Lecture Notes in Differential Equations - Bruce E. Shapiro

266 LESSON 28. SERIES SOLUTIONS
Substituting equation (28.89) into equation (28.90)
rk(k + 1)C k+1 = k(k − 1)C k − MC k−1 r + M [kC k + C k−1 ] r + MC k r 2
Dividing by rk(k + 1)C k gives
C k+1
C k
=
k(k − 1) + Mkr + Mr2
rk(k + 1)
(28.91)
(28.92)
Multiplying by t = t k+1 /t k and taking the limit as k → ∞
∣ lim
C k+1 t k+1 ∣∣∣ k→∞
∣ C k t k = lim
k(k − 1) + Mkr + Mr 2
k→∞
∣
t
rk(k + 1) ∣ (28.93)
Dividing the numerator an denominator by k 2 ,
∣ lim
C k+1 t k+1 ∣∣∣ ∣ C k t k = |t| lim
1 − 1/k + Mr/k + Mr 2 /k 2
k→∞
∣ r + r/k ∣ = |t|
|r|
k→∞
(28.94)
Therefore by the ratio test ∑ ∞
k=0 C kt k converges for |t| < r, and hence
by the comparison test ∑ ∞
k=0 c kt k also converges. Therefore there is an
analytic solution to any second order homogeneous ordinary differential
equation with analytic coefficients. The coefficients of the power series
are given by c 0 = y 0 , c 1 = y 1 (by Taylor’s theorem) and the recursion
relationship (28.83) for c 2 , c 3 , ...
Many of the best-studied differential equations of mathematical physics
(see table 28.1) are most easily solved using the method of power series
solutions, or the related method of Frobenius that we will discuss in the
next section. A full study of these equations is well beyond the scope of
these notes; many of their most important and useful properties derive from
the fact that many of them arise from their solutions as boundary value
problems rather than initial value problems. Our interest in these functions
is only that they illustrate the method of power series solutions.
Example 28.4. Find the general solutions to Airy’s Equation:
y ′′ = ty (28.95)
Using our standard substitutions,
∞∑
y = a k t k (28.96)
y ′′ =
k=0
∞∑
k(k − 1)t k−2 (28.97)
k=0