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,

$f(x) = 1 - 3x^2.$

2 comments

June 6, 2020 at 7:23 pm

Leland says:

Big mistake here… the remaining terms do not ALL turn to zero after a4. Every other term will be zero from that point on, given that the recursive relation is between m and m+2… this leaves m+1, m+3, etc… to be nonzero.

- June 15, 2020 at 4:26 pm
  
  Anonymous says:
  
  I don´t know if I understood your point, but the terms become zero as a_1=0 so a_3,a_5,… are zero and a_4=0 so a_6,a_8,… become zero too.

Stumbling Robot

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

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