# Ask Uncle Colin: about Chebyshev's Equation

Dear Uncle Colin,

I’ve been asked to solve Chebyshev’s equation using a series expansion:

$(1-x)\diffn{2}{y}{x} - x\dydx + p^2 y = 0$

assuming $y=C_0 + C_1 x + C_2 x^2 + …$.

I end up with the relation $C_{N+2} = \frac{C_N \left(N^2 -p^2\right)}{(N+2)(N+1)}$, but the given answer has a + on top. Where have I gone wrong?

- Can’t Have Everything, But Your Signs Have Erroneously Varied

Hi, CHEBYSHEV, and thanks for your message!

The TL;DR here is, I think your answer is correct and the given answer wrong. I presume this is from a Physics course, where the odd minus sign is seen as an occupational hazard.

The trickiest part (for me) with series solutions is keeping the indices and subscripts straight, so I generally write down a few terms for each of the parts to make sure I have them right before doing anything clever with $\Sigma$s.

Here, I’d start with the various derivatives:

- $y=C_0 + C_1 x + C_2 x^2 + … + C_n x^n + …$
- $y’ = C_1 + 2C_2 x + 3 C_3 x^2 + … + n C_n x^{n-1} + …$
- $y’’ = 2C_2 + 6C_3 x + 12 C_4 x^2 + … + n(n-1) C_n x^{n-2} + …$

And then look at the four (yes, four) terms of the differential equation:

- $y’’ = \sum_{n=2}^\infty C_n n(n-1) x^{n-2}$
- $-x^2 y’’ = \sum_{n=2}^\infty -C_n n(n-1) x^n$
- $-xy’ = \sum_{n=1}^\infty -n C_n x^n$
- $p^2 y = \sum_{n=0}^\infty C_n x^n p^2$

All of those have $x^n$ terms in except for the first one, which has $x^{n-2}$. If I change the variable so $N = n-2$, it becomes:

$y’’ = \sum_{N=0}^\infty C_{N+2} (N+2)(N+1) x^N$

And we want the coefficients for each power of $x$ to sum to 0. (With a little sleight of hand, I’m going to turn those placeholder $N$s back into $n$s.)

That gives us:

$C_{n+2} (n+2)(n+1) - C_n (n-1)n - n C_n + C_n p^2 = 0$, for $n \ge 2$

Rearranging, $C_{n+2} = C_n \frac{ n(n-1)+n - p^2}{(n+2)(n+1)}$, which simplifies to

$C_{n+2} = C_n \frac{n^2 - p^2}{(n+2)(n+1)}$, as you have.

As a belt-and-braces approach, I also looked it up on MathWorld, which agrees with our answers.

Physicists, eh?

Hope that helps,

- Uncle Colin