Question 11.1.3: Solve the equation x²y′ = y - x - 1.

We make the usual substitutions y = \sum{c_{n} x^{n}}   and   y^{′} = \sum{n c_{n} x^{n-1}}, which yield

x² \sum\limits_{n=1}^{\infty}{n c_{n} x^{n-1}} = -1 – x + \sum\limits_{n=0}^{\infty}{c_{n}x^{n}}

so that

\sum\limits_{n=1}^{\infty}{n c_{n} x^{n+1}} = -1 – x + \sum\limits_{n=0}^{\infty}{c_{n}x^{n}}.

Because of the presence of the two terms -1 and -x on the right-hand side, we need to split off the first two terms, c_{0} + c_{1} x, of the series on the right for comparison. If we also shift the index of summation on the left by -1 (replace n = 1 with n = 2 and n with n – 1), we get

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

Because the left-hand side contains neither a constant term nor a term containing x to the first power, the identity principle now yields c_{0} = 1, c_{1} = 1,   and   c_{n} = (n – 1)c_{n-1 }  for   n ≧ 2. It follows that

c_{2} = 1 · c_{1} = 1!,       c_{3} = 2 · c_{2} = 2!,     c_{4} = 3 · c_{3} = 3!,

and, in general, that

c_{n} = (n – 1)!          for n ≧ 2:

Thus we obtain the power series

y(x) = 1 + x + \sum\limits_{n=2}^{\infty}{(n – 1)! x^{n}}.

But the radius of convergence of this series is

ρ = \underset{n → \infty}{\lim} \frac{(n – 1)!}{n!} = \underset{n → \infty}{\lim} \frac{1}{n} = 0.

so the series converges only for x = 0. What does this mean? Simply that the given differential equation does not have a (convergent) power series solution of the assumed form y = \sum{c_{n}x^{n}}. This example serves as a warning that the simple act of writing y = \sum{c_{n}x^{n}} involves an assumption that may be false.