In the series
n=0∑∞bnxn you find that
b1=−1 but
b2 ultimately depends on what
b0 is. In fact, every
bn for
n≥2 depends on what
b0 is. Once you set that, every other coefficient gets fixed in place.
This coefficient is an arbitrary choice for us, because we have no restriction for it. Any suitable value of it will yield every other
bn value and hence give us a solution for
y2. If that makes sense to you, then the next step for you to understand is that
b0=0 is the simplest case for us pick, so just do that.
If you pick a non-zero value then you're going to get the same sum but expressed differently.
You do the same thing when you determine the solution for
y1=n=0∑∞anxn. You had the choice of picking arbitrary
a0 but you picked
a0=1 since the obtained recurrence relation must satisfy
an=n2(1−n)an−1−an−2so
a1=0 (from comparing
x−1 terms) and
a2=−4a0. Similarly,
a3 depends on
a0, and so on. You fixed
a0=1 and obtained all the coefficients you see in
y1.
No. Once you substitute in a series into this particular ODE, you should notice that we always begin with
x−1, then we have the
x0 (constant) terms, and then
x1, and so on... but for my explanation I'm only considering
x−1 and
x0.
The
x−1 coefficients turn out to be
a1 (when you substitute in
y1) and
b1+1 (when you substitute in
y2) which yields us
a1=0 and
b1=−1. But once again I stress the fact that we have no restrictions on
a0,b0 hence we choose them arbitrarily so we just pick
a0=1 and
b0=0.
I say they're arbitrary, for you shouldn't (for instance) pick
a0=0 because then the entire series for
y1 is just zero -- a trivial solution.