Question

In each of Problems 1 through 6 , determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist.

1. (t−3)y'+(lnt)y=2t, y(1)=2
2. t(t−4)y'+y=0, y(2)=1
3. y'+(tant)y=sint,y(π)=0
4. (4−t²)y'+2ty=3t² , y(−3)=1 5.
5. (4-t²) y' + 2ty = 3t², y(1) = -36.
6. (lnt)y'+y=cot t,y(2)=3

Answer 1

The question relates to determining the intervals of existence for solutions to initial value problems in differential equations, guided by the existence and uniqueness theorem.

Step-by-step explanation:

The question involves finding intervals where the solution of initial value problems is guaranteed to exist, without actually solving the problems.

This concept is linked to the existence and uniqueness theorem for differential equations, which states that if the functions are continuous and satisfy certain conditions, there will be an interval around the initial value where the solution to the differential equation exists.

To determine these intervals, one generally looks at the coefficients of the y', the function y, and the independent variable in the equation to ensure they are not undefined or discontinuous within the interval around the initial value.