Figure 6.1 We are able to think, move, and eat because of collections of neurons in our bodies. This photograph shows the neurons of a ground squirrel.

Preview

“Among all the mathematical disciplines the theory of differential equations is the most important ... It furnishes the explanation of all those elementary manifestations of nature which involve time.”

Sophus Lie, Leipziger Berichte, 47 (1805)

Equations containing one independent and one dependent variable, as well as the derivative of the dependent variable with respect to the independent variable, are known as first-order ordinary differential equations (or first-order ODEs for short). An example of a first-order ODE where the independent variable is x and the dependent variable y is:

The right hand side of this differential equation depends on t and y and, consequently, is known as a non-autonomous differential equation. When the right hand side only depends on the dependent variable, the equation is autonomous. In this Chapter, we develop quantitative and qualitative methods to understand solutions of these equations. A solution is a function y(t) that satisfies the equation in question over a specified interval of time.

Despite their moniker, ODEs are anything but “ordinary.” They have been used to describe extraordinary things such as world population growth, the dynamics of nerve impulses, and the spread of diseases. For instance, in Example 5 of Section 6.6, we use differential equations to explore how populations of neurons can store memories. In this chapter, we study these extraordinary equations three ways. First, after introducing some basic terminology and models, we will derive analytical solutions for special types of ODEs using integration techniques. Second, as ODEs often cannot be solved explicitly, we will introduce techniques that shed light on the qualitative behavior of ODEs. Third, we will use technology to generate and visualize numerical solutions to these ODEs. To this end, we will discuss a numerical method, namely Euler's method, to generate specific solutions to ODEs. By no means will this discussion be exhaustive. This chapter and Chapter 8 provide only a taste of this powerful mathematical formulation, which has been used extensively to understand the dynamics of biological systems.

6.1 A Modeling Introduction to Differential Equations

Differential equations describe how quantities change continuously over time. One of the first applications of differential equations to biology was to population growth: understanding how the sizes of populations–whether they be viruses, plants, or animals–vary in time. We begin our study of differential equations by examining some of these applications.

Exponential population growth and decay

At the beginning of the twentieth century, several notable biologists, including Georgyi F. Gause and T. Carlson, studied population growth of yeast. Both biologists grew yeast under constant environmental conditions in test tubes and flasks as illustrated in Figure 6.2, and they regularly monitored the densities. The resulting data from Carlson's experiment are shown in Figure 6.3.

Figure 6.2 Density of yeast growing in a test tube a and a flask b configured so that the same amount of medium can have two different levels of exposure to the air.

Data Source: G. F. Gause, “The Struggle for Existence.” This is a book published in 1934 by Williams and Wilkins, Baltimore.

Figure 6.3 Carlson yeast data plotted hourly over an eighteen-hour period.

Data Source: “Über Geschwindigkeit und Grösse der Hefevermehrung in Würze,” Biochem. Z.57 (1913): 313–334.

The principle entia non sunt multiplicanda praeter necessitatem (i.e., entities must not be multiplied beyond necessity) is attributed to Franciscan friar William of Ockham (ca. 1287–1347).

One of the goals of this section is to come up with a model that describes Carlson's yeast data. In developing the model, we adhere to the principle of parsimony, an operational principle used in science that requires beginning with the simplest model and only add more complexity as needed. It is also known as Occam's razor or more simply as the KISS principle–Keep It Simple, Stupid! Thus, we begin by formulating and analyzing the simplest possible model and introduce elaborations only as necessary. We start by modeling the initial growth phase of a relatively simple population: the number of cells in a yeast culture.

A solution of the equation, = RN, is a function N such that

We will analyze the solution of this differential equation qualitatively and analytically. A qualitative analysis involves discovering the qualitative behavior of solutions. In other words, it involves determining whether the solutions are increasing, decreasing, remaining constant, or even oscillating, without worrying about the exact form of the solution.

The three qualitative cases in this example correspond to three regimes of population behavior:

Constancy: R = 0 and the per capita birth rate b equals the per capita death rate d. The population neither grows nor declines.

Growth: R > 0 and the per capita birth rate b exceeds the per capita death rate d. The population increases over time.

Decay: R < 0 and the per capita death rate d exceeds the per capita birth rate b. The population decreases over time.

We made all of these qualitative predictions by looking at the sign of the right-hand side of N′ = RN. General qualitative methods for making these predictions are discussed further in Sections 6.4 and 6.5.

In contrast to a qualitative analysis, an analytical approach involves finding explicit solutions to differential equation models. For this constant per capita growth rate model, finding an analytical solution means finding a function N(t) such that its derivative is R times itself; that is, N′(t) = RN(t).

Using the solution N(t) = Ce^Rt, we can determine how well our simple model N′(t) = RN fits the initial phase of Carlson's yeast data. We can either fit the model directly or use a semi-log plot to fit a linear model through the data since, by taking logarithms, our exponential growth model N(t) = N(0)e^rt can be written as

where k = ln C = ln N(0).

Figure 6.6 Semi-log regression plot (solid line) of exponential growth model through the first eight hours of the Carlson yeast data (dots).

Using the exponential model of growth, we can estimate the doubling time for a yeast population during the early stages of colony growth.

As you will see in Problem Set 6.1 and in Section 6.3, the exponential growth model dN/dt = RN is used to model radioactive decay, decay of drugs in the bloodstream, and decay of the number of viral particles in the blood of an individual treated with antiviral drugs.

Logistic growth

Although the exponential model provides a reasonable fit for the initial growth of the yeast population considered in the previous example, it substantially overestimates the population density during the seventh and eighth hours. Moreover, beyond eight hours, the data plotted in Figure 6.7 indicate that the yeast culture asymptotically approaches a density of around 660, while the exponential growth model exhibits unbounded growth. This phenomenon of decreasing per capita growth rate with increasing population density was first elaborated by Thomas Malthus (1766–1834) in his 1798 treatise “An Essay on the Principle of Population Growth.” Malthus recognized that as populations get larger, their per capita growth rate declines due to limited resources and interference among individuals. To deal with these limitations, we modify our model using the principle of parsimony.

Figure 6.7 Per capita growth rate as a function of density for the Carlson yeast data.

Using the Carlson data in Table 6.1, we estimated the per capita growth rate, R, of yeast as a function of population density, N. These estimates are plotted in Figure 6.7 and show that the per capita growth rate is a decreasing function. The exact form of R as a function of N is not uniquely determined. Following the parsimony principle, we begin with the simplest decreasing function of N with positive intercept on the R axis, which is the linear function. Let K denote the horizontal intercept and r the vertical intercept of this linear function. In other words, we choose the per capita growth rate R(N) in the model

to be the linear function

This equation is the logistic equation, which is arguably the single most important equation in population ecology. From the data plotted in Figure 6.7, we might guess that the vertical intercept r lies somewhere between 0.5 and 0.6 and that the horizontal intercept K lies somewhere between 600 and 700. For reasons that become clearer in the next example, the value N = K is called the carrying capacity for the population. The parameter r is called the intrinsic growth rate.

What can we say about the behavior of the solutions to the logistic equation? We partially answer this question with a qualitative analysis. Finding explicit solutions have to wait until the next section.

In the next section we derive the solution to the logistic equation that we can compare directly to the data. Problem Set 6.1 has examples in which the logistic model describes the spread of AIDS and the use of the iPhone in the United States.

