Assume y′′ = xy has a power-series solution and determine the coefficient an

Assume that the differential equation

$y'' = xy$

has a power-series solution $y = \sum a_n x^n$ and find a formula for the coefficient $a_n$ .

First, we have

$\begin{align*} && y &= \sum_{n=0}^{\infty} a_n x^n \\[9pt] \implies && y' &= \sum_{n=1}^{\infty} n a_n x^{n-1} \\[9pt] \implies && y'' &= \sum_{n=2}^{\infty}n (n-1) a_n x^{n-2} = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n. \end{align*}$

Therefore, we have

$\begin{align*} && y'' &= xy \\[9pt] \implies && \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n &= x \sum_{n=0}^{\infty} a_n x^n \\[9pt] \implies && \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n &= \sum_{n=0}^{\infty} a_n x^{n+1} \\[9pt] \implies && 2a_2 + \sum_{n=1}^{\infty} (n+2)(n+1)a_{n+2}x^n &= \sum_{n=0}^{\infty} a_n x^{n+1} \\[9pt] \implies && 2a_2 + \sum_{n=0}^{\infty} (n+3)(n+2) a_{n+3} x^{n+1} &= \sum_{n=0}^{\infty} a_n x^{n+1}. \end{align*}$

Equating like powers of $x$ , we have a recursive relation for $a_n$ when $n > 2$ given by

$a_{n+3} = \frac{a_n}{(n+3)(n+2)},$

Furthermore, we have that $a_2 = 0$ and that $a_0$ and $a_1$ are arbitrary constants, say $c_0$ and $c_1$ , respectively. Then by induction we establish

$\begin{align*} a_{3n} &= \frac{a_0}{(2 \cdot 3)(5 \cdot 6) \cdots ((3n-1)\cdot(3n))} \\[9pt] a_{3n+1} &= \frac{a_1}{(3 \cdot 4)(6 \cdot 7) \cdots ((3n)\cdot(3n+1))} \\[9pt] a_{3n+2} &= 0. \end{align*}$

Therefore,

$y = c_0 \left( 1 + \sum_{n=1}^{\infty} \frac{x^{3n}}{(2 \cdot 3)(5 \cdot 6) \cdots ((3n-1)\cdot(3n))} \right) + c_1 \left( x + \sum_{n=1}^{\infty} \frac{x^{3n+1}}{(3 \cdot 4)(6 \cdot 7) \cdots ((3n)\cdot (3n+1))} \right).$

Stumbling Robot

Assume y′′ = xy has a power-series solution and determine the coefficient a_n

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply