Respuesta :
QUESTION IS INCOMPLETE.
Nevertheless, I will explain how to find, without solving, the longest interval in which an initial value problem is certain to have a unique twice differentiable solution.
Step-by-step explanation:
Consider the Existence and uniqueness theorem:
Let p(t) , q(t) and r(t) be continuous on an interval a ≤ t ≤ b, then the differential equation given by:
y''+ p(t) y' +q(t) y = r(t) ;
y(t_0) = y_0, y'(t_0) = y'_0
has a unique solution defined for all t in the stated interval.
Example:
Consider the differential equation
ty'' + 9y = t
y(1) = y_0,
y'(1) = v_0
ty'' + 9y = t .................................(1)
First, write the differential equation (1) in the form:
y'' + p(t)y' + q(t)y = r(t) ..................(2)
by dividing (1) by t
So
y''+ (9/t)y = 1 ....................................(3)
Comparing (3) with (2)
p(t) = 0
q(t) = 9/t
r(t) = 1
For t = 0, p(t) and r(t) are continuous, but q(t) is undefined.
q(t) is continuous everywhere apart from the point t = 0.
We say (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous.
But t = 1, which is contained in the initial conditions y(1) = y_0 and y'(1) = v_0 is found in (0,∞).
So, we conclude that this interval is the longest interval in which the initial value problem has a unique twice differentiable solution.
