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Assume y′′ + xy′ + y = 0 has a power-series solution and determine the coefficient an

Assume that the differential equation

    \[ y'' + xy' + y = 0 \]

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}. \end{align*}

So, from the given differential equation we have

    \begin{align*}  && y'' + xy' + y &= 0 \\[9pt]  \implies && \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + x \sum_{n=1}^{\infty} n a_n x^{n-1} + \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt]  \implies && \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} x^n + \sum_{n=1}^{\infty} n a_n x^n + \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt]  \implies && 2a_2 + \sum_{n=1}^{\infty} (n+2)(n+1)a_{n+2} x^n + \sum_{n=1}^{\infty} na_n x^n + a_0 + \sum_{n=1}^{\infty} a_n x^n &= 0 \\[9pt]  \implies && 2a_2 + a_0 + \sum_{n=1}^{\infty} ((n+2)(n+1) a_{n+2} + na_n + a_n)x^n &= 0 \\[9pt]  \implies && 2a_2 + a_0 + \sum_{n=1}^{\infty} ((n+2)(n+1)a_{n+2} + (n+1)a_n) x^n &= 0. \end{align*}

Since each coefficient of x^n must equal 0 for this equation to hold we have

    \begin{align*}  2a_2 + a_0 &= 0 \\[9pt]  (n+2)(n+1)a_{n+2} +(n+1)a_n &= 0 & \implies && a_{n+2} = \frac{-1}{n+2} a_n. \end{align*}

By induction we then have

    \begin{align*}  a_{2n} &= \frac{(-1)^n}{2 \cdot 4 \cdot \cdots \cdot (2n)} \\[9pt]  a_{2n+1} &= \frac{(-1)^{n+1}}{1 \cdot 3 \cdot \cdots \cdot (2n-1)}. \end{align*}

The coefficients a_0 and a_1 are arbitrary and we denote them by c_0 and c_1 respectively. Then we have

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

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