External influences on populations

In addition to understanding the feedbacks of populations on their own growth, it is useful to account for external influences on these populations. To do this, we need to extend the models to incorporate elements not included in the initial model. In the 1970s the biomathematician Colin W. Clark at the University of British Columbia, building on the work of others, invented a new field of research at the intersection of mathematical population biology and economics, which he called mathematical bio-economics.* His work extended the logistic equation to account for the economics of harvesting biological populations. The most important applications were in the whaling and fisheries industries. Clark's analysis is based on logistically growing populations that are harvested at a rate h(N) over time:

Two cases are of particular interest are:

• Constant harvesting: In this case, the harvest rate is h(N) = h for all N > 0. If harvesting drives N to 0, as it has in some real populations, then h(N) = 0 for N = 0.
• Proportional harvesting: In this case, the harvest rate is h(N) = vN, where the constant of proportionality v > 0 is called the harvesting effort variable.

Level 1 DRILL PROBLEMS

Write a differential equation to model the situation in Problems 1 to 8. Do not solve.

1. The number of bacteria in a culture grows at a rate that is proportional to the number of bacteria present.

2. A sample of radium decays at a rate that is proportional to the amount of radium present in the sample.

3. In Section 6.3, we will introduce Newton's law of cooling. Newton's law states that the rate at which the temperature of a body changes is proportional to the difference between the body's temperature T and the ambient temperature A.

4. In Section 6.3, we will study the von Bertalanffy growth equation. As part of that study, we will formulate a differential equation which states that the rate at which the mass, M, of a healthy critter grows through absorption of food is directly proportional to its surface area L² and declines through respiration at a rate proportional to its mass L³. If M is proportional to L³—that is, M = kL³ for some positive constant k—then write the model in terms of the unknown function M.

5. According to Benjamin Gompertz (1779–1865), the growth rate of a population is proportional to the number of individuals present, where the factor of proportionality is an exponentially decreasing function of time.

6. When a person is asked to recall a set of N facts, the rate at which the facts are recalled is proportional to the number of relevant facts in the person's memory that have not yet been recalled.

7. The rate at which an epidemic spreads through a community of P susceptible people is proportional to the product of the number of people y who have caught the disease and the number P − y who have not.

8. The rate at which people are implicated in a government scandal is proportional to the product of the number N of people already implicated and the number of people involved who have not yet been implicated.

For Problems 9 to 14 find the time for the population to double or halve its initial level, as appropriate.

15. How long does it take for a population to quadruple if its growth over years is modeled by

16. How long does it take for a population to quintuple if its growth over years is modeled by

17. How long does it take for the level of radioactivity to decay to 10% of its current level, if the level of radioactivity over years is modeled by

18. How long does it take for the level of radioactivity to decay to 1% of its current level, if the level of radioactivity over years is modeled by

A population model for Problems 19 to 22 is given by

where P(t) denotes population density at time t.

19. For what values is the population at equilibrium?

20. For what values is > 0?

21. For what values is < 0?

22. Describe how the fate of the population depends on the initial density.

A population model for Problems 23 to 26 is given by

where P(t) denotes population density at time t.

23. For what values is the population at equilibrium?

24. For what values is > 0?

25. For what values is < 0?

26. Describe how the fate of the population depends on the initial density.

27. Find the equilibria to the model

Level 2 APPLIED AND THEORY PROBLEMS

Radioactive decay: Certain types of atoms (e.g., carbon-14, xenon-133, lead-210) are inherently unstable. They exhibit random transitions to a different atom while emitting radiation in the process. Based on experimental evidence, the number, N, of atoms in a radioactive substance can be described by the equation

where t is measured in years and λ > 0 is known as the decay constant. The decay constant is found experimentally by measuring the half-life, τ, of the radioactive substance (i.e., the time it takes for half of the substance to decay). Use this information in Problems 28 to 32.

28. Find a solution to the decay equation assuming that N(0) = N₀.

29. For xenon-133, the half-life is five days. Find λ. Assume t is measured in days.

30. For carbon-14, the half-life is 5,568 years. Find the decay constant λ. Assume t is measured in years.

31. How old is a piece of human bone which contains just 60% of the amount of carbon-14 expected in a sample of bone from a living person? Assume the half-life of carbon-14 is 5,568 years.

32. The Dead Sea Scrolls were written on parchment at about 100 BC. Given that the half-life of carbon-14 is 5,568 years, what percentage of carbon-14 originally contained in the parchment remained when the scrolls were discovered in 1947?

33. King Arthur's Round Table. In Winchester castle there hangs a wooden round table, 18 feet in diameter and divided into twenty-five sections, one for the king and twenty-four for the knights. Some speculate that the Winchester round table is King Arthur's round table from the fifth century.* We know that the round table has been at Winchester since the fifteenth century. John Harding says in his chronical (1484) that the round table “ended at Winchester, and there it hangs still.” To put an end to the speculation regarding the Winchester round table, in 1976 it was taken down from the wall and tests were employed to determine the date of origin. The rate of decay of carbon-14 in the table (i.e., in dead wood) was found to be 6.08 atoms per minute per gram of sample. Estimate the age of the table to determine whether the Winchester table was King Arthur's round table. Hint: Use the facts that the half-life of carbon-14 in dead wood is 5,568 years and that in living wood the rate of decay of carbon-14 is 6.68 atoms per minute per gram of wood.

34.

The Shroud of Turin is a rectangular linen cloth kept in the Chapel of the Holy Shroud in the cathedral of St. John the Baptist in Turin, Italy. It shows the image of a man whose wounds correspond with the biblical accounts of the crucifixion of Jesus Christ. In 1389, Pierre d'Arcis, the Bishop of Troyes, wrote a memo to the pope accusing a colleague of passing off a certain cloth, cunningly painted, as the burial shroud of Jesus Christ. Despite this early testimony of forgery, the Shroud of Turin has survived as a famous relic. In 1988, a small sample of the Shroud of Turin was taken and scientists from Oxford University, the University of Arizona, and the Swiss Federal Institute of Technology were permitted to test it. Suppose the cloth contained 92.3% of the original amount of carbon. Use this information to determine the age of the shroud.

35. Consider the queen conch logistic growth model presented in Example 8 with a general harvesting function h(N):

1. Describe the qualitative behavior of solutions to this equation (i.e., the long-term abundance of the population) when h(N) = 25,000 for all N.
2. Describe the qualitative behavior of solutions to this equation (i.e., the long-term abundance of the population) when h(N) = 5N for all N.
3. Describe the qualitative behavior of solutions to this equation (i.e., the long-term abundance of the population) when h(N) = 12N for all N.

36. The cane toad (Bufo marinus) was introduced to Australia by the sugar cane industry to control two pests of sugar cane: the grey-backed cane beetle and the frenchie beetle.* Just over a hundred toads were brought to Gordonvale (near Edmonton, North Queensland) in 1935 and released into the cane fields. Unfortunately, due to an asynchrony between the life cycles of the cane toad and the sugar cane pests, the cane toad did not help suppress the cane beetle and the frenchie beetle. However, the cane toads ate almost everything else and grew at a tremendous pace. Now the cane toad is a major pest in Australia. The data in Table 6.2 describe the extent of the area occupied by the cane toads as a function of time.

A simple model to describe these data is given by

where A(t) is the area occupied at time t(years).

1. Find a solution to this model such that A(0) = 32,800 and A(10) = 73,600. Estimate the parameter R.
2. Estimate the area occupied by cane toads in 2004.
3. Modify this model to account for removing cane toads at a rate Hkm²/yr beginning in the year 2004. Determine how large H needs to be to ensure that A starts decreasing.

