after a little algebra, simplifies to

0 = a0

[

r2 − 6r + 9]

x0 +∞∑

n=1

[

an(n + r − 3)2 − an−12(n + r − 3)]

xn . (⋆)


0 = r2 − 6r + 9︸︷︷︸

(r−3)2

.

So the equation’s exponents are given by r1 = r2 = r = 3 .


Plugging r = r1 = 3 into equation (⋆), we get

0 = a0

[

32 − 6 · 3 + 9]

x0 +∞∑

n=1

[

an(n + 3 − 3)2 − an−12(n + 3 − 3)]

xn

= a0 [0] x0 +∞∑

n=1

[

ann2 − an−12n]

xn

Since each term must be zero, we must have, for n ≥ 1 ,

ann2 − an−12n = 0 an = 2

n)an−1 .

So the recursion formula is

ak = 2

kak−1 for k ≥ 1 .

Applying the above:

a1 = 2

1a1−1 = 2

1a0 ,

a2 = 2

2a2−1 = 2

2a1 = 2

2· 2

1a0 = 22

2 · 1a0 ,

a3 = 2

3a3−1 = 2

3a2 = 2

3· 22

2 · 1a0 = 23

3!a0 ,

a5 = 2

5a5−1 = 2

5a4 = 2

5· 24

4!a0 = 25

5!a0 ,

...

Clearly,

ak = 2k

k!a0 for k = 0, 1, 2, 3 . . . .

Thus,

y(x) = xr1

∞∑

k=0

ak xk

= x3[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + · · ·]

= x3

[

a0 + 2

1a0x + 22

2!a0 + 23

3!a0x3 + 24

4!a0x4 + 25

5!a0x5 + · · ·

]

= a0 y1(x)

where

y1(x) = x3

[

1 + 2

1x + 22

2!+ 23

3!x3 + 24

4!x4 + 25

5!x5 + · · ·

]

= x3∞∑

k=0

2k

k!xk = x3

∞∑

k=0

1

k!(2x)k = x3e2x .

Since there is only one value of r , no other series solution can be found by the basic

Frobenius method.


32.4 k. The equation is already in desired form.


y(x) = (x − 0)r

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 = x2 y′′ −(

x + x2)

y′ + 4xy

= x2 y′′ − xy − x2 y′ + 4xy

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

(−1)ak(k + r)xk+r

+∞∑

k=0

(−1)ak(k + r)xk+r+1 +∞∑

k=0

4ak xk+r+1 .

Dividing out xr and continuing:

0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

(−1)ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

(−1)ak(k + r)xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

4ak xk+1

︸︷︷︸

n=k+1

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

(−1)an(n + r)xn

+∞∑

n=1

(−1)an−1(n − 1 + r)xn +∞∑

n=1

4an−1xn

=

[

a0(0 + r)(0 + r − 1)x0 +∞∑

n=1

an(n + r)(n + r − 1)xn

]

+

[

−a0(0 + r)x0 +∞∑

n=1

(−1)an(n + r)xn

]

+∞∑

n=1

(−1)an−1(n − 1 + r)xn +∞∑

n=1

4an−1xn


= a0 [r(r − 1) − r] x0

+∞∑

n=0

[

an [(n + r)(n + r − 1) − (n + r)] + an−1 [−(n − 1 + r) + 4]]

xn ,

which simplifies to

0 = a0

[

r2 − 2r]

x0 +∞∑

n=0

[

an(n + r)(n + r − 2) − an−1(n + r − 5)]

xn . (⋆)


0 = r2 − 2r︸︷︷︸

r(r−2)

.

So the equation’s exponents are r1 = 2 and r2 = 0 .

Plugging the larger exponent, r = r1 = 2 , into equation (⋆), we get

0 = a0

[

22 − 2 · 2]

x0 +∞∑

n=0

[

an(n + 2)(n + 2 − 2) − an−1(n + 2 − 5)]

xn

= a0 [0] x0 +∞∑

n=0

[

an(n + 2)n − an−1(n − 3)]

xn

So, for n ≥ 1 ,

an(n + 2)n − an−1(n − 3) an = n − 3

(n + 2)nan−1 .

That is, the recursion formula is

ak = k − 3

(k + 2)kak−1 for k ≥ 1 .

