) Let y(1) = y0, y 0 (1) = v0. Solve the initial value problem. What is the longest interval on which the initial value problem is certain to have a unique twice differentiable solution?

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.

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

step-by-step explanation:

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11-1 > 1-1+a

-12 > a