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Chapter 2: Problem 2

For the following first-order differential equations, find the general solution, solving for the dependent variable. See Appendix \(A .3\) if you need to revise how to solve first-order differential equations. (a) \(\frac{d y}{d t}=-3 y\), (b) \(\frac{d C}{d t}=3 C-1\), (c) \(\frac{d y}{d t}=3 y t^{-1}\).

Short Answer

Expert verified
(a) \(y(t) = Ce^{-3t}\); (b) \(C(t) = -\frac{1}{3} + C_2e^{3t}\); (c) \(y(t) = Ct^3\).

Step by step solution

01

Recognize the Equation Type - Part (a)

The equation \(\frac{dy}{dt} = -3y\) is a separable first-order differential equation. It can be rearranged to separate the variables.
02

Separate Variables - Part (a)

Rewrite the equation by dividing both sides by \(y\) and multiplying both sides by \(dt\): \(\frac{1}{y} dy = -3 dt\).
03

Integrate Both Sides - Part (a)

Integrate both sides: \(\int \frac{1}{y} dy = \int -3 dt\), leading to \(\ln |y| = -3t + C_1\).
04

Solve for y - Part (a)

Exponentiate both sides to solve for \(y\): \(y = e^{-3t+C_1} = Ce^{-3t}\), where \(C = e^{C_1}\) is a constant. This is the general solution: \(y(t) = Ce^{-3t}\).
05

Recognize the Equation Type - Part (b)

The equation \(\frac{dC}{dt} = 3C - 1\) is a linear first-order differential equation. It can be rewritten in standard form.
06

Rewrite in Standard Form - Part (b)

Rearrange the equation: \(\frac{dC}{dt} - 3C = -1\). The left hand is the standard linear form \(\frac{dC}{dt} + pC = g(t)\).
07

Find the Integrating Factor - Part (b)

The integrating factor \(\mu(t)\) is \(e^{\int -3 dt} = e^{-3t}\). Multiply the entire differential equation by this integrating factor.
08

Solve the Differential Equation - Part (b)

Multiply through by the integrating factor: \(e^{-3t} \frac{dC}{dt} - 3e^{-3t}C = -e^{-3t}\). Recognize the left hand as the derivative of \((e^{-3t}C)\), set it up as \((e^{-3t}C)' = -e^{-3t}\), and integrate both sides to solve.
09

Integrate and Solve for C - Part (b)

Integrate: \(e^{-3t}C = \int -e^{-3t} dt + C_2\). Solve the integral: \(e^{-3t}C = \frac{1}{3}e^{-3t} + C_2\), and solve for \(C\): \(C(t) = -\frac{1}{3} + C_2e^{3t}\).
10

Recognize the Equation Type - Part (c)

The equation \(\frac{dy}{dt} = 3 \frac{y}{t}\) is a separable first-order differential equation.
11

Separate Variables - Part (c)

Rearrange the equation as \(\frac{1}{y} dy = 3 \frac{1}{t} dt\).
12

Integrate Both Sides - Part (c)

Integrate both sides: \(\int \frac{1}{y} dy = \int 3 \frac{1}{t} dt\), resulting in \(\ln |y| = 3 \ln |t| + C_3\).
13

Solve for y - Part (c)

Exponentiate both sides: \(y = e^{3 \ln |t| + C_3} = Ct^3\) where \(C = e^{C_3}\). This is the general solution: \(y(t) = Ct^3\).