Applying this:

a1 = 1 − 3

(1 + 2)1a1−1 = −2

(3)1a0 ,

a2 = 2 − 3

(2 + 2)2a2−1 = −1

(4)2a1 = −1

(4)(2)· −2

(3)1a0 = (−1)(−2)

(4 · 3)(2 · 1)a0 ,

a3 = 3 − 3

(3 + 2)3a3−1 = 0

(5)3a2 = 0 ,

a4 = 4 − 3

(4 + 2)4a4−1 = 1

(6)4a3 = 1

(6)4· 0 = 0 ,

...

Clearly, ak = 0 for k ≥ 3 . Thus,

y(x) = xr1

∞∑

k=0

ak xk

= x2[

a0 + a1x + a2x2 + a3x3 + a4x4 + · · ·]


= x2

[

a0 + −2

(3)1a0x + (−1)(−2)

(4 · 3)(2 · 1)a0x2 + 0x3 + 0x4 + · · ·

]

= a0 y1(x)

where

y1(x) = x2 − 2

3x3 + 1

12x4 .

Next, we use r = r2 = 0 in equation (⋆), obtaining

0 = a0

[

02 − 2 · 0]

x0 +∞∑

n=0

[

an(n + 0)(n + 0 − 2) − an−1(n + 0 − 5)]

xn

= a0[0]x0 +∞∑

n=0

[

ann(n − 2) − an−1(n − 5)]

xn .

So,

ann(n − 2) − an−1(n − 5) an = n − 5

n(n − 2)an−1 .

That is, the recursion formula is

ak = k − 5

k(k − 2)ak−1 for k ≥ 1 .

Applying this:

a1 = 1 − 5

1(1 − 2)a1−1 = −4

1(−1)a0

and

a2 = 2 − 5

2(2 − 2)a2−1 = −3

2 · 0a1 = −3

2 · 0· −4

1(−1)a0 = “∞” .

Since the recursion formula corresponding to r2 “blows up” at k = 2 , there can be no

solution of the form

y(x) = xr2

∞∑

k=0

ak xk .

32.4 m. The equation is already in desired form.


y(x) = (x − 0)r

∞∑

k=0

ak(x − 0)k =∞∑

k=0

ak xk+r .


y′ =∞∑

k=0

ak(k + r)xk+r−1 , y′′ =∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2

and

0 = x2 y′′ +(

x − x4)

y′ + 3x3 y

= x2 y′′ + xy − x4 y′ + 3x3 y


= x2∞∑

k=0

ak(k + r)(k + r − 1)xk+r−2 + x

∞∑

k=0

ak(k + r)xk+r−1

− x4∞∑

k=0

ak(k + r)xk+r−1 + 3x3∞∑

k=0

ak xk+r

=∞∑

k=0

ak(k + r)(k + r − 1)xk+r +∞∑

k=0

ak(k + r)xk+r

+∞∑

k=0

(−1)ak(k + r)xk+r+3 +∞∑

k=0

3ak xk+r+3 .


0 =∞∑

k=0

ak(k + r)(k + r − 1)xk

︸︷︷︸

n=k

+∞∑

k=0

ak(k + r)xk

︸︷︷︸

n=k

+∞∑

k=0

(−1)ak(k + r)xk+3

︸︷︷︸

n=k+3

+∞∑

k=0

3ak xk+3

︸︷︷︸

n=k+3

=∞∑

n=0

an(n + r)(n + r − 1)xn +∞∑

n=0

an(n + r)xn

+∞∑

n=3

(−1)an−3(n − 3 + r)xn +∞∑

n=3

3an−3xn

=[

a0(0 + r)(0 + r − 1)x0 + a1(1 + r)(1 + r − 1)x1 + a2(2 + r)(2 + r − 1)x2

+∞∑

n=3

an(n + r)(n + r − 1)xn

]

+[

a0(0 + r)x0 + a1(1 + r)x1 + a2(2 + r)x2 +∞∑

n=3

an(n + r)xn

]

+∞∑

n=3

(−1)an−3(n − 3 + r)xn +∞∑

n=3

3an−3xn

= a0 [r(r − 1) + r] x0 + a1 [(1 + r)r + (1 + r)] x1

+ a2 [(2 + r)(1 + r) + (2 + r)] x2

+∞∑

n=3

[

an [(n + r)(n + r − 1) + (n + r)] + an−3 [(−1)(n − 3 + r) + 3]]

xn ,


which simplifies to

0 = a0r2x0 + a1(1 + r)2x1 + a2(2 + r)2x2

+∞∑

n=3

[

an(n + r)2 − an−3(n + r − 6)

]

xn(⋆)


r2 = 0 .

which immediately tells us that the exponents of the equation are given by r1 = r2 = r = 0 .

