Problem 8 - Calculus

Question

(a) Find the equilibrium solution to the differential equation $$\frac{d y}{d t}=0.5 y-250$$ (b) Find the general solution to this differential equation. (c) Sketch the graphs of several solutions to this differential equation, using different initial values for \(y.\) (d) Is the equilibrium solution stable or unstable?

Step-by-Step Solution

Answer

The equilibrium solution is \( y=500 \), it's unstable. The general solution is \( y(t) = 500 + Ce^{0.5t} \).

Step 1: Determine the Equilibrium Solution

The equilibrium solution of a differential equation is found by setting the derivative to zero and solving for the variable. For the given differential equation \( \frac{d y}{d t} = 0.5y - 250 \), set \( \frac{d y}{d t} = 0 \). This yields: \[ 0.5y - 250 = 0 \] Solving for \( y \) gives: \[ 0.5y = 250 \] \[ y = \frac{250}{0.5} = 500 \] Therefore, the equilibrium solution is \( y = 500 \).

Step 2: Find the General Solution to the Differential Equation

To find the general solution of the differential equation \( \frac{d y}{d t} = 0.5y - 250 \), first rewrite it in standard form: \( \frac{d y}{d t} - 0.5y = -250 \). This is a first-order linear differential equation. The integrating factor \( \mu(t) \) is given by \( e^{\int -0.5 \, dt} = e^{-0.5t} \). Multiply throughout by the integrating factor: \[ e^{-0.5t} \frac{d y}{d t} - 0.5 e^{-0.5t}y = -250 e^{-0.5t} \] The left side is a derivative: \[ \frac{d}{dt}(e^{-0.5t}y) = -250e^{-0.5t} \] Integrate both sides: \[ e^{-0.5t}y = \int -250e^{-0.5t} \, dt \] Calculate the integral: \[ e^{-0.5t}y = \frac{-250}{-0.5} e^{-0.5t} + C \] \[ e^{-0.5t}y = 500e^{-0.5t} + C \] Solve for \( y \): \[ y = 500 + Ce^{0.5t} \] Hence, the general solution is \( y(t) = 500 + Ce^{0.5t} \).

Step 3: Sketch Graphs of Solutions with Different Initial Conditions

To sketch the graphs of solutions, choose various initial values for \( y \) and plot them according to the general solution \( y(t) = 500 + Ce^{0.5t} \). The constant \( C \) will depend on the initial value at \( t = 0 \). Consider initial conditions, e.g., \( y(0) = 400, 500, \) and \( 600 \). - For \( y(0) = 400 \), \( C = -100 \) which yields \( y(t) = 500 - 100e^{0.5t} \).- For \( y(0) = 500 \), \( C = 0 \) which yields \( y(t) = 500 \).- For \( y(0) = 600 \), \( C = 100 \) which yields \( y(t) = 500 + 100e^{0.5t} \).These functions illustrate the stability of the equilibrium by showing how solutions behave over time.

Step 4: Determine Stability of the Equilibrium Solution

The equilibrium solution \( y = 500 \) is considered stable if solutions approach it as \( t \to \infty \). In \( y(t) = 500 + Ce^{0.5t} \), the term \( e^{0.5t} \) increases with time for nonzero \( C \). This implies that solutions move away from 500 over time, indicating that the equilibrium is unstable.

Key Concepts

Equilibrium Solution

An equilibrium solution in the context of differential equations refers to a constant solution where the derivative of the function is zero. It represents a state of balance where there is no change over time. For the differential equation \( \frac{d y}{d t} = 0.5y - 250 \), we find the equilibrium by setting \( \frac{d y}{d t} = 0 \). Thus, solving \( 0.5y - 250 = 0 \) gives us \( y = 500 \). This means that when \( y(t) = 500 \), the rate of change of \( y \) with respect to time is zero, and \( y \) remains constant, creating an unchanging state or equilibrium.

General Solution

The general solution of a differential equation provides a formula that encapsulates all possible solutions based on initial conditions. For the given first-order linear differential equation \( \frac{d y}{d t} = 0.5y - 250 \), it can be rewritten as \( \frac{d y}{d t} - 0.5y = -250 \). To solve this, we use the integrating factor method. The integrating factor, \( \mu(t) \), is calculated as \( e^{\int -0.5 \, dt} = e^{-0.5t} \). This converts the left side into a derivative of a product:

• Multiply both sides by the integrating factor to obtain \( \frac{d}{dt}(e^{-0.5t}y) = -250e^{-0.5t} \)
• Integrating both sides gives \( e^{-0.5t}y = 500e^{-0.5t} + C \)
• Solving for \( y \) provides the general solution: \( y(t) = 500 + Ce^{0.5t} \)

This expression accounts for any initial condition with \( C \) as a constant determined by the specific initial value of \( y \).

Stability Analysis

Stability analysis involves studying how the solutions of a differential equation behave over time—specifically, whether they converge to an equilibrium solution or diverge from it. In our example, the equilibrium \( y = 500 \) is tested for stability using the general solution \( y(t) = 500 + Ce^{0.5t} \). Here, the term \( Ce^{0.5t} \) dictates whether the solution approaches or moves away from the equilibrium:

• If \( C = 0 \), \( y(t) = 500 \) remains constant, sustaining the equilibrium.
• If \( C eq 0 \), the term \( Ce^{0.5t} \) grows over time, indicating that solutions diverge from \( y = 500 \).

Therefore, this equilibrium is unstable, as perturbations do not settle back to \( y = 500 \). Instead, they move indefinitely away as time progresses.

First-Order Linear Differential Equation

First-order linear differential equations are an essential class of differential equations characterized by derivatives of the first order. They take the form \( \frac{d y}{d t} + P(t)y = Q(t) \). In our example \( \frac{d y}{d t} - 0.5y = -250 \), \( P(t) = -0.5 \) and \( Q(t) = -250 \). Solving these types of equations often involves applying the method of integrating factors.Steps to solve include:

• Identify and calculate the integrating factor \( \mu(t) = e^{\int P(t) \, dt} \).
• Multiply the entire differential equation by this integrating factor.
• Recognize and leverage that the left-hand side becomes the derivative of a product \( \frac{d}{dt} (\mu(t)y) \).
• Integrate both sides to find a general solution that includes the constant of integration \( C \).

These principles are vital when addressing this type of differential equation, making solutions systematic and attainable.