Table 6.2 Area occupied (in square kilometers) by the cane toad in Australia

Year	Area
1939	32,800
1944	55,800
1949	73,600
1954	138,000
1964	257,000
1969	301,000
1974	584,000

37. Consider the following problem of historical curiosity. The percentage of U.S. households that own a videocassette recorder (VCR) rose steadily from the time of their introduction in the late 1970s to the point at which other technologies displaced them. Let y(t) denote the percentage of U.S. households with a VCR where t is measured in years from 1980 to 1991.

Assume that

can be used to describe the data.

1. Use the first and third data points and the approximation ≈ ry when y is small compared with K to estimate r.
2. Use the fact that the data are saturating to estimate the value of K.

38. In the previous example, compare an estimate obtained for r using growth from 1981 to 1982 versus 1981 to 1984.

39. The Ohio Department of Health released the following data tallying the number of newly diagnosed cases of AIDS in the state from the initial stages of the epidemic to the early 1990s when the first antiretroviral drugs began to become widely available: (Cincinnati Enquirer, December 11, 1994).

Let y(t) denote the number of AIDS cases in Ohio in year t. Assume that

can be used to describe the data.

1. Using the first few data points and the fact that ≈ ry when y is small, estimate r.
2. Estimate the value of K.

40. Hyperthyroidism is caused by growth of tumor-like cells that secrete thyroid hormones in excess of normal amounts. If untreated, an individual with hyperthyroidism may experience extreme weight loss, anorexia, muscle weakness, heart disease, and intolerance to stress and eventually may die. The most successful and least invasive treatment option is generally considered to be radioactive iodine-131 therapy. This involves the injection of a small amount of radioactive iodine into the body. For the type of hyperthyroidism called Graves' disease, it is usual for about 40% to 80% of the administered radioactive iodine to concentrate in the thyroid gland. For functioning adenomas (“hot nodules”), the uptake is closer to 20% to 30%. Excess iodine-131 is excreted rapidly by the kidneys. The quantity of radioiodine used to treat hyperthyroidism is not enough to injure any tissue except the thyroid, which slowly shrinks over a matter of weeks to months. Radioactive iodine is either swallowed in a capsule or sipped in solution through a straw. A typical dose is 5 to 15 millicuries. The half-life of iodine-131 is eight days.

1. Suppose that it takes forty-eight hours for a shipment of iodine-131 to reach a hospital. How much of the initial amount shipped is left once it arrives at the hospital?
2. Suppose a patient is given a dose of 10 millicuries, of which 30% concentrates in the thyroid gland. How much is left one week later?
3. Suppose a patient is given a dose of 10 millicuries, of which 30% concentrates in the thyroid gland. How much is left thirty days later?

6.2 Solutions and Separable Equations

In this chapter, we consider differential equations in which the right-hand side may explicitly depend on both the dependent and independent variables. If t is the independent variable and y is the dependent variable, then the equations we are considering have the general form

where h is an expression involving t and y. Although this moves us into the realm of functions of two variables, we postpone a more formal development of the theory for such functions to Chapter 8. Even though h(t, y) is a function of two variables, the solution y(t) to the differential equation is a function of the single variable t.

In this section, we discuss what it means to be a solution to these differential equations and develop an important method, separation of variables, for solving differential equations of the form . Using this method, we explore super-exponential growth and logistic growth of populations.

Solutions to differential equations

A function y(t) is a solution to a differential equation if, when the function y(t) is substituted into both sides of the differential equation, the differential equation is satisfied.

Separation of variables

As with the case of integration, solving differential equations requires specialized techniques, and there is no guarantee that in general we can find an elementary solution. A special class of differential equations for which we often can find solutions are separable equations–differential equations that can be written in this form:

To solve such an equation on an interval of time for which g(y) ≠ 0, we separate the variables to obtain

and then integrate both sides separately to obtain

The expression we derived for integration by substitution in Section 5.5 implies that the left-hand side can be expressed purely in terms of y (i.e., without reference to t) to obtain the equation

which we can then integrate to solve for y in terms of t, as illustrated in the next example.

Notice the treatment of constants in the solution to Example 4. Because all constants can be combined into a single constant, it is customary not to write C = 2(C₁ − C₂), but rather to simply replace all the arbitrary constants in the problem by a single arbitrary constant after the last integral is found.

Sometimes separation of variables leads to integrals we cannot compute or leads to expressions for which y is only implicitly defined.

Super-exponential and logistic population growth

In Section 1.4 we modeled human population growth in the United States during the nineteenth century with an exponential growth model. Although this model worked reasonably well, the human population on the entire globe appears to be growing faster than exponentially. In the next example, we develop a model of this super-exponential growth.

We know that populations cannot increase to arbitrarily large sizes. The logistic equation, which we introduced in Section 6.1, accounts for limits on population growth. Recall, the logistic equation is

where N is the population abundance, r is the intrinsic rate of growth, and K is the carrying capacity. For many data sets, we were able to estimate the parameters r and K. In Example 4 of Section 6.1, we estimated r ≈ 0.53 and K ≈ 660 for the yeast data set of Carlson. Using the method of separation of variables, we can find an analytic solution to this equation and see how well it describes the yeast data of Carlson.

Note that if N(0) had exceeded 660 in Example 8, the equation to solve for N(t) would be ln N − ln(N − 660) = 0.53t + C, rather than ln N − ln (660 − N) = 0.53t + C as was done above.

Level 1 DRILL PROBLEMS

Verify in Problems 1 to 8 that if y satisfies the prescribed relationship with t

Determine whether the functions given in Problems 9 to 12 are solutions of

Determine whether the functions given in Problems 13 to 16 are solutions of

Solve the differential equations in Problems 17 to 28.

Find the solutions to Problems 29 to 36.

37. Create a differential equation of the form

such that y(t) = e^t is a solution.

38. Create a differential equation of the form

such that y(t) = sin t is a solution.

Level 2 APPLIED AND THEORY PROBLEMS

39. Doomsday prediction. In 1960, three electrical engineers at the University of Illinois published a paper in Science titled “Doomsday.” Based on world population growth data from 1000 AD to 1960 AD, the engineers found that population growth was faster than proportional to the population size. Using the data, they modeled the growth of the population as

where P is the population size in billions and t is centuries after 1000 AD. Solve this differential equation and sketch the solution. What year is doomsday according to their model? Compare to the prediction found in Example 7.*

40. The logistic equation did a remarkable job in describing the number of new cases of AIDS in the United States from 1980 until the early 1990s, as seen in Figure 6.15. (see the website http://www.nlreg.com/aids.htm). Let y(t) denote the number of new cases t years after 1980.

Nonlinear regression techniques used to fit the data resulted in the equation

1. Find the solution to this differential equation.
2. Plot this solution. What happens as t → ∞?
3. Check the Web to see how this compares with the incidence of AIDS in the United States today. What do you conclude and how do you explain the discrepancy? (There is no right or wrong answer to this part.)

Figure 6.15 New cases of AIDS in the United States Permission of Phill Sherrod, http://www.nlreg.com and http://www.dtreg.com

41. A model for tumor growth is the Gompertz function that is a solution to the differential equation

where y is the weight of tumor in milligrams, t is measured in days, a is a constant, and K is the limiting size of the tumor. Assume that a = 0.5 and K = 100.

1. Find a solution to this differential equation that satisfies y(0) = 1 mg.
2. Plot this solution.

42. The 1984 U.S. census recorded a population of 15,757,000 Hispanics; in 1990, the size was 16,098,000. Assuming that the rate of population growth is proportional to the population, predict the population of Hispanics in the United States in the year 2000. Use the Web to find the actual Hispanic population in 2000. How does your prediction compare with the actual number? What do you think can account for any differences?

