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Sketch solution curves with a variety of initial values for the differential equations. You do not need to find an equation for the solution. \(\frac{d y}{d t}=\alpha-y,\) where \(\alpha\) is a positive constant.

Short Answer

Expert verified
Solution curves approach the horizontal line \(y = \alpha\), rising if below and falling if above.

Step by step solution

01

Understand the Differential Equation

The differential equation given is \(\frac{d y}{d t} = \alpha - y\). This states that the rate of change of \(y\) with respect to \(t\) depends on the difference between the constant \(\alpha\) and the current value \(y\).
02

Determine Equilibrium Point

To find the equilibrium (steady state) where the rate of change is zero, set \(\frac{d y}{d t} = 0\). This gives \(\alpha - y = 0\), so the equilibrium point is at \(y = \alpha\).
03

Analyze Behavior Near Equilibrium

For \(y < \alpha\), \(\frac{d y}{d t} > 0\), indicating \(y\) increases. For \(y > \alpha\), \(\frac{d y}{d t} < 0\), indicating \(y\) decreases. Hence, \(y\) tends towards \(\alpha\) regardless of its initial value.
04

Sketch Solution Curves

Consider different initial values: for \(y(0) < \alpha\), the solutions increase towards \(\alpha\). For \(y(0) > \alpha\), the solutions decrease towards \(\alpha\). On a graph of \(y\) versus \(t\), these solutions will be curves that either rise or fall towards the horizontal line \(y = \alpha\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Points
In differential equations, equilibrium points play a crucial role in understanding system behavior. An equilibrium point is where the rate of change of the system's state is zero. For the differential equation given, \( \frac{d y}{d t} = \alpha - y \), we find the equilibrium point by setting the equation to zero: \( \alpha - y = 0 \). Solving for \( y \), we discover that the equilibrium point is at \( y = \alpha \).

This point represents a steady state where the system does not change over time. Given that the equation describes how \( y \) varies over time, the equilibrium point is where \( y \) will settle in the absence of external influences or perturbations. Identifying equilibrium points helps us predict the long-term behavior of the system.
Solution Curves
Solution curves provide a visual representation of how a system evolves over time for a set of initial conditions. In our example, understanding solution curves means seeing how different starting points for \( y(t) \) lead to different paths over time.

When sketching these curves, we take note of how they behave relative to the equilibrium point \( y = \alpha \). Solutions that start with \( y(0) < \alpha \) will rise towards the equilibrium point. Conversely, those that start with \( y(0) > \alpha \) will descend towards it.

This behavior means that regardless of where \( y \) starts, it eventually stabilizes at \( y = \alpha \). Plotting these solutions would show curves converging onto the line \( y = \alpha \), visually confirming the system's tendency to reach equilibrium.
Rate of Change
The rate of change in a differential equation indicates how a variable changes with respect to another, usually time. In the equation \( \frac{d y}{d t} = \alpha - y \), it tells us how \( y \) changes over time \( t \).

This rate depends on the difference between the constant \( \alpha \) and the current value \( y \). If \( y < \alpha \), the rate \( \frac{d y}{d t} > 0 \) is positive, indicating an increase. Conversely, if \( y > \alpha \), the rate \( \frac{d y}{d t} < 0 \) is negative, indicating a decrease.

This dynamic ensures that \( y \) moves towards the equilibrium point \( \alpha \), displaying a stabilizing behavior indicative of many natural systems, such as temperature adjustments or population modeling.
Initial Values
Initial values are the starting points that dictate the trajectory of the system described by a differential equation. In this case, setting \( y(0) \) determines the shape and behavior of the solution curve over time.

Choosing different initial values allows us to explore the behavior of the system under various conditions. For \( y(0) < \alpha \), the system starts below the equilibrium and the solution curve shows an upward trend as it approaches \( \alpha \). For \( y(0) > \alpha \), the initial state is above the equilibrium, and the solution will demonstrate a downward trend until it stabilizes at \( \alpha \).

Understanding the role of initial values is crucial when predicting how systems react to different conditions or disturbances. It highlights the predictability and the influence of starting conditions on the eventual outcome.

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Most popular questions from this chapter

(a) A cup of coffee is made with boiling water and stands in a room where the temperature is \(20^{\circ} \mathrm{C}\) If \(H(t)\) is the temperature of the coffee at time \(t,\) in minutes, explain what the differential equation $$\frac{d H}{d t}=-k(H-20)$$ says in everyday terms. What is the sign of \(k ?\) (b) Solve this differential equation. If the coffee cools to \(90^{\circ} \mathrm{C}\) in 2 minutes, how long will it take to cool to \(60^{\circ} \mathrm{C}\) degrees?

Many organ pipes in old European churches are made of tin. In cold climates such pipes can be affected with tin pest, when the tin becomes brittle and crumbles into a gray powder. This transformation can appear to take place very suddenly because the presence of the gray powder encourages the reaction to proceed. The rate of the reaction is proportional to the product of the amount of tin left and the quantity of gray powder, \(p,\) present at time \(t .\) Assume that when metallic tin is converted to gray powder, its mass does not change. (a) Write a differential equation for \(p .\) Let the total quantity of metallic tin present originally be \(B\) (b) Sketch a graph of the solution \(p=f(t)\) if there is a small quantity of powder initially. How much metallic tin has crumbled when it is crumbling fastest? (c) Suppose there is no gray powder initially. (For example, suppose the tin is completely new.) What does this model predict will happen? How do you reconcile this with the fact that many organ pipes do get tin pest?

Give an example of: A slope field for a differential equation where the formula for \(d y / d x\) depends on \(y\) but not \(x\)

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of \(0.002 \mathrm{m}^{3} / \mathrm{min}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

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