Solve the initial value problem by using power series:
$y'' + xy' - 2y = 0$,
$y(0) = 1$,
$y'(0) = 0$
Assume the form
$y(x) = \sum_{n=0}^{\infty} c_n x^n$.
Then
$y'(x) = \sum_{n=1}^{\infty} n c_n x^{n-1}$
$y''(x) = \sum_{n=2}^{\infty} n(n-1) c_n x^{n-2}$
$y''(x) = \sum_{n=0}^{\infty} (n+1)(n+2) c_{n+2} x^n$ (Note: shift of index of summation must be used here)
$xy'(x) = \sum_{n=1}^{\infty} n c_n x^n$
$-2y(x) = \sum_{n=0}^{\infty} -2 c_n x^n$
Then $y'' + xy' - 2y$
$= \sum_{n=0}^{\infty} [(n+2)(n+1) c_{n+2} + \boxed{n} c_{n+1} + \boxed{-2} c_n]x^n$
Requiring that the individual terms of this series for the left side of the differential equation vanish gives the recurrence relation
$c_{n+2} = \boxed{-\frac{n}{ (n+2)(n+1)}} c_{n+1} + \boxed{\frac{2}{(n+2)(n+1)}} c_n$ for $n = 0, 1, 2, ...$