43. Consider a chemical reaction involving two reactants, A and B, that form a product C. Let [A], [B], and [C] denote the concentrations of A, B, and C. If a molecule of A encounters molecules of B at a rate proportional to their concentration, then the law of mass action states that

where k is a positive constant. If the initial concentration of A is a, the initial concentration of B is b, and we set y = [C], then [A] = a − [C], [B] = b − [C], and

Assume that a = b. Find and plot the solution to this differential equation satisfying y(0) = 0.

44. Populations may exhibit seasonal growth in response to seasonal fluctuations in resource availability. A simple model accounting for seasonal fluctuations in the abundance N of a population is

where R is the average per capita growth rate and t is measured in years.

1. Assume R = 0 and find a solution to this differential equation that satisfies N(0) = N₀. What can you say about N(t) as t → ∞?
2. Assume R = 1 (more generally R > 0) and find a solution to this differential equation that satisfies N(0) = N₀. What can you say about N(t) as t → ∞?
3. Assume R = −1 (more generally R < 0) and find a solution to this differential equation that satisfies N(0) = N₀. What can you say about N(t) as t → ∞?

6.3 Linear Models in Biology

An important class of models is described by the linear differential equation

where the constants a and c are model parameters that have specific physical or biological interpretations. For example, in Section 6.1 we saw how models with a = 0 and c = r were used to describe exponential population growth (c = R) and radioactive decay (c = −λ). In this section, we discuss more applications where the constant coefficient a is nonzero.

Mixing models

Mixing models formulated on the premise that the density of individuals or concentration of molecules, which are generically characterized in terms of a number of objects per unit area or volume, form a homogeneous pool. The flow of objects into the pool is controlled by an external constant rate, while the flow of objects out of the pool is in proportion to the density of objects in the pool. This latter assumption implies that the greater the density of objects in the pool, the faster the total flow of objects out of the pool.

Our first application of mixing models is to obtain a better understanding of within-host pathogen dynamics of HIV. Human immunodeficiency virus-type 1 (HIV-1) has many puzzling quantitative features. For instance, most persons with HIV undergo a ten-year period during which concentration of the virus in plasma is very low. It is only after this quiescent period that a person experiences the onset of AIDS. The reason for this quiescent period is unknown, and it had been presumed that during this period the virus was relatively inactive. Using models, as described in the next example, Perelson and colleagues quantified viral levels in the blood of infected individuals during this quiescent period.*

Hospital patients often receive drugs by intravenous infusion. For drugs to be administered effectively and safely, correct infusion rates must be determined. Differential equation models are a basic tool used by doctors to determine these infusion rates. These models are known as pharmacokinetics or biopharmaceutics models. (Check out http://www.boomer.org/c/p1/index.html for a whole course on this topic.)

In the solution to Example 2b, we found the general solution to the mixing model, which we restate for future reference.

In Example 1, we determined the clearance rate constant when the half-life of the viral particles in the patient's blood was known. In Example 2, we determined the infusion rate necessary to maintain a desired concentration of drug in the patient's blood. Next, in Example 3, we determine the clearance rate constant when the half-life of the viral particles in the patient's blood is not known.

Another application of the mixing model is the following analysis of the concentration of pollution in a lake.

Newton's cooling law and forensic medicine

In forensic medicine, determining the time of death of a victim can be critical in convicting the murderer. Mathematics can aid this process by using Newton's law of cooling. This law states that the rate at which the temperature of a body changes is proportional to the difference between the body's temperature T and the ambient temperature A. Mathematically, this statement corresponds to the following differential equation:

where k a positive constant proportionate to the thermal conductivity of the body. A large k means the body readily conducts heat and quickly adjusts to the ambient temperature. A small k means the body is well insulated and slowly adjusts to the ambient temperature. Notice that this differential equation is also a mixing model where y = T, a = kA, and b = k. Hence, the general solution of this cooling model is

where C is an arbitrary constant.

Organismal growth

The study of growth involves determining body size as a function of age. Various measurements of body size exist, including weight, length, and girth. A famous equation, the von Bertalanffy growth equation, which describes growth of an organism, can be derived from first principles using scaling laws. To derive this equation, consider a cubical critter with length L as illustrated in Figure 6.18.

Figure 6.18 Cubical critter

The surface area of this critter is 6L² and the volume is L³. If we assume that length is measured in centimeters and that the critter is mostly made of water, which has a density of 1 g/cm³, then the critter's mass M is L³ grams. If a critter ingests food at a rate proportional to its surface area and respires at a rate proportional to its mass, then

where a and b are positive proportionality constants. Since M = kL³, where the value of k depends on units of measurement and so can be set to 1 without affecting the form of the solution (the values of a and b would be adjusted according to the selected units), we obtain

Combining the previous two equations yields

Defining k = b/3 and L_∞ = a/b yields the two-parameter von Bertalanffy growth equation

The next example clarifies why biologists use the notation L_∞ for one of the parameters.

Level 1 DRILL PROBLEMS

In Example 1 we modeled HIV levels using the data of Perelson and colleagues. In that example we assumed that the half-life of the viral particles was 0.2 days (or 4.8 hours) and that the mean plasma viral level was 2.16 · 10⁵ viral particles per milliliter (ppmL). Estimate the clearance rate constant b for the half-life given in Problems 1 to 6 and then estimate the daily rate of production of HIV particles for the specified mean plasma viral level.

1. 2.4 h; 1.89 · 10⁵ viral ppmL

2. 3 h; 2.15 · 10⁵ viral ppmL

3. 4 h; 2.25 · 10⁵ viral ppmL

4. 5 h; 2.35 · 10⁵ viral ppmL

5. 6 h; 3.15 · 10⁵ viral ppmL

6. 7.2 h; 2.75 · 10⁵ viral ppmL

Using the information from Example 2, determine the length of time it takes for the concentration of theophylline to be the quantity given in Problems 7 to 12.

7. 5 mg/L

8. 7 mg/L

9. 12 mg/L

10. 14 mg/L

11. 14.5 mg/L

12. 14.99 mg/L

Find the amount of drug in a patient's body as a function of time t, given the infusion rate and the concentration of the drug one hour later, as given in Problems 13 to 18. Assume the patient has 5 L of blood.

13. 10 mg/h; 1.6 mg/L

14. 12 mg/h; 1 mg/L

15. 12 mg/h; 2 mg/L

16. 20 mg/h; 1 mg/h

17. 20 mg/h; 2 mg/h

18. 20 mg/h; 3 mg/h

Rework Example 4 using all of the given information and changing only the lake size and outflow as shown in Problems 19 to 24.

19. Lake size of 50 km³ and outflow of 23 km³/year

20. Lake size of 100 km³ and outflow of 62 km³/year

21. Lake size of 100 km³ and outflow of 23 km³/year

22. Lake size of 120 km³ and outflow of 10 km³/year

23. Lake size of 50 km³ and outflow of 18 km³/year

24. Lake size of 80 km³ and outflow of 10 km³/year

Level 2 APPLIED AND THEORY PROBLEMS

In Problems 25 to 28, set up an appropriate model to answer the given question. These problems use a special case of the linear model in which the relative rate of change remains constant.

25. In 1990 the gross domestic product (GDP) of the United States was $5,464 billion. Suppose the growth rate from 1989 to 1990 was 5.08%. Predict the GDP in 2003. Check your answer by finding the actual 2003 GDP.

26. In 1980 the gross domestic product (GDP) of the United States in constant 1972 dollars was $1,481 billion. Suppose the growth rate from 1980 to 1984 was 2.5% per year. Predict the GDP in 2003. Check your answer by finding the actual 2003 GDP.

