Question 11.1.4: Solve the equation y″ + y = 0.
Solve the equation y″ + y = 0.
If we assume a solution of the form
y = \sum\limits_{n=0}^{\infty}{c_{n}x^{n}},
we find that
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}}.
Substitution for y and y″ in the differential equation then yields
\sum\limits_{n=2}^{\infty}{n(n – 1)c_{n}x^{n-2}} + \sum\limits_{n=0}^{\infty}{c_{n}x^{n-2}} =0.
We shift the index of summation in the first sum by +2 (replace n = 2 with n = 0 and n with n + 2). This gives
\sum\limits_{n=0}^{\infty}{(n + 2)(n + 1)c_{n+2}x^{n}} + \sum\limits_{n=0}^{\infty}{c_{n}x^{n-2}} =0.
The identity (n + 2)(n + 1)c_{n+2} + c_{n} now follows from the identity principle, and thus we obtain the recurrence relation
c_{n+2} = – \frac{c_{n}}{(n + 1)(n + 2)} (27)
for n ≧ 0. It is evident that this formula will determine the coefficients c_{n} with even subscript in terms of c_{0} and those of odd subscript in terms of c_{1}; c_{0} and c_{1} are not predetermined and thus will be the two arbitrary constants we expect to find in a general solution of a secondorder equation. When we apply the recurrence relation in (27) with n = 0, 2, and 4 in turn, we get
c_{2} = – \frac{c_{0}}{2!}, c_{4} = \frac{c_{0}}{4!}, and c_{6} = – \frac{c_{0}}{6!}.
Taking n = 1, 3, and 5 in turn, we find that
c_{3} = – \frac{c_{1}}{3!}, c_{5} = \frac{c_{1}}{5!}, and c_{7} = – \frac{c_{1}}{7!}.
Again, the pattern is clear; we leave it for you to show (by induction) that for k ≧ 1,
c_{2k} = \frac{(-1)^{k} c_{0}}{(2k)!} and c_{2k + 1} = \frac{(-1)^{k} c_{1}}{(2k + 1)!}.
Thus we get the power series solution
y(x) = c_{0} (1 – \frac{x²}{2!} + \frac{x^{4}}{4!} – \frac{x^{6}}{6!} + · · · ) + c_{1} (x – \frac{x^{3}}{3!} + \frac{x^{5}}{5!} – \frac{x^{7}}{7!} + · · · );
that is, y(x) = c_{0} \cos x + c_{1} \sin x. Note that we have no problem with the radius of convergence here; the Taylor series for the sine and cosine functions converge for all x.