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a class of first-order differential equations that can be easily manipulated to separate the variables. This means that all terms involving the dependent variable can be moved to one side of the equation, while all terms involving the independent variable can be put on the other side. This is particularly useful, because it allows us to independently integrate both sides with respect to their respective variables.
For instance, consider the differential equation \( rac{dy}{dt} = -3y\). We can move the terms involving \(y\) to the left and terms involving \(t\) to the right, ending up with \( rac{1}{y} dy = -3 dt\). By doing this separation, we position ourselves to integrate both sides effectively.
Problems (a) and (c) from our exercise are excellent examples of separable equations. It鈥檚 worth noting that separable differential equations are some of the most straightforward types to solve, thanks to this neat separation and independent integration.
Linear Differential Equations
Linear differential equations form another crucial category in first-order differential equations. These equations are characterized by their linearity in the dependent variable and its derivatives. That means no powers, products, or combinations of the dependent variable and its derivative are present.
Let's look at the problem (b) \( rac{dC}{dt} = 3C - 1\), which can be rewritten in linear form, \( rac{dC}{dt} - 3C = -1\). In this form, we observe it's a linear first-order equation resembling the typical format: \(\frac{dC}{dt} + pC = g(t)\).
A key step in solving these equations is finding an integrating factor, \(\mu(t)\), which simplifies the equation into an easily integrable form. The technique of using an integrating factor is essential, as it helps convert the left-hand side of the equation into the derivative of a product, ready to be integrated seamlessly.
General Solution
The general solution of a first-order differential equation provides a family of solutions that include an arbitrary constant, representing infinitely many specific solutions. For separable or linear equations, finding the general solution involves performing integration and then solving for the dependent variable.
In our exercises, the general solution to each problem was derived after integrating both sides of the equation and applying exponentiation as needed.
For example, in part (a), after integrating the separated components, the general solution obtained is \(y(t) = Ce^{-3t}\), where \(C\) is the arbitrary constant. This represents a whole family of solutions where different initial values determine specific solutions.
The arbitrary constant is crucial because it accounts for any initial conditions you might have, allowing the solution to fit various scenarios depending on the given initial value.
Integration Techniques
Integration is at the heart of solving first-order differential equations, whether they are linear or separable. Knowing how to handle the integrals is paramount to finding solutions.
To solve a separable differential equation like \( rac{dy}{dt} = 3 \frac{y}{t}\), one separates and integrates: \(\int \frac{1}{y} dy = \int 3 \frac{1}{t} dt\). This results in natural logarithms: \( \ln |y| = 3 \ln |t| + C_3 \).
Conversely, in linear equations such as in part (b), the integration involves using an integrating factor. This approach modifies the equation into the form of a derivative of a product, \((e^{-3t}C)'\), which we then integrate with respect to \(t\).
Understanding these integration strategies allows one to not only tackle these differential equations but also apply similar principles to a broad array of mathematical and physical problems. Mastery of integration techniques is an invaluable tool in the mathematician's toolkit.

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Most popular questions from this chapter

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.

