Solutions in terms of Power Series

Consider a linear second order equation of the type

$$y^{\prime\prime} + a(x) y^\prime + b(x) y = 0.$$ (9.2.1)

Let $a$ and $b$ be analytic around the point $x_0 = 0.$ In such a case, we may hope to have a solution $y$ in terms of a power series, say

$$y = \sum_{k=0}^\infty c_k x^k.$$ (9.2.2)

In the absence of any information, let us assume that (9.2.1) has a solution $y$ represented by (9.2.2). We substitute (9.2.2) in Equation (9.2.1) and try to find the values of $c_k$ 's. Let us take up an example for illustration.

EXAMPLE 9.2.1   Consider the differential equation

$$y^{\prime\prime} + y = 0$$ (9.2.3)

Here $a(x) \equiv 0, \; b(x) \equiv 1,$ which are analytic around $x_0 = 0.$
Solution: Let

$$y = \sum_{n=0}^\infty c_n x^n.$$ (9.2.4)

Then $y^\prime = \sum\limits_{n=0}^\infty n c_n x^{n-1}$ and $y^{\prime\prime} = \sum\limits_{n=0}^\infty n (n-1) c_n x^{n-2}.$ Substituting the expression for $y, \;y^\prime$ and $y^{\prime\prime}$ in Equation (9.2.3), we get

$$\sum_{n=0}^\infty n (n-1) c_n x^{n-2} + \sum_{n=0}^\infty c_n x^n = 0$$

or, equivalently

$$0 = \sum_{n=0}^\infty (n+2)(n+1) c_{n+2} x^{n} + \sum_{n=0}^\infty c_n x^n = \sum_{n=0}^\infty \{ (n+1)(n+2) c_{n+2} + c_n\} x^{n}.$$

Hence for all $n = 0, 1, 2, \ldots,$

$$(n+1)(n+2) c_{n+2} + c_n = 0 \; {\mbox{ or }} \; c_{n+2} = - \frac{c_n}{(n+1)(n+2)}.$$

Therefore, we have

$$\begin{array}{ll} c_2 = - \frac{c_0}{2!}, & \hspace{1.5in} c_... ...pace{1.5in} c_{2n+1} = (-1)^n \frac{c_1}{(2n+1)!}. \end{array}$$

Here, $c_0$ and $c_1$ are arbitrary. So,

$$y = c_0 \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} + c_1 \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

or $y = c_0 \cos (x) + c_1 \sin (x)$ where $c_0$ and $c_1$ can be chosen arbitrarily. For $c_0 = 1$ and $c_1 = 0,$ we get $y = \cos (x).$ That is, $\cos (x)$ is a solution of the Equation (9.2.3). Similarly, $y = \sin (x)$ is also a solution of Equation (9.2.3).

EXERCISE 9.2.2   Assuming that the solutions $y$ of the following differential equations admit power series representation, find $y$ in terms of a power series.

1. $y^\prime = - y,$ (centre at $x_0 = 0$ ).
2. $y^\prime = 1 + y^2,$ (centre at $x_0 = 0$ ).
3. Find two linearly independent solutions of
   1. $y^{\prime\prime} - y = 0,$ (centre at $x_0 = 0$ ).
   2. $y^{\prime\prime} + 4 y = 0,$ (centre at $x_0 = 0$ ).