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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

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  1. 4 July, 11:34
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    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.
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