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23 April, 20:00

Solve each of the following initial value problems and plot the solutions for several values of y0. Then describe in a few words how the solutions resemble, and differ from, each other.

(a) dy/dt = - y + 5, y (0) = y0

(b) dy/dt = - 2y + 5, y (0) = y0

(c) dy/dt = - 2y + 10, y (0) = y0

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  1. 23 April, 20:54
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    Step-by-step explanation:

    Answer:

    a) y-8 = (y₀-8), b) 2y - 5 = (2y₀-5)

    Explanation:

    To solve these equations the method of direct integration is the easiest.

    a) the given equation is

    dy / dt = and - 8

    dy / y-8 = dt

    We change variables

    y-8 = u

    dy = du

    We replace and integrate

    ∫ du / u = ∫ dt

    Ln (y-8) = t

    We evaluate at the lower limits t = 0 for y = y₀

    ln (y-8) - ln (y₀-8) = t-0

    Let's simplify the equation

    ln (y-8 / y₀-8) = t

    y-8 / y₀-8 =

    y-8 = (y₀-8)

    b) the equation is

    dy / dt = 2y - 5

    u = 2y - 5

    du = 2 dy

    du / 2u = dt

    We integrate

    ½ Ln (2y-5) = t

    We evaluate at the limits

    ½ [ln (2y-5) - ln (2y₀-5) ] = t

    Ln (2y-5 / 2y₀-5) = 2t

    2y - 5 = (2y₀-5)

    c) the equation is very similar to the previous one

    u = 2y - 10

    du = 2 dy

    ∫ du / 2u = dt

    ln (2y-10) = 2t

    We evaluate

    ln (2y-10) - ln (2y₀-10) = 2t

    2y-10 = (2y₀-10)
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