Question 11.1.1: Solve the equation y′ + 2y = 0.

We substitute the series

y = \sum\limits_{n=0}^{\infty}{c_{n}x^{n}}      and   y^{′} =\sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}}

and obtain

\sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}} + 2 \sum\limits_{n=0}^{\infty}{c_{n} x^{n}} = 0.                       (21)

To compare coefficients here, we need the general term in each sum to be the term containing x^{n}. To accomplish this, we shift the index of summation in the first sum. To see how to do this, note that

\sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}} = c_{1} + 2c_{2} x + 3c_{3} x² + · · · = \sum\limits_{n=0}^{\infty}{(n + 1)c_{n+1} x^{n}}.

Thus we can replace n with n + 1 if, at the same time, we start counting one step lower; that is, at n = 0 rather than at n = 1. This is a shift of +1 in the index of summation. The result of making this shift in Eq. (21) is the identity

\sum\limits_{n=0}^{\infty}{(n + 1)c_{n+1}x^{n}} +  2 \sum\limits_{n=0}^{\infty}{c_{n}x^{n}} = 0;

that is,

\sum\limits_{n=0}^{\infty}{[(n + 1)c_{n+1} + 2c_{n}]x^{n}} = 0.

If this equation holds on some interval, then it follows from the identity principle that (n + 1)c_{n+1} + 2c_{n} = 0 for all n ≧ 0; consequently,

c_{n+1} = – \frac{2c_{n}}{n + 1}                 (22)

for all n≧0. Equation (22) is a recurrence relation from which we can successively compute c_{1}, c_{2}, c_{3}, . . .  in terms of c_{0}; the latter will turn out to be the arbitrary constant that we expect to find in a general solution of a first-order differential equation. With n = 0, Eq. (22) gives

c_{1} = – \frac{2c_{0}}{1}.

With n = 1, Eq. (22) gives

c_{2} = – \frac{2c_{1}}{2} = + \frac{2²c_{0}}{1 · 2} = \frac{2²c_{0}}{2!}.

With n = 2, Eq. (22) gives

c_{3} = – \frac{2c_{2}}{3} = – \frac{2³c_{0}}{1 · 2 · 3} = – \frac{2³c_{0}}{3!}.

By now it should be clear that after n such steps, we will have

c_{n} = (-1)^{n} \frac{2^{n}c_{0}}{n!},           n ≧ 1.

(This is easy to prove by induction on n.) Consequently, our solution takes the form

y(x) = \sum\limits_{n=0}^{\infty}{c_{n} x^{n}}  = \sum\limits_{n=0}^{\infty}{(-1)^{n} \frac{2^{n}c_{0}}{n!}x^{n}} = c_{0}  \sum\limits_{n=0}^{\infty}{\frac{(-2x)^{n}}{n!}} = c_{0}e^{-2x}.

In the final step we have used the familiar exponential series in Eq. (5) to identify our power series solution as the same solution y(x) = c_{0}e^{-2x} we could have obtained by the method of separation of variables.

e^{x} = \sum\limits_{n=0}^{\infty}{\frac{x^{n}}{n!}} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + · · · ;                      (5)