Power series solution (Ordinary point)

Consider the 2nd-order linear differential equation with variable coefficients

$$y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = R(x).$$

If $P(x), Q(x)$, and $R(x)$ area analytic at $a$, then we call $a$ ordinary point.

Theorem 4..6 (Existence and uniqueness of initial value problem)   For the 2nd-order linear differential equation

$$y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = R(x),$$

if $a$ is an ordinary point, then for any $b_{0}, b_{1}$, the initial value problem

$$y(a) = b_{0}, \ \ y^{\prime}(a) = b_{1}$$

has a unique solution $y(x) = \sum_{n=0}^{\infty}c_{n}(x - a)^{n}$.

Example 4..2   Find the power series solution around $x = 0$

$$y^{\prime\prime} + xy' -4y = 0.$$

SOLUTION Since $P(x) = x, Q(x) = -4, R(x) = 0$, $x = 0$ is an ordinary point. Now let

$$y(x) = \sum_{n=0}^{\infty}c_{n}x^{n}$$

be a solution. Then，

$$y^{\prime} = \sum_{n=1}^{\infty}nc_{n}x^{n-1}$$

$$y^{\prime\prime} = \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}$$

Substitute these back into the given differential equation.

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=0}^{\infty}nc_{n}x^{n} - 4\sum_{n=0}^{\infty}c_{n}x^{n} = 0$$

Now let the power of $x$ be the least number $n-2$.，

$$\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2} + \sum_{n=3}^{\infty}(n-2)c_{n-2}x^{n-2} - 4\sum_{n=2}^{\infty}c_{n-2}x^{n-2} = 0 .$$

Then let the index be the largest number $n=3$. Then

$$2c_{2} - 4c_{0} + \sum_{n=3}^{\infty}[n(n-1)c_{n} + (n-6)c_{n-2}]x^{n-2} = 0.$$

Since the right-hand side is 0，by the term-by-term differentiation，the coefficients of $x^{n}$ are all 0. Thus

$$2c_{2} - 4c_{0} = 0, \ n(n-1)c_{n} + (n-6)c_{n-2} = 0 , \ n \geq 3$$

or

$$c_{2} = 2c_{0}, \ c_{n} = -\frac{(n-6)c_{n-2}}{n(n-1)}, \ n \geq 3.$$

Here, $c_{0},c_{1}$ can be decided by the initial conditions $y(0), y^{\prime}(0)$. Thus, we can think of these as constants. Now we find $c_{2},c_{3},\ldots$. Then

$$c_{2} = 2c_{0}, \ c_{3} = \frac{-c_{1}}{2}, \cdots$$

and

$$c_{2n} = \frac{c_{2n}}{c_{2n-2}}\frac{c_{2n-2}}{c_{2n-4}}\frac{c_{2n-4}}{c_{2n-6}}c_{2n-6}, n \leq 2 \ c_{2n} = 0, n \geq 3$$

or

$$c_{2n+1} = \frac{c_{2n+1}}{c_{2n-1}}\frac{c_{2n-1}}{c_{2n-3}}\cdo... ...{c_{3}}{c_{1}}c_{1} = \frac{3(-1)^{n}c_{1}}{2^n n! (2n+1)(2n-1)(2n-3)}x^{2n+1}$$

Thus

$$y = \sum_{n=0}^{\infty}c_{n}x^{n}$$

$$= c_{0} + 2c_{0}x^{2} - \frac{c_{0}}{3}x^4 - c_{1}\sum_{n=2}^{\infty} \frac{3(-1)^{n}}{2^n n! (2n+1)(2n-1)(2n-3)}x^{2n+1} \ensuremath{\ \blacksquare}$$