/*! 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 14 In Section \(2.7\), we also deve... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 14

In Section \(2.7\), we also developed a model to describe the levels of antihistamine and decongestant in a patient taking a course of cold pills: $$ \begin{aligned} &\frac{d x}{d t}=I-k_{1} x, \quad x(0)=0, \\ &\frac{d y}{d t}=k_{1} x-k_{2} y, \quad y(0)=0 . \end{aligned} $$ Here \(k_{1}\) and \(k_{2}\) describe rates at which the drugs move between the two sequential compartments (the GI-tract and the bloodstream) and \(I\) denotes the amount of drug released into the GI-tract in each time step. The levels of the drug in the GI-tract and bloodstream are \(x\) and \(y\), respectively. By solving the equations sequentially show that the solution is $$ x(t)=\frac{I}{k_{1}}\left(1-e^{-k_{1} t}\right), \quad y(t)=\frac{I}{k_{2}}\left[1-\frac{1}{k_{2}-k_{1}}\left(k_{2} e^{-k_{1} t}-k_{1} e^{-k_{2} t}\right)\right] . $$

Short Answer

Expert verified
The solutions are \( x(t) = \frac{I}{k_1}(1 - e^{-k_1 t}) \) and \( y(t) = \frac{I}{k_2}\left(1-\frac{1}{k_2-k_1}(k_2 e^{-k_1 t} - k_1 e^{-k_2 t})\right) \).

Step by step solution

01

Solve the First Differential Equation

The equation \( \frac{d x}{d t}=I-k_{1} x \) is a first-order linear differential equation that we can solve using the integrating factor method. The integrating factor \( \mu(t) = e^{k_1 t} \) simplifies the equation to \( \frac{d}{dt}(xe^{k_1 t}) = Ie^{k_1 t} \). Integrating both sides with respect to \( t \) gives \( x(t)e^{k_1 t} = \frac{I}{k_1}e^{k_1 t} + C \). Solving for \( x(t) \) yields \( x(t) = \frac{I}{k_1} + Ce^{-k_1 t} \). Using the initial condition \( x(0) = 0 \), we find that \( C = -\frac{I}{k_1} \), leading to the solution \( x(t) = \frac{I}{k_1} (1 - e^{-k_1 t}) \).
02

Understand the Second Differential Equation Setup

Given that \( \frac{d y}{d t}=k_{1} x-k_{2} y \) with an initial condition \( y(0) = 0 \), and knowing \( x(t) = \frac{I}{k_1} (1 - e^{-k_1 t}) \), substitute \( x(t) \) into the equation to express \( \frac{d y}{d t} = I(1 - e^{-k_1 t}) - k_2 y \). This sets up the equation for \( y(t) \) in terms of a known expression for \( x(t) \).
03

Solve the Second Differential Equation

This equation \( \frac{d y}{d t} + k_2 y = I - Ie^{-k_1 t} \) can be solved using the integrating factor method, with integrating factor \( \mu(t) = e^{k_2 t} \). Multiplying through by \( \mu(t) \), we have \( e^{k_2 t} \frac{d y}{d t} + k_2 e^{k_2 t} y = I e^{k_2 t} - I e^{(k_2 - k_1)t} \). Integrate both sides: \( \int d(ye^{k_2 t}) = I \int e^{k_2 t} dt - I \int e^{(k_2 - k_1)t} dt \). Solving these integrals leads to \( y(t) = \frac{I}{k_2} (1 - \frac{1}{k_2-k_1}(k_2 e^{-k_1 t} - k_1 e^{-k_2 t})) \).

