Thread: Example of Fundamental Solutions of Linear Homogeneous Equations

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!

Originally Posted by posix_memalign

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

We want to divide through by in order to do that we need to know the function is non-zero. We also have that needs to be contained in this interval by the initial value condition. Therefore, it is the interval .