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

by RoRi

April 24, 2016

Assume that the differential equation

$y' = \alpha y$

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

First, we have

$y = \sum_{n=0}^{\infty} a_n x^n \quad \implies \quad y' = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=0}^{\infty} (n+1)a_{n+1} x^n.$

Therefore, we have

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

Equating like powers of $x$ we obtain the recursive relation

$a_{n+1} = \frac{\alpha \cdot a_n}{n+1}, \qquad a_0 = 1.$

By induction, we then have

$a_n = \frac{\alpha}{n!}$

Therefore,

$y = \sum_{n=0}^{\infty} \frac{\alpha}{n!} x^n.$

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