With this value of r , equation (⋆) reduces to

0 = a1x1 + a24x2 +∞∑

n=3

[

ann2 − an−3(n − 6)]

xn

Since each term on the right must be zero, we have a1 = 0 , a2 = 0 and, for n ≥ 3 ,

ann2 − an−3(n − 6) = 0 an = n − 6

n2an−3 .

So the recursion formula is

ak = k − 6

k2ak−3 for k ≥ 3 .

Applying the above:

a1 = 0 ,

a2 = 0 ,

a3 = 3 − 6

32a3−3 = −3

32a0 = −1

3a0 ,

a4 = 4 − 6

42a4−3 = −2

42a1 = −2

42· 0 = 0 ,

a5 = 5 − 6

52a5−3 = −1

52a2 = −1

52· 0 = 0 ,

a6 = 6 − 6

62a6−3 = 0

62a3 = 0

62· −1

3a0 = 0 ,

...

Since a4 = a5 = a6 = 0 , it should be clear that the recursion formula will yield ak = 0 for

k > 6 . Hence, we have

y(x) = xr1

∞∑

k=0

ak xk

= x0[

a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + · · ·]

= a0 + 0x + 0x2 − 1

3a0x3 + 0

= a0 y1(x)

with y1(x) = 1 − 1

3x3 .

And since r2 = r1 , the basic Frobenius method does not yield any othe solutions.


32.5 a. When x = 3 the first coefficient is zero and the the last coefficient is nonzero. So x0 = 3

is a singular point. Moreover, we can easily get the equation into quasi-Euler form by

multiplying through by x − 3 :

(x − 3)2 y′′ + (x − 3)2 y′ + (x − 3)y = 0

So x0 = 3 is a regular singular point.

To simplify further computations, we let

Y (X) = y(x) with X = x − 3 .

The above differential equation then becomes

X2Y ′′ + X2Y ′ + XY = 0 .

Since X0 = x0 − 3 = 3 − 3 = 0 , we set

Y (X) = (X − 0)r

∞∑

k=0

ak(X − 0)k =∞∑

k=0

ak Xk+r .

Differentiating this and plugging the results into the last differential equation, we get

Y ′ =∞∑

k=0

ak(k + r)Xk+r−1 , Y ′′ =∞∑

k=0

ak(k + r)(k + r − 1)Xk+r−2

and

0 = X2Y ′′ + X2Y ′ + XY

= X2∞∑

k=0

ak(k + r)(k + r − 1)Xk+r−2 + X2∞∑

k=0

ak(k + r)Xk+r−1 + X

∞∑

k=0

ak Xk+r

=∞∑

k=0

ak(k + r)(k + r − 1)Xk+r +∞∑

k=0

ak(k + r)Xk+r+1 +∞∑

k=0

ak Xk+r+1

Dividing out the Xr and continuing:

0 =∞∑

k=0

ak(k + r)(k + r − 1)Xk

︸︷︷︸

n=k

+∞∑

k=0

ak(k + r)Xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

ak Xk+1

︸︷︷︸

n=k+1

=∞∑

n=0

an(n + r)(n + r − 1)Xn +∞∑

n=1

an−1(n − 1 + r)Xn +∞∑

n=1

an−1 Xn

=

[

a0r(r − 1)x0 +∞∑

n=1

an(n + r)(n + r − 1)Xn

]

+∞∑

n=1

an−1(n − 1 + r)Xn +∞∑

n=1

an−1 Xn ,


which simplifies to

0 = a0

[

r2 − r]

x0 +∞∑

n=1

[

an(n + r − 1) + an−1

]

(n + r)Xn (⋆)


0 = r2 − r︸︷︷︸

r(r−1)

,

which tells us that the equation’s exponents are r1 = 1 and r2 = 0 .

Letting r = r1 = 1 in equation (⋆) yields:

0 = a0 [0] x0 +∞∑

n=0

[

ann + an−1

]

nXn .

Since each term must be zero, we have, for n ≥ 1

[

ann + an−1

]

n = 0 an = −1

nan−1 .

So our recursion formula is

ak = −1

kak−1 for k ≥ 1 .

