An example task from Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" sounds as follow (p. 145, Example 1):
"Find the longest interval in which the solution of the initial value problem
(t^2 - 3t)y'' + ty' - (t + 3)y = 0, y(1) = 2, y'(1) = 1
is certain to exist.
If the given differential equation is written in the form of Eq. (4), then p(t) = 1/(t-3), q(t) = -(t+3)/t(t-3), and g(t) = 0.
The example does not stop at this point and I think I understand the rest of it, however I don't understand how the values of p(t) and q(t) are determined, how is this accomplished?
The Eq. (4) which is refered to is part of Theorem 3.2.1. which sounds as follows:
"Consider the initial value problem
y'' + p(t)y' + q(t)y = g(t), y(t0) = y0, y'(t0) = y'0, (4)
where q, p, and g are continuous on an open interval I that contains the point t0. Then there is
exactly one solutions y = ø(t) of this problem, and the solution exists throughout the interval I."
So again my question is, how are p(t) and q(t) determined?
Thanks in advance!