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olga55
12 days ago
7

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
Mathematics
1 answer:
Inessa [9K]12 days ago
4 0

Response:

Step-by-step clarification:

Reply:

a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)

Clarification:

To address these equations, using direct integration is the simplest approach.

a) The equation provided is

          dy / dt = -y + 8

         dy / (y-8) = dt

We substitute variables

          y-8 = u

         dy = du

Substituting and integrating gives us

           ∫ du / u = ∫ dt

           Ln (y-8) = t

Evaluating at the lower limits t = 0 for y = y₀

          ln (y-8) - ln (y₀-8) = t-0

Simplifying the equation results in

           ln (y-8 / y₀-8) = t

           y-8 / y₀-8 =

            y-8 = (y₀-8)

b) The equation here is

            dy / dt = 2y -5

            u = 2y -5

            du = 2 dy

            du / 2u = dt

Integrating now

             ½ Ln (2y-5) = t

Evaluating at limits gives

            ½ [ln (2y-5) - ln (2y₀-5)] = t

            Ln (2y-5 / 2y₀-5) = 2t

            2y -5 = (2y₀-5)

c) The equation here bears a strong resemblance to the previous one

             u = 2y -10

             du = 2 dy

             ∫ du / 2u = dt

             ln (2y-10) = 2t

We evaluate this to get

             ln (2y-10) –ln (2y₀-10) = 2t

               2y-10 = (2y₀-10)

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