530
8.6. POWER SERIES
in the form x k+1 . Explain why
∞
k=2
(k − 1)ka k x
k−2 =
∞
k=0
(k + 1)(k + 2)a k+2 x
k .
(8.30)
(d) Now show that
∞
k=0
a k x
k+1 =
∞
k=1
a k−1 x
k .
(8.31)
(e) We can now substitute (8.30) and (8.31) into (8.29) to obtain
∞
n=0
(n + 1)(n + 2)a n+2 x
n +
∞
n=1
a n−1 x
n = 0.
(8.32)
Combine the like powers of x in the two series to show that our solution must
satisfy
2a 2 +
∞
k=1
[(k + 1)(k + 2)a k+2 + a k−1 ] x
k = 0.
(8.33)
(f) Use equation (8.33) to show the following:
(i) a 3k+2 = 0 for every positive integer k,
(ii) a 3k =
1
(2)(3)(5)(6)···(3k−1)(3k) a 0 for k ≥ 1,
(iii) a 3k+1 =
1
(3)(4)(6)(7)···(3k)(3k+1) a 1 for k ≥ 1.
(g) Use the previous part to conclude that the general solution to the Airy equation
(8.28) is
y = a 0
1 +
∞
k=1
x 3k
(2)(3)(5)(6) · · · (3k − 1)(3k)
+a 1
x +
∞
k=1
x 3k+1
(3)(4)(6)(7) · · · (3k)(3k + 1)
.
Any values for a 0 and a 1 then determine a specific solution that we can
approximate as closely as we like using this series solution.
8.6. POWER SERIES
in the form x k+1 . Explain why
∞
k=2
(k − 1)ka k x
k−2 =
∞
k=0
(k + 1)(k + 2)a k+2 x
k .
(8.30)
(d) Now show that
∞
k=0
a k x
k+1 =
∞
k=1
a k−1 x
k .
(8.31)
(e) We can now substitute (8.30) and (8.31) into (8.29) to obtain
∞
n=0
(n + 1)(n + 2)a n+2 x
n +
∞
n=1
a n−1 x
n = 0.
(8.32)
Combine the like powers of x in the two series to show that our solution must
satisfy
2a 2 +
∞
k=1
[(k + 1)(k + 2)a k+2 + a k−1 ] x
k = 0.
(8.33)
(f) Use equation (8.33) to show the following:
(i) a 3k+2 = 0 for every positive integer k,
(ii) a 3k =
1
(2)(3)(5)(6)···(3k−1)(3k) a 0 for k ≥ 1,
(iii) a 3k+1 =
1
(3)(4)(6)(7)···(3k)(3k+1) a 1 for k ≥ 1.
(g) Use the previous part to conclude that the general solution to the Airy equation
(8.28) is
y = a 0
1 +
∞
k=1
x 3k
(2)(3)(5)(6) · · · (3k − 1)(3k)
+a 1
x +
∞
k=1
x 3k+1
(3)(4)(6)(7) · · · (3k)(3k + 1)
.
Any values for a 0 and a 1 then determine a specific solution that we can
approximate as closely as we like using this series solution.