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Use the method of undetermined coefficients to solve (1-x2)y′′ – 2xy′ + 6y = 0

Consider the differential equation

    \[ (1-x^2)y'' - 2xy' + 6y = 0. \]

The solution to this differential equation has a power-series expansion

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \qquad \text{with} \qquad f(0) = 1, \quad f'(0) = 0. \]

Using the method of undetermined coefficients obtain a recursion formula relating the terms a_{n+2} to the terms a_n. Give an explicit formula for a_n for each n and find the sum of the series.


First, we differentiate twice,

    \begin{align*}  f(x) &= \sum_{n=0}^{\infty} a_n x^n \\  f'(x) &= \sum_{n=1}^{\infty} na_n x^{n-1} \\  f''(x) &= \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}. \end{align*}

From the initial conditions f(0) = 1 and f'(0) = 0 we have

    \begin{align*}  f(0) &= \sum_{n=0}^{\infty} a_n x^n = a_0 = 1 \\  f'(0) &= \sum_{n=1}^{\infty} na_n x^{n-1} = a_1 = 0. \end{align*}

Now, we plug the expressions for y, \ y', and y'' back into the given differential equation,

    \begin{align*}  && (1-x^2)y'' - 2xy' + 6y &= 0 \\[9pt]  \implies && (1-x)^2 \left(\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} \right) - 2x \sum_{n=1}^{\infty} na_n x^{n-1} + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt]  \implies && \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \sum_{n=2}^{\infty} n(n-1)a_n x^n - 2 \sum_{n=1}^{\infty} na_n x^n + 6 \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=2}^{\infty} n(n-1)a_n x^n - 2 \sum_{n=1}^{\infty} na_n x^n + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\[9pt]  \implies && \left(\sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + 6a_n \right)x^n\right) + 2a_2 + 6a_3 x -2a_1 x + 6a_0 + 6a_1 x &= 0. \end{align*}

Then, we use the fact from above that a_0 = 1 and a_1 = 1 to get

    \begin{align*}  && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + 6a_n \right) \right) + 2a_2 + 6a_3 x + 6 &= 0 \\[9pt]  \implies && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - a_n (n^2 + n - 6) \right) \right) + 6a_3 x + 2a_2 + 6 &= 0 \\  \implies && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - (n+3)(n-2)a_n \right)\right) + 6a_3 x + 2a_2 + 6 &= 0. \end{align*}

Since this sum is equal to 0, we know that every coefficient of every power of x must be equal to 0. First, we solve for a_2 and a_3,

    \begin{align*}  2a_2 + 6 &= 0 & \implies && a_2 &= -3 \\  6a_3 &= 0 & \implies && a_3 &= 0. \end{align*}

Then, we establish the recursive relationship between a_n and a_{n+2},

    \begin{align*}  && (n+2)(n+1)a_{n+2} - (n+3)(n-2)a_n &= 0 \\  \implies && (n+2)(n+1) a_{n+2} &= (n+3)(n-2)a_n \\  \implies && a_{n+2} &= \frac{(n+3)(n-2)}{(n+2)(n+1)} a_n \end{align*}

for all n \geq 0. Then since a_3 = 0 we have a_{2n+1} = 0 for all n (since for every odd integer n the formula for a_{n+2} has is multiplied by a_n, but each of these will be 0). For the even terms we have for n=2,

    \[ a_4 = \frac{(n+3)(n-2)}{(n+2)(n+1)} a_2 = \frac{(5)(0)}{(4)(3)} (-3) = 0.\]

This means all of the remaining even terms will be 0 as well. So we have

    \[ a_0 = 1, \quad a_1 = 0, \quad a_2 = -3, \quad a_3 = a_4 = \cdots = 0. \]

Hence,

    \[ f(x) = 1 - 3x^2. \]

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