Solve each of the following initial value problems and plot the solutions for several values of y0. (a computer algebra system is recommended.) (a) dy/dt = y−8, y(0) = y0(b) dy/dt = 2y−5, y(0) = y0(c) dy/dt = 2y−10, y(0) = y0

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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