Home » Blog » Use the method of undetermined coefficients to solve (1-x2)y′′ – 2xy′ + 12y = 0

Use the method of undetermined coefficients to solve (1-x2)y′′ – 2xy′ + 12y = 0

Consider the differential equation

    \[ (1-x^2)y'' - 2xy' + 12y = 0. \]

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

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \qquad \text{with} \qquad f(0) = 0, \quad f'(0) = 2. \]

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) = 0 and f'(0) = 2 we have

    \begin{align*}  f(0) &= \sum_{n=0}^{\infty} a_n 0^n = a_0 = 0 \\  f'(0) &= \sum_{n=1}^{\infty} a_n 0^{n-1} = a_1 = 2. \end{align*}

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

    \begin{align*}  &&(1-x^2)\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} - 2x \sum_{n=1}^{\infty} na_n x^{n-1} + 12 \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 + 12\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 + 12 \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 + 12a_n \right)x^n \right) + 2a_2 + 6a_3 x- 2a_1 x + 12a_0 + 12a_1 x &= 0. \end{align*}

Then, we substitute in the values we obtained above a_0 = 0 and a_1 = 2 to obtain

    \begin{align*}  && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} - n(n-1)a_n - 2na_n + 12a_n \right)x^n \right) + 2a_2 + 6a_3 x - 4x + 24x &= 0 \\[9pt]  \implies && \left( \sum_{n=2}^{\infty} \left( (n+2)(n+1)a_{n+2} -a_n (n^2 +n -12) \right) x^n \right) +2a_2 + (6a_3 + 20)x &= 0. \end{align*}

Since this sum is equal to 0 for all x, we know that the coefficients must all be equal to 0. First, we solve for a_2 and a_3,

    \begin{align*}  2a_2 &=0 &\implies && a_2 &= 0 \\  6a_3 +20 &=0 &\implies && a_3 &= -\frac{10}{3}. \end{align*}

Then, we obtain the recursion relation,

    \[ (n+2)(n+1)a_{n+2} - (n+4)(n-3)a_n = 0 \quad \implies \quad a_{n+2} = \frac{(n+4)(n-3)}{(n+2)(n+1)} a_n. \]

For the even terms, since a_2 = 0 we have a_4 = a_6 = \cdots = 0. For the odd terms we start with n =3 and have

    \[ a_5 = \frac{(7)(0)}{(5)(4)} a_3 = 0. \]

Therefore, the remaining odd terms a_5 = a_7 = \cdots = 0. Therefore, the have the following coefficients

    \[ a_0 = a_2 = \cdots = 0, \qquad a_1 = 2, \quad a_3 = -\frac{10}{3}, \quad a_5 = a_7 = \cdots = 0. \]

So the sum is

    \[ f(x) = 2x - \frac{10}{3}x^3. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):