) 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 t

Question

) 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?

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Iris 3 months 2021-10-20T01:16:06+00:00 1 Answer 0 views 0

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    2021-10-20T01:17:23+00:00

    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.

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