/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Each of the following differenti... [FREE SOLUTION] | 91Ó°ÊÓ

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Each of the following differential equations has only one equilibrium solution. Find that equilibrium solution and determine if it is stable or unstable? (a) \(\frac{d y}{d t}=y-1\). (b) \(\frac{d C}{d t}=\frac{F}{V} c_{i}-\frac{F}{V} C\), where \(F, V, c_{i}\) are positive constants.

Short Answer

Expert verified
(a) Equilibrium at \(y = 1\) is unstable; (b) Equilibrium at \(C = c_i\) is stable.

Step by step solution

01

Understand the Concept of Equilibrium Solutions

Equilibrium solutions occur when the rate of change is zero. For a differential equation \(\frac{dy}{dt} = f(y)\), the equilibrium solution satisfies \(f(y) = 0\).
02

Solve for Equilibrium Solution (Part a)

For the differential equation \(\frac{dy}{dt} = y - 1\), set \(y - 1 = 0\) to find the equilibrium. Solving this gives \(y = 1\).
03

Determine Stability (Part a)

To determine the stability, check the sign of \(f'(y)\) at the equilibrium point. Here, \(f(y) = y - 1\) means \(f'(y) = 1\). Since \(f'(1) > 0\), the equilibrium \(y = 1\) is unstable.
04

Solve for Equilibrium Solution (Part b)

The differential equation \(\frac{dC}{dt} = \frac{F}{V} c_{i} - \frac{F}{V} C\) can be set to zero: \(\frac{F}{V} c_{i} - \frac{F}{V} C = 0\). Simplifying gives \(C = c_{i}\).
05

Determine Stability (Part b)

The equation can be rewritten as \(\frac{dC}{dt} = -\frac{F}{V} (C - c_{i})\). The derivative with respect to \(C\) is \(-\frac{F}{V}\), which is negative. Therefore, the equilibrium \(C = c_{i}\) is stable.

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Key Concepts

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

Equilibrium Solutions in Differential Equations
Equilibrium solutions are a fundamental concept in differential equations. They occur when the system is in a state where no changes happen over time. In mathematical terms, for a given differential equation \( \frac{dy}{dt} = f(y) \), the equilibrium solution is found by setting \( f(y) = 0 \). This translates to finding the values of \( y \) where the system is "at rest".
For example, consider the equation \( \frac{dy}{dt} = y - 1 \). The equilibrium solution is determined by solving \( y - 1 = 0 \), leading to \( y = 1 \). This means that when \( y \) is 1, there is no change in the system, marking it as an equilibrium point. Similarly, for the equation \( \frac{dC}{dt} = \frac{F}{V} c_{i} - \frac{F}{V} C \), setting it to zero gives \( C = c_{i} \), indicating that at this point, the concentration remains constant over time.
Equilibrium solutions provide valuable insights into the behavior of systems modeled by differential equations, revealing conditions under which the system remains constant.
Stability Analysis of Equilibrium Solutions
Stability analysis is a crucial step to understand how a system behaves when disturbed from its equilibrium state. Once we determine the equilibrium solution, the next step is to check this solution's stability by analyzing the derivative of the function.
To do this, we find the derivative \( f'(y) \) of the original function \( f(y) \). The nature of this derivative at the equilibrium point can reveal whether the solution is stable or unstable:
  • If \( f'(y) < 0 \), the equilibrium is stable, meaning the system returns to equilibrium after a small disturbance.
  • If \( f'(y) > 0 \), the equilibrium is unstable, meaning the system moves away from the equilibrium after a disturbance.
In Part (a) of our example, the function is \( f(y) = y - 1 \), thus \( f'(y) = 1 \). Since \( f'(1) > 0 \), the equilibrium \( y = 1 \) is unstable, indicating that any small deviation will cause the system to diverge from this point.
In Part (b), the rewritten equation shows \( f'(C) = -\frac{F}{V} \). Since this is negative, the equilibrium solution \( C = c_{i} \) is stable, meaning if the system is disturbed, it will naturally return to its equilibrium state.
The Role of Mathematical Modelling in Differential Equations
Mathematical modelling is the process of using mathematical structures and concepts to represent real-world systems. Differential equations often play a key role in these models due to their ability to describe how systems change over time.
For example, the differential equation \( \frac{dC}{dt} = \frac{F}{V} c_{i} - \frac{F}{V} C \) models how the concentration \( C \) changes with time based on the inflow and outflow rates in a vessel. Here, \( F \), \( V \), and \( c_{i} \) are constants reflecting the system's properties. This model helps predict how the concentration will adjust based on different parameters and conditions.
Similarly, the equation \( \frac{dy}{dt} = y - 1 \) can model simple biological processes or physical systems where growth or decay moves towards a particular value (in this case, 1).
  • Through mathematical modelling, complex processes become analyzable and predictable.
  • It aids in decision-making and system optimization by providing insights into behavior under various scenarios.
These models are crucial for scientists and engineers as they simplify understanding complex systems and allow for effective predictions and solutions.