Applying the above:

a1 = −1

1a1−1 = −1

1a0 ,

a2 = −1

2a2−1 = −1

2a1 = −1

2· −1

1a0 = (−1)2

2 · 1a0 ,

a3 = −1

3a3−1 = −1

3a2 = −1

3· (−1)2

2 · 1a0 = (−1)3

3!a0 ,

a4 = −1

4a4−1 = −1

4a3 = −1

4· (−1)3

3!a0 = (−1)4

4!a0 ,

...

Clearly,

ak = (−1)k

k!a0 for k = 0, 1, 2, . . . ,

Hence, corresponding to r1 , we have

Y (X) = Xr1

∞∑

k=0

ak Xk = X1∞∑

k=0

(−1)k

k!a0 Xk = a0Y1(X)

where

Y1(X) = X[

1 − 1

2X + 1

2 · 1X2 − 1

3 · 2 · 1X3 + · · ·

]

= X

∞∑

k=0

(−1)k

k!Xk = X

∞∑

k=0

1

k!(−X)k = Xe−X .


Recalling that y(x) = Y (X) with X = x − 3 , we see that the corresponding solution to

our original differential equation is y(x) = a0 y1(x) where

y1(x) = Y1(x − 3)

= (x − 3)

[

1 − 1

2(x − 3) + 1

2 · 1(x − 3)2 − 1

3 · 2 · 1(x − 3)3 + · · ·

]

= (x − 3)

∞∑

k=0

(−1)k

k!(x − 3)k

= (x − 3)e−(x−3) .

Next we seek the solutions corresponding to r = r2 = 0 . Using this value for r in

equation (⋆) yields

0 = a0

[

02 − 0]

x−1 +∞∑

n=1

[

an(n + 0 − 1) + an−1

]

(n + 0)Xn

=∞∑

n=1

[

an(n − 1) + an−1

]

nXn

So, for n ≥ 1 , we must have

[

an(n − 1) + an−1

]

n = 0 an = −1

n − 1an−1 ,

and the recursion formula is

ak = −1

k − 1ak−1 for k = 1, 2, 3, . . . .

Unfortunately, this recursion formula blows up with its first usage,

a1 = −1

1 − 1a1−1 = −1

0a0 .

Hence, the basic method of Frobenius does not yield the second set of solutions.

32.5 c. It should be clear that x0 = 1 is a singular point. To apply theorem 32.2, we compute

limx→x0

(x − x0)b(x)

a(x)= lim

x→1(x − 1)

0

4= 0

and

limx→x0

(x − x0)2 c(x)

a(x)= lim

x→1(x − 1)2 (4x − 3)/(x − 1)2

4= lim

x→1

(4x − 3)

4= 1

4.

Since both limits are finite, theorem 32.2 assures us that x0 = 1 is a regular singular point.

To simplify further computations, we let

Y (X) = y(x) with X = x − 1 .

The differential equation in question then becomes

4Y ′′ + 4(X + 1) − 3

X2Y = 0 .


Doing a little algebra, we get this to desired form:

4X2Y ′′ + (4X + 1)Y = 0 .

Since X0 = x0 − 1 = 1 − 1 = 0 , we set

Y (X) = (X − 0)r

∞∑

k=0

ak(X − 0)k =∞∑

k=0

ak Xk+r .


Y ′ =∞∑

k=0

ak(k + r)Xk+r−1 , Y ′′ =∞∑

k=0

ak(k + r)(k + r − 1)Xk+r−2

and

0 = 4X2Y ′′ + (4X + 1)Y

= 4X2∞∑

k=0

ak(k + r)(k + r − 1)Xk+r−2 + (4X + 1)

∞∑

k=0

ak Xk+r

=∞∑

k=0

4ak(k + r)(k + r − 1)Xk+r +∞∑

k=0

4ak Xk+r+1 +∞∑

k=0

ak Xk+r .

Dividing out Xr and continuing:

0 =∞∑

k=0

4ak(k + r)(k + r − 1)Xk

︸︷︷︸

n=k

+∞∑

k=0

4ak Xk+1

︸︷︷︸

n=k+1

+∞∑

k=0

ak Xk

︸︷︷︸

n=k

=∞∑

n=0

4an(n + r)(n + r − 1)Xn +∞∑

k=1

4an−1 Xn +∞∑

n=0

an Xn

=

[

a04(0 + r)(0 + r − 1)X0 +∞∑

n=1

4an(n + r)(n + r − 1)Xn

]

+∞∑

k=1

4an−1 Xn

+

[

a0 X0 +∞∑

n=1

an Xn

]

= a0 [4r(r − 1) + 1] X0 +∞∑

n=1

[

an [4(n + r)(n + r − 1) + 1] + 4an−1

]

Xn ,

which simplifies to

0 = a0

[

4r2 − 4r + 1]

X0 +∞∑

n=1

[

an [4(n + r)(n + r − 1) + 1] + 4an−1

]

Xn . (⋆)


0 = 4r2 − 4r + 1 ,


whose solution is

r = −(−4) ±√

(−4)2 − 4 · 4 · 2

2 · 4= 1

2.

