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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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Answer:

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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