Series Solution of ODE: Finding Non-Zero Coefficients for y(x) Expansion

[hadroneater]

Homework Statement

(1 - x)y'' + xy' + xy = 0

Find the first 3 nonzero coefficients of the power series expansion about x = 0 if y(0) = -1 and y'(0) = 0

Homework Equations

The Attempt at a Solution

$y = \sum^{∞}_{n = 0}c_{n}x^{n}$

From above, I can say that $y(0) = 1 = c_{0}$ and $y'(0) = 0 = c_{1}$

$L_{y} = (1 - x)\sum^{∞}_{n = 2}c_{n}n(n-1)x^{n-2} + \sum^{∞}_{n = 1}c_{n}nx^{n} + \sum^{∞}_{n = 0}c_{n}x^{n + 1} = 0$

Some indices manipulation so all the power series are for $x^{m-2}$...

$L_{y} = \sum^{∞}_{n = 3}[c_{m}m(m-1) - c_{m - 1}(m - 1)(m - 2) + c_{m - 2}(m - 2) + c_{m - 3}]x^{m - 2} + 2c_{2} = 0$

$c_{2} = arbitrary number$ but I can't assume that it's 0, right?

$c_{m}m(m-1) - c_{m - 1}(m - 1)(m - 2) + c_{m - 2}(m - 2) + c_{m - 3} = 0 for m ≥ 3$

m = 3:
$6c_{3} - 2c_{2} = -1$

m = 4:
$12c_{4} - 6c_{3} + 2c_{2} = 0$

$c_{4} = -1/12$

recurrence equation:
$c_{m} = \frac{c_{m-1}(m - 1)(m - 2) - c_{m-2}(m - 2) - c_{m-3}}{m(m - 1)}$

I can't seem to find the 3rd non-zero constant...

 

[LCKurtz]

$L_{y} = \sum^{∞}_{n = 3}[c_{m}m(m-1) - c_{m - 1}(m - 1)(m - 2) + c_{m - 2}(m - 2) + c_{m - 3}]x^{m - 2} + 2c_{2} = 0$

$c_{2} =$ arbitrary number but I can't assume that it's 0, right?

You don't have to assume $c_2$ is zero. You have proved it is by equating the constant terms on both sides of that equation.

 

[hadroneater]

Ah, right. Got it.