So the exponents of the equation are given by

r1 = r2 = 1

2.

Plugging r = r1 = 1

2into equation (⋆), we get

0 = a0 [0] X0 +∞∑

n=1

[

an

[

4(

n + 1

2

) (

n + 1

2− 1

)

+ 1]

+ 4an−1

]

Xn

=∞∑

n=1

[

4ann2 + 4an−1

]

Xn ,

telling us that, for n ≥ 1 ,

4ann2 + 4an−1 = 0 an = −1

n2an−1 .

So,

ak = −1

k2ak−1 for k = 1, 2, 3, . . .

is our recursion formula, and

a1 = −1

12a1−1 = −1

12a0 ,

a2 = −1

22a2−1 = −1

22a1 = −1

22· −1

12a0 = (−1)2

(2 · 1)2a0 ,

a3 = −1

32a3−1 = −1

32a2 = −1

32· −1

(2 · 1)2a0 = (−1)3

(3 · 2 · 1)2a0 = (−1)3

(3!)2a0 ,

a4 = −1

42a4−1 = −1

42a3 = −1

42· −1

(3!)2a0 = (−1)4

(4!)2a0 ,

...

It’s easy to see that, in fact,

ak = (−1)k

(k!)2a0 for k = 0, 1, 2, 3, . . . .

So,

Y (X) = Xr1

∞∑

k=0

ak Xk = X1/2

∞∑

k=0

(−1)k

(k!)2a0 Xk = a0Y1(X)

where

Y1(X) = X1/2

∞∑

k=0

(−1)k

(k!)2Xk

= X1/2

∞∑

k=0

[

1 − X + 1

22X2 − 1

(3!)2X3 − 1

(4!)2X4 + · · ·

]

.


But the corresponding solutions to the original differential equation are given by y(x) =Y (X) where X = x − 1 . Hence, (at least for x > 1 ), y(x) = a0 y1(x) where

y1(x) = Y1(x − 1)

=√

x − 1

∞∑

k=0

(−1)k

(k!)2(x − 1)k

=√

x − 1

[

1 − (x − 1) + 1

22(x − 1)2 − 1

(3!)2(x − 1)3 − 1

(4!)2(x − 1)4 + · · ·

]

.

Since r1 = r2 , no other solutions can be found by the basic method of Frobenius.

32.6 a. Applying the product rule, theorem 29.6 on page 572 on the differentiation of power series,

and basic power series algebra:

d

dx

[

xr∞∑

k=0

ak xk

]

= dxr

dx

∞∑

k=0

ak xk + xr d

dx

∞∑

k=0

ak xk

= rxr−1∞∑

k=0

ak xk + xr

∞∑

k=0

akkxk−1

=∞∑

k=0

akrxk+r−1 +∞∑

k=0

akkxk+r−1

=∞∑

k=0

[akr + akk] xk+r−1

=∞∑

k=0

[ak(k + r)xk+r−1 .

32.6 b. The formula rigorously obtained above for

d

dx

[

xr

∞∑

k=0

ak xk

]

is the same as the one naively obtained by differentiating term-by-term.

32.7 a. For the first equation:

dy∗

dx= d

dx

[

(u + iv)∗]

= d

dx

[

u − iv]

= du

dx− i

dv

dx=

(du

dx+ i

dv

dx

)∗=

(dy

dx

)∗.

The second equation is verified in the same way, replacing the first derivatives with second

derivatives.


32.7 b. Taking the complex conjugate of

a(x)d2 y

dx2+ b(x)

dy

dx+ c(x)y = 0 ,

and using what was just derived along with basic complex algebra and the fact that a , b and

c are all real-valued functions (hence, a∗ = a , b∗ = b and c∗ = c ):

(

a(x)d2 y

dx2+ b(x)

dy

dx+ c(x)y

)∗= 0∗

→֒(

a(x)d2 y

dx2

)∗+

(

b(x)dy

dx

)∗+ (c(x)y)∗ = 0

→֒ a∗(x)

(

d2 y

dx2

)∗+ b∗(x)

(dy

dx

)∗+ c∗(x) (y)∗ = 0

→֒ a(x)d2 y∗

dx2+ b(x)

dy∗

dx+ c(x)y∗ = 0 .