(From Borelli and Coleman (1996).) Olduvai Gorge, in Kenya, cuts through volcanic flows, tuff (volcanic ash), and sedimentary deposits. It is the site of bones and artefacts of early hominids, considered by some to be precursors of man. In 1959, Mary and Louis Leakey uncovered a fossil hominid skull and primitive stone tools of obviously great age, older by far than any hominid remains found up to that time. Carbon-14 dating methods being inappropriate for a specimen of that age and nature, dating had to be based on the ages of the underlying and overlying volcanic strata. The method used was that of potassium-argon decay. The potassium-argon clock is an accumulation clock, in contrast to the \({ }^{14} \mathrm{C}\) dating method. The potassium-argon method depends on measuring the accumulation of 'daughter' argon atoms, which are decay products of radioactive potassium atoms. Specifically, potassium- \(40\left({ }^{40} \mathrm{~K}\right)\) decays to argon \(\left({ }^{40} \mathrm{Ar}\right)\) and to Calcium- 40 \(\left({ }^{40} \mathrm{Ca}\right)\) by the branching cascade illustrated below in Figure 2.17. Potassium decays to calcium by emitting a \(\beta\) particle (i.e. an electron). Some of the potassium atoms, however, decay to argon by capturing an extra-nuclear electron and emitting a \(\gamma\) particle. The rate equations for this decay process may be written in terms of \(K(t), A(t)\) and \(C(t)\), the potassium, argon and calcium in the sample of rock: $$ \begin{aligned} &K^{\prime}=-\left(k_{1}+k_{2}\right) K \\ &A^{\prime}=k_{1} K \\ &C^{\prime}=k_{2} K \end{aligned} $$ where $$ k_{1}=5.76 \times 10^{-11} \text { year }^{-1}, \quad k_{2}=4.85 \times 10^{-10} \text { year }^{-1} \text { . } $$ (a) Solve the system to find \(K(t), A(t)\) and \(C(t)\) in terms of \(k_{1}, k_{2}\), and \(k_{3}=k_{1}+k_{2}\), using the initial conditions \(K(0)=k_{0}, A(0)=C(0)=0 .\) (b) Show that \(K(t)+A(t)+C(t)=k_{0}\) for all \(t \geq 0 .\) Why would this be the case? (c) Show that \(K(t) \rightarrow 0, A(t) \rightarrow k_{1} k_{0} / k_{3}\) and \(C(t) \rightarrow k_{2} k_{0} / k_{3}\) as \(t \rightarrow \infty\). (d) The age of the volcanic strata is the current value of the time variable \(t\) because the potassiumargon clock started when the volcanic material was laid down. This age is estimated by measuring the ratio of argon to potassium in a sample. Show that this ratio is $$ \frac{A}{K}=\frac{k_{1}}{k_{3}}\left(e^{k_{3} t}-1\right) $$ (e) Now show that the age of the sample in years is $$ \frac{1}{k_{3}} \ell \mathrm{n}\left[\left(\frac{k_{3} A}{k_{1} K}\right)+1\right] $$ (f) When the actual measurements were made at the University of California at Berkeley, the age of the volcanic material (and thus the age of the bones) was estimated to be \(1.75\) million years. What was the value of the measured ratio \(A / K ?\)

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?

North American lake system. Consider the American system of two lakes: Lake Erie feeding into Lake Ontario. What is of interest is how the pollution concentrations change in the lakes over time. You may assume the volume in each lake to remain constant and that Lake Erie is the only source of pollution for Lake Ontario. (a) Write a differential equation describing the concentration of pollution in each of the two lakes, using the variables \(V\) for volume, \(F\) for flow, \(c(t)\) for concentration at time \(t\) and subscripts 1 for Lake Erie and 2 for Lake Ontario. (b) Suppose that only unpolluted water flows into Lake Erie. How does this change the model proposed? (c) Solve the system of equations to get expressions for the pollution concentrations \(c_{1}(t)\) and \(c_{2}(t)\) (d) Set \(T_{1}=V_{1} / F_{1}\) and \(T_{2}=V_{2} / F_{2}\), and then \(T_{1}=k T_{2}\) for some constant \(k\) as \(V\) and \(F\) are constants in the model. Substitute this into the equation describing pollution levels in Lake Ontario to eliminate \(T_{1}\). Then show that, with the initial conditions \(c_{1,0}\) and \(c_{2,0}\), the solution to the differential equation for Lake Ontario is $$ c_{2}(t)=\frac{k}{k-1} c_{1,0}\left(e^{-t /\left(k T_{2}\right)}-e^{-t / T_{2}}\right)+c_{2,0} e^{t / T_{2}} $$ (One way of finding the solution would be to use an integrating factor. See Appendix A.4.) (e) Compare the effects of \(c_{1}(0)\) and \(c_{2}(0)\) on the solution \(c_{2}(t)\) over time.

Solving differential equations. For the differential equation $$ \frac{d y}{d t}=2 y $$ (a) Show that \(y(t)=c e^{2 t}\), where \(c\) is any real valued constant, satisfies the differential equation by substituting into both sides of the equation. Is there any value of \(c\) that isn't a solution? (b) Find the one solution that corresponds to the initial condition \(y(0)=5 .\) See Appendix A.1 if you need any revision of the meaning of differential equations.
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