Question 11.2.5: Find the general solution in powers of x of (x² - 4)y″ + 3xy...
Find the general solution in powers of x of
(x² – 4)y″ + 3xy^{′} + y = 0. (7)
Then find the particular solution with y(0) = 4, y^{′}(0) = 1.
The only singular points of Eq. (7) are ±2, so the series we get will have radius of convergence at least 2. (See Problem 35 for the exact radius of convergence.) Substitution of
y = \sum\limits_{n=0}^{\infty}{c_{n} x^{n}} , y^{′} = \sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}}, and y″ = \sum\limits_{n=2}^{\infty}{n(n – 1)c_{n} x^{n-2}}
in Eq. (7) yields
\sum\limits_{n=2}^{\infty}{n(n – 1)c_{n} x^{n } – 4} \sum\limits_{n=2}^{\infty}{n(n – 1)c_{n} x^{n-2}} + 3 \sum\limits_{n=1}^{\infty}{n c_{n} x^{n}} + \sum\limits_{n=0}^{\infty}{c_{n} x^{n}} = 0.
We can begin the first and third summations at n = 0 as well, because no nonzero terms are thereby introduced. We shift the index of summation in the second sum by + 2, replacing n with n + 2 and using the initial value n = 0. This gives
\sum\limits_{n=0}^{\infty}{n(n – 1)c_{n} x^{n} } – 4 \sum\limits_{n=0}^{\infty}{(n + 2)(n + 1)c_{n+2} x^{n}} + 3 \sum\limits_{n=0}^{\infty}{n c_{n} x^{n}} + \sum\limits_{n=0}^{\infty}{c_{n} x^{n}} = 0.After collecting coefficients of c_{n} and c_{n+2}, we obtain
\sum\limits_{n=0}^{\infty}{[(n² + 2n + 1)c_{n} – 4(n + 2)(n + 1)c_{n+2}]x^{n} }= 0.
The identity principle yields
(n + 1)²c_{n} – 4(n + 2)(n + 1)c_{n+2} = 0.
which leads to the recurrence relation
c_{n+2} = \frac{(n + 1)c_{n}}{4(n + 2)} (8)
for n ≧ 0. With n = 0, 2, and 4 in turn, we get
c_{2} = \frac{c_{0}}{4 · 2}, c_{4} = \frac{3c_{2}}{4 · 4} = \frac{3c_{0}}{4² · 2 · 4}, and c_{6} = \frac{5c_{4}}{4 · 6} = \frac{3 · 5c_{0}}{4³ · 2 · 4 · 6}.
Continuing in this fashion, we evidently would find that
c_{2n} = \frac{1 · 3 · 5 · · · (2n – 1)}{4^{n} · 2 · 4· · ·(2n)}c_{0}.
With the common notation
(2n + 1)!! = 1 · 3 · 5 · · · (2n + 1) = \frac{(2n + 1)!}{2^{n} · n!}
and the observation that 2 · 4 · 6 · · ·(2n) = 2^{n} · n!, we finally obtain
c_{2n} = \frac{(2n – 1)!!}{2^{3n} · n!}c_{0}. (9)
(We also used the fact that 4^{n} · 2^{n} = 2^{3n}.)
With n = 1, 3, and 5 in Eq. (8), we get
c_{3} = \frac{2c_{1}}{4 · 3}, c_{5} = \frac{4c_{3}}{4 · 5} = \frac{2 · 4c_{1}}{4²· 3 · 5}, and c_{7} = \frac{6c_{5}}{4 · 7} = \frac{2 · 4 · 6c_{1}}{4³ · 3 · 5 · 7}.
It is apparent that the pattern is
c_{2n+1} = \frac{2 · 4 · 6 · · ·(2n)}{4^{n} · 1 · 3 · 5· · ·(2n + 1)}c_{1} = \frac{n!}{2^{n} · (2n + 1)!!}c_{1}. (10)
The formula in (9) gives the coefficients of even subscript in terms of c_{0}; the formula in (10) gives the coefficients of odd subscript in terms of c_{0}. After we separately collect the terms of the series of even and odd degree, we get the general solution
y(x) = c_{0}(1 + \sum\limits_{n=1}^{\infty}{\frac{(2n – 1)!!}{2^{3n} · n!} x^{2n}}) + c_{1}(x + \sum\limits_{n=1}^{\infty}{\frac{n!}{2^{n} · (2n + 1)!!}x^{2n+1}}) (11)
Alternatively,
y(x) = c_{0}(1 + \frac{1}{8} x² + \frac{3}{128}x^{4} + \frac{5}{1024}x^{6} + · · ·) + c_{1}(x + \frac{1}{6}x^{3} + \frac{1}{30}x^{5} + \frac{1}{140}x^{7} + · · ·). (11′)
Because y(0) = c_{0} and y^{′}(0) = c_{1}, the given initial conditions imply that c_{0} = 4 and c_{1} = 1. Using these values in Eq. (11′), the first few terms of the particular solution satisfying y(0) = 4 and y^{′}(0) = 1 are
y(x) = 4 + x + \frac{1}{2}x^{2} + \frac{1}{6} x^{3} + \frac{3}{32}x^{4} + \frac{1}{30}x^{5} + · · ·. (12)