Post on 05-Jun-2018
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
Worked Solutions 297
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),
298 Modified Power Series Solutions and the Basic Method of Frobenius
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)
Worked Solutions 299
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
300 Modified Power Series Solutions and the Basic Method of Frobenius
=∞∑
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 .
Worked Solutions 301
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 · · ·]
302 Modified Power Series Solutions and the Basic Method of Frobenius
= 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 ,
Worked Solutions 303
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.
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′ =∞∑
k=0
ak(k + r)xk+r−1 , y′′ =∞∑
k=0
ak(k + r)(k + r − 1)xk+r−2
304 Modified Power Series Solutions and the Basic Method of Frobenius
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 .
Dividing out the xr and continuing:
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 .
Worked Solutions 305
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 .
So the recursion formula when r = 2 is
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, . . . .
306 Modified Power Series Solutions and the Basic Method of Frobenius
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 = “ ∞ ” .
Worked Solutions 307
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 .
Differentiating this and plugging the results into the 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 =(
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 .
Dividing out the xr and continuing:
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
308 Modified Power Series Solutions and the Basic Method of Frobenius
=∞∑
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 ,
Worked Solutions 309
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 .
Differentiating this and plugging the results into the 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 = 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 .
310 Modified Power Series Solutions and the Basic Method of Frobenius
Dividing out the xr and continuing:
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 . (⋆)
From the first term, we get the indicial equation
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 .
Worked Solutions 311
So the recursion formula when r = 1 is
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 .
312 Modified Power Series Solutions and the Basic Method of Frobenius
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.
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′ =∞∑
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
Worked Solutions 313
=∞∑
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 . (⋆)
From the first term, we get the indicial equation
0 = r2 − 6r + 9︸ ︷︷ ︸
(r−3)2
.
So the equation’s exponents are given by r1 = r2 = r = 3 .
314 Modified Power Series Solutions and the Basic Method of Frobenius
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.
Worked Solutions 315
32.4 k. 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′ =∞∑
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
316 Modified Power Series Solutions and the Basic Method of Frobenius
= 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 . (⋆)
From the first term, we get the indicial equation
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 + · · ·]
Worked Solutions 317
= 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.
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′ =∞∑
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
318 Modified Power Series Solutions and the Basic Method of Frobenius
= 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 .
Dividing out the xr and continuing:
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 ,
Worked Solutions 319
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(⋆)
From the first term, we get the indicial equation
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.
320 Modified Power Series Solutions and the Basic Method of Frobenius
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 ,
Worked Solutions 321
which simplifies to
0 = a0
[
r2 − r]
x0 +∞∑
n=1
[
an(n + r − 1) + an−1
]
(n + r)Xn (⋆)
From the first term, we get the indicial equation
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 .
322 Modified Power Series Solutions and the Basic Method of Frobenius
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 .
Worked Solutions 323
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 .
Differentiating this and plugging the results into the 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 = 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 . (⋆)
From the first term, we get the indicial equation
0 = 4r2 − 4r + 1 ,
324 Modified Power Series Solutions and the Basic Method of Frobenius
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 + · · ·
]
.
Worked Solutions 325
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.
326 Modified Power Series Solutions and the Basic Method of Frobenius
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 .
Worked Solutions 327
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
328 Modified Power Series Solutions and the Basic Method of Frobenius
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 .