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Most popular questions from this chapter

(Adapted from Borelli and Coleman (1996).) In a biochemical laboratory radioactive phosphorus \(\left({ }^{32} \mathrm{P}\right)\) was used as a tracer. (A tracer, through its radioactive emission, allows the course followed by a substance through a system to be tracked, which otherwise would not be visible.) \({ }^{32} \mathrm{P}\) decays exponentially with a half-life of \(14.5\) days and its quantity is measured in curies (Ci). (Although it is not necessary for the calculations, one curie is the quantity of a radioactive isotope undergoing \(3.7 \times 10^{-5}\) disintegrations per second.) After the experiment the biochemists needed to dispose of the contents, but they had to store them until the radioactivity had decreased to the acceptably safe level of \(1 \times 10^{-5} \mathrm{Ci} .\) The experiment required 8 Ci of \(^{32}\) P. Using a simple model of exponential decay, establish how long they had to store the contents of the experiment before it could be disposed of safely.

If an archaeologist uncovers a seashell which contains \(60 \%\) of the \({ }^{14} \mathrm{C}\) of a living shell, how old do you estimate that shell, and thus that site, to be? (You may assume the half-life of \({ }^{14} C\) to be 5,568 years.

Continuous compounding for invested money can be described by a simple exponential model, \(M^{\prime}(t)=0.01 r M(t)\), where \(M(t)\) is the amount of money at time \(t\) and \(r\) is the percent interest compounding. Business managers commonly apply the Rule of 72, which says that the number of years it takes for a sum of money invested at \(r \%\) interest to double, can be approximated by \(72 / r .\) Show that this rule always overestimates the time required for the investment to double.

(Adapted from Fulford et al. (1997).) A public bar opens at \(6 \mathrm{p.m}\). and is rapidly filled with clients of whom the majority are smokers. The bar is equipped with ventilators that exchange the smoke-air mixture with fresh air. Cigarette smoke contains \(4 \%\) carbon monoxide and a prolonged exposure to a concentration of more than \(0.012 \%\) can be fatal. The bar has a floor area of \(20 \mathrm{~m}\) by \(15 \mathrm{~m}\), and a height of \(4 \mathrm{~m} .\) It is estimated that smoke enters the room at a constant rate of \(0.006 \mathrm{~m}^{3} / \mathrm{min}\), and that the ventilators remove the mixture of smoke and air at 10 times the rate at which smoke is produced. The problem is to establish a good time to leave the bar, that is, sometime before the concentration of carbon monoxide reaches the lethal limit. (a) Starting from a word equation or a compartmental diagram, formulate the differential equation for the changing concentration of carbon monoxide in the bar over time. (b) By solving the equation above, establish at what time the lethal limit will be reached.

In Section \(2.7\), we developed the model $$ \begin{aligned} &\frac{d x}{d t}=-k_{1} x, \quad x(0)=x_{0} \\ &\frac{d y}{d t}=k_{1} x-k_{2} y, \quad y(0)=0 \end{aligned}$$ where \(k_{1}, k_{2}>0\) determine the rate at which a drug, antihistamine or decongestant moves between two compartments in the body, the GI-tract and the bloodstream, when a patient takes a single pill. Here \(x(t)\) is the level of the drug in the GI-tract and \(y(t)\) is the level in the bloodstream at time \(t\). (a) Find solution expressions for \(x(t)\) and \(y(t)\) that satisfy this pair of differential equations, when \(k_{1} \neq k_{2}\). Show that this solution is equivalent to that provided in the text. (b) The solution above is invalid at \(k_{1}=k_{2} .\) Why is this, and what is the solution in this case? (c) For old and sick people, the clearance coefficient (that is, the rate at which the drug is removed from the bloodstream) is often much lower than that for young, healthy individuals. How does an increase or decrease in \(k_{2}\) change the results of the model? Using Maple or MATLAB to generate the time- dependent plots, check your results.

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