32.8 a. Suppose that, contrary to the claim, y1 and y2 are constant multiples of each other. Since

both are nonzero functions, there is then a nonzero constant c such that y2 = cy1 . But then

x−r2 y2(x) = x−r2 cy1(x)

→֒ x−r2 xr2

∞∑

k=0

βk xk = cx−r2 xr1

∞∑

k=0

αk xk

→֒∞∑

k=0

βk xk = cxr1−r2

∞∑

k=0

αk xk

Taking the limit as x → 0 of both sides of the last equation, and keeping in mind that

r2 > r1 (so r1 − r2 > 0 ), then gives us

1 = β0 = limx→0

∞∑

k=0

βk xk

= c limx→0

xr1−r2

∞∑

k=0

αk xk

= c limx→0

xr1−r2 · limx→0

∞∑

k=0

αk xk

= c · 0 · α0 = 0 ,

which is nonsense. Hence, our supposition that y1 and y2 could be constant multiples of

each other was wrong. So they cannot be constant multiples of each other, and, as noted in

chapter 13, this means the two solutions y1 and y2 make up a fundamental set of solutions

for the differential equation.

32.8 b. Since {y1, y2} is a fundamental set of solutions to the differential equation, solution y3

must be a linear combination of y1 and y2 .


32.8 c. As just noted, y3 must be a linear combination of y1 and y2 . So there are constants A

and B , not both zero, such that

y3 = Ay1 + By2 .

Note, also, that

y3(x) = xr2

∞∑

k=M

γk xk

= xr2

[

γM x M + γM+1x M+1 + γM+2x M+2 · · ·]

= x M+r2

∞∑

k=0

γM+k xk .

So, going through computations similar to that done above,

γM = limx→0

∞∑

k=0

γM+k xk

= limx→0

x−(M+r2)y3(x)

= limx→0

x−(M+r2) [Ay1(x) + By2(x)]

= A limx→0

x−(M+r2)y1(x) + B limx→0

x−(M+r2)y2(x)

= A limx→0

x−(M+r2)xr1

∞∑

k=0

αk xk + B limx→0

x−(M+r2)xr2

∞∑

k=0

βk xk

= A limx→0

xr1−(M+r2)α0 + B limx→0

x−Mβ0 .

Since α0 = β0 = γ0 = 1 , the above reduces to

1 = A limx→0

x (r1−r2)−M + B limx→0

x−M

Remember, both r1 − r2 and M are positive, and at least A or B is nonzero. Now, observe

that, if M > r1 − r2 , then both of the above limits are infinite, which, clearly, is impossible.

If, on the other hand, M < r1 − r2 , then

1 = A limx→0

x (r1−r2)−M + B limx→0

x−M = A · 0 + B · (±∞)

which is also clearly impossible. This leaves M = r1 − r2 as the only possibility. But then

1 = A limx→0

x M−M + B limx→0

x−M = A · 1 + B · (±∞)

which obviously means that A = 1 and B = 0 . So, M = r1 − r2 and

y3 = Ay1 + By2 = 1 · y1 + 0 · y2 = y1 .

32.8 d. First of all, it follows from part a that if r1 and r2 are the equation’s exponents and

r1 > r > 2 , then any two solutions of the form

y1(x) = xr1

∞∑

k=0

αk xk with α0 = 1


and

y2(x) = xr2

∞∑

k=0

βk xk with β0 = 1

make up a fundamental set of solutions. Hence, for y2 , we can choose any such modified

series solution. If, in our first choice, βM 6= 0 with M = r1−r2 (equivalently, r1 = r2+M ),

then another valid solution is

y4(x) = y2(x) − βM y1(x) = · · · = xr2

∞∑

k=0

ak xk ,

where

ak =

{

βk − 0 if k < M

βk − βMαM−k if k ≥ M

}

=

βk if k < M

0 if k = M

βk − βMαM−k if k > M

,

which is simply the second solution obtained in the basic Frobenius method when M =r1 − r2 is a positive integer and the method actually does yield a solution corresponding to

r2 .

Chapter 32: Modiﬁed Power Series Solutions and the...

Documents

Transcript of Chapter 32: Modiﬁed Power Series Solutions and the...