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.

Solution:

If the given differential equation is written in the form of y'' + p(t)y' + q(t)y = g(t), y(t_0) = y_0, y'(t_0) = y'_0 then p(t) = 1/(t - 3), q(t) = -(t + 3)/t(t - 3) and g(t) = 0. The only points of discontinuity of the coefficients are t = 0 and t = 3. Therefore, the longest open interval, containing the initial point t = 1, in which all the coefficients are continuous is 0 < t < 3. Thus this is the longest interval in which the theorem guarantees that the solution exists.

Taken from an example from Boyce & DiPrima's book on differential equations.

Question:

Why is it that p(t) = 1/(t - 3)? Why is not p(t) = t?