Lecture Notes in Differential Equations - Bruce E. Shapiro

268 LESSON 28. SERIES SOLUTIONS
in the differential equation gives
∞∑
∞∑
∞∑
k(k − 1)a k t k−2 = t a k t k = a k t k+1 (28.98)
k=0
Renumbering the index to j = k − 2 (on the left) and j = k + 1 (on the
right), and recognizing that the first two terms of the sum on left are zero,
gives
∞∑
∞∑
(j + 2)(j + 1)a j+2 t j = a j−1 t j (28.99)
Rearranging,
2a 2 +
j=0
k=0
j=1
k=0
∞∑
[(j + 2)(j + 1)a j+2 − a j−1 ] t j = 0 (28.100)
j=1
By linear independence, a 2 = 0 and the remaining a j satisfy
If we let k = j + 2 and solve for a k ,
a k =
a k−3
k(k − 1)
(j + 2)(j + 1)a j+2 = a j−1 (28.101)
(28.102)
The first two coefficients, a 0 and a 1 , are determined by the initial conditions;
all other coefficients follow from the recursion relationship.. In particular,
since a 2 = 0, every third successive coefficient a 2 = a 5 = a 8 = · · · = 0.
Starting with a 0 and a 1 we can begin to tabulate the remaining ones:
and
a 3 = a 0
3 · 2
a 6 = a 3
6 · 5 = a 0
6 · 5 · 3 · 2
a 9 = a 6
9 · 8 = a 0
9 · 8 · 6 · 5 · 3 · 2
.
a 4 = a 1
4 · 3
a 7 = a 4
7 · 6 = a 1
7 · 6 · 4 · 3
a 10 = a 7
10 · 9 = a 1
10 · 9 · 7 · 6 · 4 · 3
.
(28.103)
(28.104)
(28.105)
(28.106)
(28.107)
(28.108)