Question 3 of 5.2 in Boyce's Differential Calculus asks
a) Seek power series solutions around x_0 and find the recurrence relation
b) Find the first four terms in y1 and y2
c) Show that y1 and y2 are a fundamental set of solutions (through evaluating W(y1,y1)(x0)
d) Find the general term if possible in each solution
y" - xy' - y = 0, x_0 = 1
The textbook gives the answer to a) as (n+2)a_n+2 - a_n+1 - a_n = 0 but I can't figure out for the life of me how I would distribute the x = 1 + (x-1) properly to get that answer.
$$y''-(1-(x-1))y' - y = 0$$
$$y''-y'-(x-1)y' - y = 0$$
The goal is to center this solution around $x_o=1$, so we use power series in the form
$y=\sum_{n=0}^{\infty} a_n (x-1)^n$
$y'=\sum_{n=1}^{\infty} a_n n(x-1)^{n-1} $
$\ = \sum_{n=0}^{\infty} a_{n+1} (n+1)(x-1)^{n}$
$y''=\sum_{n=2}^{\infty} a_n n(n-1)(x-1)^{n-2} $
$\ = \sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)(x-1)^{n}$
$\ $
The trick above is to reindex your sums. This gives us a few forms of $y,y',y''$ to play with. To reindex, think "if I scroll my index back by 1, I need to set every occurrence of the index in the summand ahead by 1." Hence changing index $n=1$ to $n=0$ changes the term $n$ to $n+1$. We'll find that some of the above forms of $y,y',y''$ will be useful, while otherwise not so much. Carefully follow the choice of substitutions that will follow.
Remember, we want all the powers of $(x-1)$ to be the same. After substitutions, the differential equation becomes:
$\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)(x-1)^{n}-\sum_{n=0}^{\infty} a_{n+1} (n+1)(x-1)^{n}$
$\ -(x-1)\sum_{n=1}^{\infty} a_n n(x-1)^{n-1} - \sum_{n=0}^{\infty} a_n (x-1)^n = 0 $
$\ $
Okay, did you notice the sleight of hand? We used two different forms for $y'$. The second form has power $(x-1)^{n-1}$, but it can gobble up the $(x-1)$ in front of the summation to make it the desired power $n$.
Once we do that and collect terms, we find
$\sum_{n=0}^{\infty} [a_{n+2}(n+2)(n+1)- a_{n+1}(n+1)- a_n](x-1)^{n}-\sum_{n=1}^{\infty} a_n n(x-1)^{n}= 0$
$\ $
Now we're close. We want all of our indices to be "on the same page." The left sum is at $n=0$ the right is at $n=1$. Here we do not reindex. Rather, we just take the sum for $n=0$.
$2a_2-a_1-a_0 + $
$\ \sum_{n=1}^{\infty} [a_{n+2}(n+2)(n+1)- a_{n+1}(n+1)- a_n](x-1)^{n}-\sum_{n=1}^{\infty} a_n n(x-1)^{n}= 0$
$\ $
$2a_2-a_1-a_0 + $
$\ \sum_{n=1}^{\infty} [a_{n+2}(n+2)(n+1)- a_{n+1}(n+1)- (n+1) a_n](x-1)^{n}= 0$
$\ $
So the recurrence relation is
$$a_{n+2}(n+2)(n+1)- a_{n+1}(n+1)- (n+1) a_n = 0$$
$$a_{n+2}(n+2)- a_{n+1}- a_n = 0$$