27. According to the Department of Health and Human Services, the annual growth rate in the number of divorces per year in 1990 in the United States was 4.7% and there were 1,175,000 divorces that year. How many divorces will there be in 2004 if the annual growth rate in the number of divorces per year remains constant?

28. According to the Department of Health and Human Services, the annual growth rate in the number of marriages per year in 1990 in the United States was 9.8% and there were 2,448,000 marriages that year. How many marriages will there be in 2004 if the annual growth rate in the number of marriages per year is constant?

29. The rate at which a drug is absorbed into the blood system is given by

where x(t) is the concentration of the drug in the bloodstream at time t. What does x(t) approach in the long run (that is, as t → ∞)? At what time is x(t) equal to half this limiting value? Assume x(0) = 0.

30. Calculate the infusion rate in milligrams per hour required to maintain a long-term drug concentration of 50 mg/L (i.e., the rate of change of drug in the body equals zero when the concentration is 50 mg/L). Assume that the half-life of the drug is 3.2 hours and that the patient has 5 L of blood.

31. Calculate the infusion rate in milligrams per hour required to maintain a desired drug concentration of 2 mg/L. Assume the patient has 5.6 L of blood and the half-life of the drug is 2.7 h.

32. Calculate the infusion rate required to achieve a desired drug concentration of 2 mg/L in 1 h. Assume the elimination rate constant of the drug is 5 mg/L per hour and the patient has 6 L of blood.

33. Calculate the infusion rate required to achieve a desired drug concentration of 12 mg/L in 20 min. Assume the clearance rate constant is 2 mg/L per hour and the patient has 5 L of blood.

34. A drug is given at an infusion rate of 50 mg/h. The drug concentration value determined at 3 h after the start of the infusion is 8 mg/L. Assuming the patient has 5 L of blood, estimate the half-life of this drug.

35. A drug is given at an infusion rate of 250 mg/h. The drug concentration determined at 4 h after the start of the infusion is 50 mg/L. Assuming the patient has 5.5 L of blood, estimate the elimination rate constant of this drug.

36. A lake with a constant volume of 10,000 m³ is initially clean and pristine. Water flows into the lake from two streams, Babbling Brook and Raging Rapids, at rates of 250 m³ per day and 750 m³ per day, respectively. At time t = 0, road salt from a nearby road contaminates Babbling Brook with concentration of 2 kg/m³. Find an equation that describes the amount of salt in the lake for all t ≥ 0 and find the limiting amount of salt in the lake.

37. After one hydrodynamic experiment, a tank contains 300 L of a dye solution with a dye concentration of 2 g/L. To prepare for the next experiment, the tank is to be rinsed with water flowing in at a rate of 2 L/min, with the well-stirred solution flowing out at the same rate. Write an equation that describes the amount of dye in the container. Be sure to identify variables and their units.

38. At midnight the coroner was called to the scene of the brutal murder of Casper Cooly. The coroner arrived and noted that the air temperature was 70°F and Cooly's body temperature was 85°F. At 2 A.M., she noted that the body had cooled to 76°F. The police arrested Cooly's business partner Tatum Twit and charged her with the murder. She has an eyewitness who said she left the theater at 11:00 P.M. Does her alibi help?

39. A cup of coffee at a cafe is served at 95°C and left on the counter. The cafe is air-conditioned with an ambient temperature of 20°C. After five minutes, the coffee's temperature is 45°C. Determine how long before the coffee loses its taste quality; that is, it cools down to the temperature of 22°C.

40. In 1986 the Chernobyl nuclear disaster in the Soviet Union contaminated the atmosphere. The buildup of radioactive material in the atmosphere satisfies the differential equation

where M = mass of radioactive material in the atmosphere after time (in years); k is the rate at which the radioactive material is introduced into the atmosphere; r is the annual decay rate of the radioactive material. Find the solution, M(t), of this differential equation in terms of k and r.

41. The von Bertalanffy curve was used to examine growth patterns in both body length and mass of female and male polar bears (Ursus maritimus) captured live near Svalbard, Norway (see Figure 6.20). A longer growth period in males resulted in pronounced sexual dimorphism in both body length and mass. Males were 1.16 times longer and 2.10 times heavier than females. For females, L_∞ = 194 cm, k = 0.75/year, and t₀ = −0.27 is the theoretical age at which the polar bear would have no length (L₀ = 0). For males, L_∞ = 225 cm, k = 0.537/year, and t₀ = −0.395 is the theoretical age at which the polar bear would have no length. Use the von Bertalanffy curve to determine at what age males and females achieve half of their limiting size.

Figure 6.20 The von Bertalanffy curve fitted to age and body length: data for female (blue dots and curve) and male (red dots and curve) polar bears.

Data Source: A. E. Derocher and Wiig, “Postnatal Growth in Body Length and Mass of Polar Bears (Ursus maritimus) at Svalbard,” J. Zool., Lond. 256 (2002): 343–349.

42. In Chapter 5 Section 8, we introduced survival-renewal equations to describe how populations (e.g. money, organisms) change when individuals arrive at rate r(t) and survive with probability s(t) for t units of time. If initially the population size is y(0), then the survival-renewal equation is

If s(t) = e^at, then verify that y(t) solves the linear differential equation:

6.4 Slope Fields and Euler's Method

Not all equations are separable, and many separable equations do not lead to explicit solutions. Furthermore, even when you find a solution, it may be so complex that it is nearly impossible to interpret. To address these issues, we discuss a qualitative method, slope fields, and a numerical method, Euler's method, for studying solutions of differential equations.

Slope fields

Consider a differential equation of the general form

where f(t, y) denotes an expression involving t and y. Since a solution y(t) to this differential equation satisfies y′(t) = f(t, y(t)), it follows that the slope of all solutions at time t are given by the right-hand side f(t, y(t)) of the differential equation. Equivalently, f(t, y(t)) is the slope of the tangent line of y(t) at time t. A qualitative way to investigate the behavior of solutions to y′(t) = f(t, y) is to sketch its slope field, a figure in the ty plane with infinitesimal line segments of slope f(t, y) at (t, y).

Consider, for example, sketching the slope field for . Since for t = 1 the slope is = 1, we draw short line segments at t = 1, each with slope 1, for different y values, as shown in Figure 6.21a. For t = −3, the slope is −1/3 and we draw short line segments at t = −3, each with slope −1/3, also shown in Figure 6.21a. If we continue to plot these slope points for different values of t, we obtain many little slope lines. The resulting graph, shown in Figure 6.21b, is the slope field for the equation y′(t) = 1/t.

We can also examine the relationship between the slope field for y′(t) = 1/t and its solutions y = ln |t| + C. If we choose particular values for C, say C = 0, C = −ln 2, or C = 2, and draw these particular solutions as shown in Figure 6.21c, we see that these solutions are tangent to the slope field.

Figure 6.21 Solution of the differential equation y′ = using a slope field

The slope field for y′(t) = 1/t was relatively straightforward to sketch, but such sketches can be more challenging for differential equations with a more complicated right-hand side. In the next example, we illustrate how to obtain a qualitative understanding using nullclines.

Slope fields can provide qualitative insights when solutions are only implicitly defined or are very complicated. In the next example, the equations are separable and an explicit solution can be found. The solution, however, requires solving for the roots of a cubic equation, yielding complicated expressions that are hard to interpret. The behavior of the solution, however, is quite easy to infer using slope fields. The example is inspired by the work of Warder Clyde Allee, an American ecologist who argued that the per capita growth rate of populations may be negative when population densities are low, despite being positive at higher densities. Reasons for this so-called Allee effect can be due to cooperative hunting or the difficulty of finding mates at low population densities.

