The differential equation is
$y'' - xy' - y = 0$ with $x_0 = 1$
Now, I know how to find the recurrence relation... and it's given by:
$a_(n+2) = [(a_(n+1) + a_(n)) / (n+2)]$
But I can't quite figure out how to go about finding the first four terms of $y_1$ and $y_2...$ for instance, the answer for y1 given in the book is
$y_1 = 1 + (1/2)(x-1)^2 + (1/6)(x-1)^3 + (1/6)(x-1)^4$
But when I work it out myself I get
$y_1 = 1 + (1/2)(x-1)^2 + (1/6)(x-1)^3 + (1/8)(x-1)^4$
The book's answer is correct. Also $y_2(x)=x-1+\frac{1}{2}(x-1)^2+\frac{1}{2}(x-1)^3+\frac{1}{4}(x-1)^4+\cdots$ is the other solution.