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Key Concepts

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

Mathematical Modelling
Mathematical modelling is a powerful tool that helps to represent real-world phenomena through mathematical formulas and structures. In our exercise, we are dealing with the dynamics of drug levels in the human body using differential equations to model this situation. This approach involves two compartments: the GI-tract and the bloodstream. We aim to understand how the drug moves and changes over time in these areas using mathematical expressions.
The equation \( \frac{d x}{d t}=I-k_{1} x \) models the rate of change of drug levels in the GI-tract, while \( \frac{d y}{d t}=k_{1} x-k_{2} y \) captures the change in the bloodstream. Each term in these equations has a clear physical meaning.
  • The constant \( I \) represents the amount of drug entering the GI-tract.
  • \( k_1 \) and \( k_2 \) are rate constants showing how fast the drug moves or is absorbed between the compartments.
By solving these differential equations, we create a model that helps predict the behavior of drug levels over time. It is an invaluable method, particularly in fields like pharmacokinetics, where understanding drug dynamics is crucial.
Pharmacokinetics
Pharmacokinetics is the study of how drugs move through the body over time. In our exercise, this involves examining how antihistamine and decongestant levels change after administering cold pills. Each drug has specific properties influencing how quickly it's absorbed, distributed, metabolized, and excreted by the body.
The equations here represent this journey through the body. The levels of drugs in the GI-tract and the bloodstream are represented as functions, \( x(t) \) and \( y(t) \), respectively. These functions are influenced by:
  • The administration rate, or the rate at which the drug is ingested (\( I \)).
  • The absorption rate into the bloodstream (\( k_1 \)) and the exit rate from the bloodstream (\( k_2 \)).
Understanding these processes helps in designing effective dosing regimens, ensuring that the drug reaches therapeutic levels without causing harm. Pharmacokinetics allows researchers and medical professionals to predict the concentrations of drugs in different body compartments over time accurately.
Integrating Factor Method
The integrating factor method is a technique used to solve linear first-order differential equations. When facing the equation \( \frac{d x}{d t}=I-k_{1} x \), we use this method to simplify and find the function \( x(t) \).
To solve it, we multiply the entire equation by an "integrating factor" which, for our exercise, is \( e^{k_1 t} \). This factor transforms the left side into the derivative of a product, making it easier to integrate. The equation quickly becomes manageable as:
\[ \frac{d}{dt}(xe^{k_1 t}) = Ie^{k_1 t} \]
After integrating both sides and using the initial condition \( x(0) = 0 \), we solve for \( x(t) \).
Similarly, the integrating factor method applies to \( \frac{d y}{d t}=k_{1} x-k_{2} y \), allowing us to find the function \( y(t) \) by using \( e^{k_2 t} \) as the integrating factor.
  • The goal is to simplify the integration process by converting the original equation into a more workable form.
  • This method is efficient in yielding exact solutions to these types of differential equations.
This approach is common in mathematical modelling and highlights the beauty of mathematics in solving real-world problems, like understanding drug kinetics in the human body.

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Most popular questions from this chapter

The Earth's atmospheric pressure \(p\) is often modelled by assuming that \(d p / d h\) (the rate at which pressure \(p\) changes with altitude \(h\) above sea level) is proportional to \(p\). Suppose that the pressure at sea level is 1,013 millibars and that the pressure at an altitude of \(20 \mathrm{~km}\) is 50 millibars. Answer the following questions and then check your calculations with Maple or MATLAB. (a) Use an exponential decay model $$ \frac{d p}{d h}=-k p $$ to describe the system, and then by solving the equation find an expression for \(p\) in terms of h. Determine \(k\) and the constant of integration from the initial conditions. (b) What is the atmospheric pressure at an altitude of \(50 \mathrm{~km}\) ? (c) At what altitude is the pressure equal to 900 millibars?

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.

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Alcohol is unusual in that it is removed (that is, metabolised through the liver) from the bloodstream by a constant amount each time period, independent of the amount in the bloodstream. This removal can be modelled by a Michaelis- Menten type function \(y^{\prime}=-k_{3} y /(y+M)\), where \(y(t)\) is the 'amount' (BAL) of alcohol in the bloodstream at time \(t, k_{3}\) is a positive constant and \(M\) a small positive constant. (a) If \(y\) is large compared with \(M\), then show that \(y^{\prime} \simeq-k_{3} .\) Solve for \(y\) in this case. (b) Alternatively, as \(y\) decreases and becomes small compared with \(M\), show that then \(y^{\prime} \simeq\) \(-k_{3} y / M .\) Solve for \(y\) in this case. (c) Now sketch the solution function for \(y^{\prime}=-k_{3} y /(y+M)\) assuming that, initially, \(y\) is much greater than \(M .\) Indicate clearly how the graph changes in character when \(y\) is small compared with \(M\), compared with when \(y\) is large compared with \(M .\) Show how the solution behaves as \(t \rightarrow \infty\). (d) When and why would this function be more suitable than simply using \(y^{\prime}=-k_{3}\) to model the removal rate?
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