Problem 59 - Calculus

Question

Approach a steady state \(y_{x}\) as \(t \rightarrow \infty\). Graph \(y(t)\) when \(y^{\prime}=3 y-12\) and \(y_{0}\) is (a) below 4 (b) equal to 4 (c) above 4

Step-by-Step Solution

Answer

The steady state \( y = 4 \) attracts all solutions, regardless of initial condition.

Step 1: Rewrite the Differential Equation

The differential equation given is \( y' = 3y - 12 \). Rewrite this equation in the form \( y' = 3(y - 4) \) to better understand its behavior as \( t \rightarrow \infty \). This reveals that the equilibrium solution (steady state) is \( y = 4 \).

Step 2: Consider the Solution for Different Initial Conditions

We are tasked with examining the behavior of solutions based on different initial conditions: \( y_0 < 4 \), \( y_0 = 4 \), and \( y_0 > 4 \). Use these initial conditions to analyze how they approach the steady state \( y = 4 \).

Step 3: Solve the Standard Linear Differential Equation

Solve the differential equation \( y' = 3(y - 4) \). The general solution to the equation is found using the integrating factor method or by recognizing it as a standard linear first-order differential equation. You can use the form \( y(t) = Ce^{3t} + 4 \) where \( C \) is determined by the initial conditions.

Step 4: Analyze Initial Condition (a) \( y_0 < 4 \)

For initial conditions below 4, e.g., \( y_0 = 3 \): \( C = y_0 - 4 = 3 - 4 = -1 \). Thus, the solution is \( y(t) = -e^{3t} + 4 \). As \( t \rightarrow \infty \), \( -e^{3t} \) exponentially decays to zero and \( y(t) \) approaches 4 from below.

Step 5: Analyze Initial Condition (b) \( y_0 = 4 \)

For initial condition \( y_0 = 4 \): \( C = y_0 - 4 = 0 \). The solution is \( y(t) = 4 \), meaning it remains constant at 4 for all \( t \), indicating that \( y \) is already at the steady state.

Step 6: Analyze Initial Condition (c) \( y_0 > 4 \)

For initial conditions above 4, e.g., \( y_0 = 5 \): \( C = y_0 - 4 = 5 - 4 = 1 \). The solution is \( y(t) = e^{3t} + 4 \). As \( t \rightarrow \infty \), \( e^{3t} \) exponentially grows towards infinity, and \( y(t) \) approaches 4 from above.

Step 7: Sketch the Graph of \( y(t) \)

Sketch the graphs for all three initial conditions. For \( y_0 < 4 \), the curve approaches 4 from below. For \( y_0 = 4 \), the graph is a horizontal line at 4. For \( y_0 > 4 \), the curve approaches 4 from above. All solutions tend toward \( y = 4 \), demonstrating that it is a global attractor.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that involves an unknown function and its derivatives. It expresses a relationship between a function and its rate of change. Differential equations are fundamental in describing various phenomena, such as motion, growth, and decay, as they provide a framework for modeling how quantities change over time or space.

In our problem, we have the differential equation \( y' = 3y - 12 \). Here, \( y' \) represents the derivative of \( y \) with respect to time, and the equation describes how \( y \) changes as a function of \( y \) itself. This specific type of equation is used to model growth processes where the rate of change is proportional to the current amount, modified by a constant offset.

Steady State Analysis

Steady-state analysis refers to evaluating the behavior of a system as it approaches equilibrium, where the system's state does not change anymore despite the passage of time. In practical terms, this is often the scenario where a system stops reacting to stimuli as it stabilizes.

In this exercise, a steady-state occurs when the derivative \( y' \) becomes zero, meaning that \( y \) no longer changes with time. From the rewritten equation \( y' = 3(y - 4) \), the steady-state solution is found by setting \( y' = 0 \), giving us \( y = 4 \). Thus, regardless of the initial condition, as time progresses towards infinity, the value of \( y \) tends to stabilize around 4. This indicates that \( y = 4 \) is an equilibrium point or the steady-state solution of the differential equation.

First-order Linear Differential Equations

First-order linear differential equations are a class of differential equations characterized by the linearity in the unknown function and its first derivative. Such equations take the general form \( y' + p(t)y = g(t) \).

The given problem \( y' = 3(y - 4) \) is a simplified form of a first-order linear differential equation after re-arranging terms. To solve it, the integrating factor method is usually employed, but given the simplicity, it can also be recognized as a separable equation. The solution form \( y(t) = Ce^{3t} + 4 \), where \( C \) is a constant based on initial conditions, exemplifies how the general and particular solutions to a linear differential equation are structured.

Graphical Solution of Differential Equations

Visualizing differential equations graphically helps to understand the behavior of solutions over time. A graph illustrates how a solution approaches steady state, moves towards stability, and responds to initial conditions.

In this exercise, the graphical solution is drawn for three initial conditions \( y_0 < 4 \), \( y_0 = 4 \), and \( y_0 > 4 \). Each corresponds to a distinct behavior:

• For \( y_0 < 4 \), the curve approaches 4 from below, converging upwards as it stabilizes.
• For \( y_0 = 4 \), the value remains constant, represented by a horizontal line at 4.
• For \( y_0 > 4 \), the curve approaches 4 from above, descending gradually to reach equilibrium.

This graphical approach underscores the power of visual tools in comprehending how different starting points can affect the trajectory of solutions in differential equations.