Any model of the form y′ = f(y) is called autonomous because the associated slope field is independent of time (i.e., the function f(y) does not explicitly depend on time). We discuss these equations in much greater detail in the next two sections. The counterparts to these autonomous models are purely time dependent equations = f(t)—that is, the slope field does not depend at all on the variable y. In this case, a solution y(t) is an antiderivative of f(t).

Using slope fields, we sometimes can quickly answer questions about the long-term behavior of solutions to a differential equation.

Euler's method

Sometimes it is not possible to solve for the solution of a differential equation analytically; nevertheless, we want more than a qualitative sense of the solution. Such situations require numerical methods. The simplest numerical method is Euler's method, which roughly corresponds to sliding short linear segments along the slope field. More precisely, Euler's method for the differential equation

involves choosing a step size h > 0 and making the approximation

Rearranging this expression gives

Therefore, if we are given y at time t, we can approximate y at time t + h. Applying this approximation iteratively yields an approximation to the solution of the differential equation.

Euler's method is illustrated in the following example.

The one thing that is obvious in comparing the solutions graphed in the left and right panels of Figure 6.28 is that approximate solutions with larger h values can be crude. They may even go horribly wrong, as in the next example. However, as the step size decreases, the time it takes to complete the calculation increases. Thus it is important, as with any numerical scheme, to have error bounds to determine how small h needs to be to get an answer to within a desired level of accuracy. The ultimate trade-off involves accuracy versus time to complete the calculation. Although we do not discuss error bounds for differential equations in this book, you can learn about them in any introductory numerical analysis course.

The moral of the previous example is that with the Euler method, as with other numerical methods for solving differential equations, one must be careful in choosing the appropriate step size h.

Level 1 DRILL PROBLEMS

Sketch at least three particular solutions for each of the slope fields shown in Problems 1 to 6.

Sketch a solution satisfying the specified initial conditions shown over the slope field in Problems 7 to 10.

7. y(0) = 0.3

8. y(0) = 2

9. y(6) = 0

10. y(8) = 2

11. Match the following four equations with the four slope fields.

12. Match the following four equations with the four slope fields.

Sketch the slope fields and sketch a few solutions for the differential equations given in Problems 13 to 18.

Sketch the slope fields and the solution passing through the specified point for the differential equations given in Problems 19 to 24.

Estimate a solution for Problems 25 to 28 using Euler's method. For each of these problems, a slope field is given with an actual solution. Superimpose the segments from Euler's method on the given slope field and assess how well your solution approximates the actual solution as drawn.

Use Euler's method to approximate the solution to = f(t, y) and sketch the approximate solution in Problems 29 to 32 over the specified interval.

29. Over the interval 0 ≤ t ≤ 2 with f(t, y) = (4 − y)(y + 2), y(0) = 0.1, h = 0.5

30. Over the interval 0 ≤ t ≤ 1 with f(t, y) = y − t, y(0) = 2, h = 0.2

31. Over the interval 0 ≤ t ≤ 3.5 with f(t, y) = sin πt − 2y, y(0) = 0, h = 0.5

32. Over the interval −1 ≤ t ≤ 0 with f(t, y) = (4 − y)(y + 2), y(−1) = 0, h = 0.2

33. Consider the differential equation

1. Verify that y(t) = ln t is a solution to this differential equation satisfying y(1) = 0.
2. Use Euler's method to approximate y(2) = ln 2 with h = 0.5.

34. Consider the differential equation

1. Verify that y(t) = e^t is a solution to this differential equation satisfying y(0) = 1.
2. Use Euler's method to approximate e with h = 0.2.

Level 2 APPLIED AND THEORY PROBLEMS

35. A patient receives a continuous drug infusion of 100 mg/h. The half-life of the drug is two hours.

1. Write a differential equation for the amount of drug in the body. (Hint: Review Example 2 in Section 6.3.)
2. Sketch the slope field for this differential equation.
3. Determine the limiting amount of the drug in the patient's body.

36. A patient receives a continuous drug infusion of 50 mg/h. The half-life of the drug is one hour.

1. Write a differential equation for the amount of drug in the body. (Hint: Review Example 2 in Section 6.3.)
2. Sketch the slope field for this differential equation.
3. Determine the limiting amount of the drug in the patient's body.

37. A population subject to seasonal fluctuations can be described by the logistic equation with an oscillating carrying capacity. Consider, for example,

Although it is difficult to solve this differential equation, it is easy to obtain a qualitative understanding.

1. Sketch a slope field over the region 0 ≤ t ≤ 5 and 0 ≤ P ≤ 200.
2. Sketch solutions that satisfy P(0) = 0, P(0) = 10, and P(0) = 200.
3. Use technology to obtain a better rendition of the slope field and solutions.
4. Comment on your solutions and compare to your work using different methods.

38. The velocity v(t) of a skydiver is governed by the equation

where m is the mass of the skydiver, g is gravitational acceleration, and k is a dampening constant (i.e., accounts for air friction).

1. Sketch the slope field for this equation assuming that m = 70 kg, g = 9.8m/s², and k= 110 kg/s.
2. Using the slope field, determine the value of v(t) for the solution v(t) satisfying v(0) = 0. Note that this limiting value is known as the terminal velocity.

39. An autocatalytic chemical reaction involves two molecules, A and B. Let a denote the concentration of A and assume that the concentration b of B remains constant throughout the experiment (e.g., B is added to the mixture in such a way to keep b constant). If A combines with a molecule of B to form two molecules of A and in a backward reaction, two molecules of A form a molecule of A and B, then

where k₁ and k₂ are positive rate constants.

1. Sketch the slope field for this equation for the case k₁ = 1, b = 1, k₂ = 0.5.
2. For the cases a(0) = 0.2 and a(0) = 3, sketch in the solutions and determine the value of a(t).

40. A population, in the absence of harvesting, exhibits the following growth

where N is abundance and t is time in years.

1. Write an equation that corresponds to harvesting the population at a rate of 0.5% per day.
2. Sketch the slope field for the differential equation you found in part a; by sketching solutions, describe how the fate of the population depends on its initial abundance.

6.5 Phase Lines and Classifying Equilibria

Now we focus on autonomous (independent of time) differential equations:

The slope fields for autonomous differential equations are time independent. Since the slope field at any point in time contains all the information about the slope field, an infinite amount of redundancy exists. In this section we remove this redundancy using phase lines, and we discuss classifying equilibria, the y values for which f(y) = 0. With this new qualitative approach to studying autonomous differential equations, we examine evolutionary games and voltage drops across cell membranes.

Phase lines

In the previous section, we sketched slope fields by determining where the slope is zero (nullcline), where it is positive, and where it is negative. In this section, we consider a phase-line diagram that collapses the two-dimensional slope field to the y axis without losing any information about the qualitative behavior of solutions to = f(y) as illustrated Figure 6.30.

Figure 6.30 Illustration of how the slope field for the logistic equation = y(1 − y), can be collapsed on the y axis by removing (or projecting down) the time axis t.

In the next example we consider competition between two clonal populations of the same species. A clonal population consists of genetically identical individuals who only have mothers (reproducing asexually) and are all descendants of the same female. Clonal populations include many kinds of plants, fungi, and bacteria, and some insects, fish, amphibians, and reptiles such as the whip-tail lizard shown in Figure 6.31.

