Lecture Notes in Differential Equations - Bruce E. Shapiro

259
Since the first two terms (corresponding to j = −2 and j = −1) in the
first sum are zero, we can start the index at j = 0 rather than j = −2.
Renaming the index back to k, and combining the two series into a single
sum,
∞∑
∞∑
0= (k + 2)(k + 1)c k+2 t k − (k + 1)c k t k (28.33)
k=0
By linear independence,
k=0
(k + 2)(k + 1)c k+2 = (k + 1)c k (28.34)
Rearranging, c k+2 = c k /(k + 2); letting j = k + 2 and including the initial
conditions (equation (28.29)), the general recursion relationship is
Therefore
c 0 = 1, c 1 = 1, c k = c k−2 /k (28.35)
c 2 = c 0
2 = 1 2
c 4 = c 2
4 = 1
4 · 2
c 6 = c 4
6 = 1
6 · 4 · 2
.
c 3 = c 1
3 = 1 3
c 5 = c 3
5 = 1
5 · 3
c 7 = c 5
7 = 1
7 · 5 · 3
(28.36)
(28.37)
(28.38)
So the solution of the initial value problem is
y=c 0 + c 1 t + c 2 t 2 + c 3 t 3 + · · · (28.39)
=1 + 1 2 t2 + 1 1
4 · 2 t4 +
6 · 4 · 2 t6 + · · · (28.40)
+t + 1 3 t3 + 1 1
5 · 3 t5 +
7 · 5 · 3 t7 + · · · (28.41)
Example 28.3. Solve the Legendre equation of order 2, given by
⎫
(1 − t 2 )y ′′ − 2ty ′ + 6y = 0⎪⎬
y(0) = 1
(28.42)
⎪⎭
y ′ (0) = 0
We note that “order 2” in the name of equation (28.42) is not related to the
order of differential equation, but to the fact that the equation is a member
of a family of equations of the form
(1 − t 2 )y ′′ − 2ty ′ + n(n + 1)y = 0 (28.43)