Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. (Enter your answer using interval notation.) (t - 5) y' + (ln t) y = 6t, y (1) = 6

0 < t < 5 is the required interval for the differential equation (t - 5) y' + (ln t) y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5) y' + (ln t) y = 6t

and the condition y (1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [ (ln t) / (t - 5) ]y = 6t / (t - 5)

(ln t) / (t - 5) is continuous on the interval (0, 5) and (5, + infinity).

6t / (t - 5) is continuous on (-infinity, 5) and (5, + infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, + infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.