Figure 6.31 Certain species of whiptail lizards are clonally reproducing; that is, all individuals are female and all progeny are identical genetic copies of their mothers, except for changes due to mutations.

In the previous example, we found phase lines from an equation, but sometimes we have a graph of f(y) and not an equation. The next example shows how to find phase lines in such a case.

The equation y = (r_a − r_A)y(1 − y) in Example 1 has a special name in the context of evolutionary game theory. It is called the replicator equation. Evolutionary games were formalized into a coherent mathematical theory in the late 1970s by the theoretical evolutionary biologist John Maynard Smith (1920–2004). (For more information about one of the world's greatest evolutionary biologists, see the in Problem Set 6.5.) Perhaps the best known of his games is the Hawk–Dove game, which describes under what conditions aggressive versus non-aggressive behaviors persist in populations.

In general, for any two inherited contrasting strategies, the growth rates r_a and r_A for genotype a (e.g., doves) and genotype A (e.g., hawks), respectively, in the replicator equation are constructed from a two-by-two table. This table is known as the payoff matrix; it tells how much payoff (benefit if positive, cost if negative) an individual gets after a pairwise interaction with another individual. The payoffs when a meets a are denoted by P_aa. Similarly, P_aA, P_Aa, and P_AA denote the payoffs when a meets A, A meets a, and A meets A, respectively. This information is summarized in Table 6.3.

Table 6.3 Payoff Matrix that specifies the payoff to individuals in the rows interacting with individuals in the columns.

If the payoff affects an individual's reproductive rate, we determine the per capita (i.e., proportional) growth rate of a genotype. We find the expected payoff by calculating the product of the chance of meeting an individual playing a particular strategy and the corresponding payoff.

For genotype a, this expected payoff is

and for genotype A the expected payoff is

Substituting these expressions for r_a and r_A into y′ = y(1 − y)(r_a − r_A), we get the two-strategy replicator equations.

John Maynard Smith and George Price used replicator equations to understand why conflicts between individuals within species rarely escalate to fights to the death. For instance, in their classic 1973 paper “The Logic of Animal Conflict,” Nature 246: 15–18, they wrote with regard to mule deer (Figure 6.35):

“In mule deer (Odocoileus hemionus ) the bucks fight furiously but harmlessly by crashing or pushing antlers against antlers, while they refrain from attacking when an opponent turns away, exposing the unprotected side of its body.”

Figure 6.35 Two male mule deer engaging in a contest for a mate. Such furious conflicts rarely lead to either individual being seriously hurt.

To model animal conflicts, they considered a population of individuals competing for a limiting resource such as mates, food, or shelter. To win this resource, individuals engage in pairwise contests and play one of two strategies, hawk or dove. Individuals playing the hawk strategy constantly escalate the intensity of the contest until either they get the resource or they get injured. Individuals playing the dove strategy leave the contest whenever their opponent escalates the conflict. This game is illustrated in the next two examples.

The equilibrium value y = 1/3 in Example 4 supports both the hawk and dove strategies in the proportion of one individual playing the dove strategy for every two individuals playing the hawk strategy. Such an equilibrium is called a polymorphic strategy, as opposed to a monomorphic strategy where all individuals play either hawk or dove. In Example 4, we can understand the growth rates of hawks and doves at low frequencies as follows. Imagine the population consists mainly of doves and only a few hawks. In this case, individuals are most likely to have a contest with a dove. For a dove this means getting a payoff of V/2 = 2 × (1/2) = 1. For a hawk this means getting a payoff of V = 2. Therefore, hawk numbers will grow at twice the rate of doves. Alternatively, consider a population consisting mostly of hawks and only a few doves. In this case, individuals are most likely to encounter a hawk. For a hawk this means an average payoff of (V − C)/2 = (2 − 3)/2 = −1/2. For a dove this means a payoff of 0. Therefore, hawk numbers will decline until a balance is reached at the polymorphic equilibrium.

Classifying equilibria

When a system starts at an equilibrium, it remains there for all time. In the real world, however, biological systems are constantly subject to environmental perturbations. Thus, if a system starting at equilibrium is slightly perturbed from equilibrium, we need to ask if it tends to return to the equilibrium or not. When the system tends to return to the equilibrium, we call the equilibrium stable. Otherwise, we call it unstable. Precise definitions follow; see Figure 6.37 for a graphical representation.

Figure 6.37 Graphical representation of the classification of equilibria

The voltage V across the membrane of a typical cell is maintained by voltage-gated (i.e., controlled) protein channels embedded in the cell membrane. These channels (see Figure 6.38) regulate the flow of positively charged potassium ions and negatively charged organic molecules out of the cell, and negatively charged chlorine ions and positively charged sodium ions into the cell.

Figure 6.38 An illustration of a voltage-gated channel depicting the electrical field gradients and ions that flow across cell membranes

Linearization

An analytical approach to classifying equilibria involves linearizing the right-hand side of the differential equation

about its equilibria. To this end, suppose y* is an equilibrium of this equation and we evaluate the derivative of f(y) at y* to obtain a = f′(y*). A linear approximation to f(y) for y near y* is given by

Hence,

for values of y near y*. As you are asked to show in Problem 37 of Problem Set 6.5, the solution to

satisfying y(0) = y₀ is

We can use this solution as a first-order approximation for the solution to = f(y) satisfying y(0) = y₀. This approximation y(t) = (y₀ − y*)e^at + y* is reasonable provided y(t) remains near y*. If a < 0, then this approximation suggests that y(t) = y* whenever y₀ is sufficiently close to y*. Alternatively, if a > 0, then this approximation suggests that y(t) = y* whenever y₀ is sufficiently close to y*. These observations can be made mathematically precise and can be used to prove the following theorem.

When one population recovers more rapidly from environmental perturbations than another population (as with P versus N in Example 7), it is said to be more resilient.

Level 1 DRILL PROBLEMS

Draw phase lines, classify the equilibria, and sketch a solution satisfying the specified initial value for the equations in Problems 1 to 10.

Draw a phase line for = f(y) for the graphs shown in Problems 11 to 14. Sketch the requested solutions.

11. y(0) = −1.1, y(0) = 1.1, y(0) = 0.9

12. y(0) = −0.1, y(0) = 0.9, y(0) = 1.1

13. y(0) = −2, y(0) = 1, y(0) = 1.5

14. y(0) = −0.1, y(0) = 1.9, y(0) = 3.9

Linearize about the equilibrium in Problems 15 to 20 and classify stability properties.

Sketch the phase line and classify the equilibria for the Hawk-Dove game with the values V and C given in Problems 21 to 24.

21. V = 2, C = 2

22. V = 4, C = 2

23. V = 3, C = 2

24. V = 2, C = 4

Sketch the phase line and classify the equilibria for the replication equations with the indicated payoffs in Problems 25 to 28.

25. P_aa = 2, P_aA = 1, P_Aa = 1, and P_AA = 2

26. P_aa = 1, P_aA = 2, P_Aa = 3, and P_AA = 4

27. P_aa = −1, P_aA = 2, P_Aa = 1, and P_AA = −1

28. P_aa = 2, P_aA = −1, P_Aa = −1, and P_AA = 3

Level 2 APPLIED AND THEORY PROBLEMS

29. Consider a pack of wolves that can either jointly hunt a stag or individually on their own hunt hares. Suppose during the course of a stag hunt, a hare comes along and an individual wolf considers either remaining with the pack to hunt the stag or going off on its own to hunt the hare. Suppose the payoffs for these two strategies (remain with the pack or go off and hunt the hare) are given by the following payoff matrix in the context of an evolutionary game:

1. Find the replicator equation, assuming that in any one hunt only one wolf in the pack can make the decision to go or not go after the hare.
2. Sketch the phase line and classify the equilibria.
3. Discuss how the outcome of the evolutionary game depends on the initial strategy composition of the population.

