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

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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