Question

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

Answer 1

To solve the initial value problems, use separation of variables to find the general form of the solution and then apply the initial condition to find the specific solution. The solutions for the given differential equations resemble each other in terms of their exponential nature, but differ in terms of the initial conditions and the behavior of the slope.

Step-by-step explanation:

For the differential equations given:

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

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

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

1. To solve these initial value problems, we can use separation of variables. We'll start with the general form y = Ce^(-kt) + c, where C and c are constants and k represents the coefficient of y.
2. For problem (a), integrating both sides of the equation will give us y = -e^(-t) + 5t + y0. The graph of the solution will start with a non-zero y-intercept and have a downward slope that levels off at zero.
3. For problem (b), integrating both sides of the equation will give us y = 2e^(-2t) + 5/2 + y0. The graph of the solution will start at zero with an upward slope that decreases in magnitude until the curve levels off.
4. For problem (c), integrating both sides of the equation will give us y = 5e^(-2t) + 5 + y0. The graph of the solution will start at zero with an upward slope that increases in magnitude until it becomes a positive constant.

The solutions resemble each other in terms of their exponential nature, but differ in terms of the initial conditions and the behavior of the slope.

Answer 2

Step-by-step explanation:

Answer:

a) y-8 = (y₀-8) , b) 2y -5 = (2y₀-5)

Step-by-step 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)