30. Consider two scenarios based on Problem 29:

• In a population of stag hunters, a few individuals decide to hunt hares.
• In a population of hare hunters, a few individuals decide to hunt stags.

Use linearization to determine in which of these scenarios the “defecting” individuals are more rapidly excluded.

31. Evolution of cooperation, part 1. Consider a population with two strategies, cooperate and defect. Individuals that cooperate provide a benefit B to their opponents and pay a cost C for providing this benefit. Defectors provide no benefits to their opponents and pay no cost. Under these assumptions, we get the following payoff matrix:

1. Write a replicator equation for this payoff matrix.
2. Assuming B > 0 and C > 0, sketch the phase line for the replicator equation.
3. Discuss the implications of your phase line.

32. Evolution of cooperation, part 2. In Problem 31, cooperation could not evolve. However, cooperation occurs in natural populations. In this problem, we investigate how individuals that interact frequently and respond to the strategy of their opponents can promote the evolution of cooperation. Imagine that each time two opponents meet they interact an average n times. Individuals can play one of two strategies: defect always or tit-for-tat, in which case an individual initially cooperates but switches to defecting if its opponent defected.

1. If each time individuals interact the payoffs are as in Problem 31, then discuss why the payoff matrix should be this:

   

2. Write a replicator equation for this game.
3. Assume B = 3 and C = 2. Sketch phase lines for n = 2, 3, 4.
4. Discuss the implications for the evolution of cooperation.

33. To account for the effect of a generalist predator (with a type II functional response) on a population, ecologists often write differential equations of the form

where N is the population abundance and t is time (in years). The first term of the equation corresponds to logistic growth and the second term corresponds to saturating predation.

1. Sketch the phase line for this system.
2. Discuss how the fate of the population depends on its initial abundance.

34. Construct the phase line for the model

and demonstrate that this equation belongs to the class of membrane voltage models presented in Example 6.

35. Use a phase line diagram to discuss the behavior of the membrane voltage models presented in Example 6 with constants k = 3, V₀ = −65 mV, V_c = 40 mV, and V_d = 40 mV. Does this membrane have the property that it is able to switch between two states when perturbed by a current?

36.

John Maynard Smith, or JMS as he was almost always known, was a professor at the University of Sussex (United Kingdom), and one of the world's great evolutionary biologists. JMS introduced mathematical modeling from game theory into the study of mathematical biology and completely revolutionized the way that biologists think about behavioral evolution. Jonathan Weiner in his book titled A Conversation with John Maynard Smith describes JMS as a classical geneticist and leading theorist in evolutionary biology. Weiner also relates how JMS applied game theory to explain jousting matches among many species, including sticklebacks, sea lions, stag beetles, and stags.

JMS received many prestigious awards. For this , research and write a few words about each of the following awards received by JMS, or, in the case of the last one, established in his honor.

1. Balzan Prize
2. Crafoord Prize
3. Kyoto Prize
4. John Maynard Smith Prize

37. Verify that the solution to

satisfying y(0) = y₀ is given by

38. Show that the linearization theorem is inconclusive when the derivative equals zero at the equilibrium by considering the stability of the equilibria to these equations:

39. Consider a population of clonally reproducing individuals consisting of two genotypes, a and A, with per capita growth rates, r_a and r_A, respectively. If N_a and N_A denote the densities of genotypes a and A, then

Also, let y = be the fraction of individuals in the population that are genotype a. Show that y satisfies

40. In the Hawk-Dove replicator equation

if the value V > 0 is specified, then find the range of values of C (in terms of V) that will ensure a polymorphism exists (i.e., find conditions that ensure the existence of an equilibrium 0 < y* < 1 that is stable).

41. Production of pigments or other protein products of a cell may depend on the activation of a gene. Suppose a gene is autocatalytic and produces a protein whose presence activates greater production of that protein. Let y denote the amount of the protein (say, micrograms) in the cell. A basic model for the rate of this self-activation as a function of y is

where a represents the maximal rate of protein production, k > 0 is a “half saturation” constant, and b ≥ 1 corresponds to the number of protein molecules required to active the gene. On the other hand, proteins in the cell are likely to degrade at a rate proportional to y, say cy. Putting these two components together, we get the following differential equation model of the protein concentration dynamics:

1. Verify that A(y) = a and A(k) = a/2.
2. Verify that y = 0 is an equilibrium for this model and determine under what conditions it is stable.

42. Consider the model of an autocatalytic gene in Problem 41 with b = 1, k > 0, a > 0, and c > 0.

1. Sketch the phase line for this model when ck > a.
2. Sketch the phase line for this model when ck < a.

43. Sketch the phase line for the autocatalytic gene model in Problem 41 with b = 2, k = 1, and c = 1.

1. Sketch the phase for this model when a = 1.
2. Sketch the phase for this model when a = 2.
3. Sketch the phase for this model when a = 2.5.
4. Discuss what you found.

6.6 Bifurcations

Biological systems can exhibit a range of dynamic behaviors that change abruptly or gradually in response to external perturbations. The term bifurcation is used in the context of difference or differential equation models to denote a change in the qualitative behavior of the solutions that occurs as a parameter in the model is varied–that is, as one of the constants in the model changes its value. In this section, we introduce bifurcation theory. We will use the notation

to represent an expression in y and a where y is a model variable and a is a parameter. The goal is to understand how the behavior of the solutions of this differential equation depends on a. More precisely, we will study how the phase line varies with the parameter a.

We will examine bifurcation theory in populations subjected to harvesting, protein dynamics associated with an autocatalytic gene, and the firing rates of neural populations.

Sudden population disappearances

Example 1 illustrates that the phase line of = f(y, a) can vary substantially as we vary the parameter a. Moreover, it shows that at certain parameter values (i.e., a = 25,000 in Example 1) there is a qualitative change in the phase line. These values are important enough to have their own name: bifurcation values. A bifurcation value is defined as the value of a parameter in an equation where either the number of equilibrium solutions changes or the stability of these equilibria undergo a transition from stable to unstable. A simple way to graphically summarize how the behavior of the system depends on a is to create a bifurcation diagram.

An important implication of the bifurcation diagram in Example 2 is that gradual changes in harvesting can bring about discontinuous changes in population abundance. More specifically, when the harvesting rate is ever so slightly increased beyond the bifurcation value (a = 25,000 in Example 1), the population begins slowly at first, but then more rapidly, to decline from high abundance to extinction. Such population disappearances have been observed in natural populations. Dramatic examples include the precipitous drop of blue pike (Stizostedion vitreum glaucum) from annual catches of 10 million pounds to less than a thousand pounds in the mid-1950s, or the unexpected collapse of the Peruvian anchovy (see Figure 6.43) population in 1973, as illustrated in Figure 6.44, and the sudden reduction of Great Britain's grey partridge (Perdix perdix) population in 1952.

Figure 6.43 The Peruvian anchovy (Engraulis ringens) once supported the biggest fishery of all time with a catch of 13.1 million metric tons in 1970.

Figure 6.44 Catch data for Peruvian anchovies in the twentieth century

Data Source: Table 3 in: Castillo, S., & Mendo, J. (1987). Estimation of unregistered Peruvian anchoveta (Engraulis ringens) in official catch statistics, 1951–1982. In D. Pauly, & I. Tsukayama (Eds.), The Peruvian anchoveta and its upwelling ecosystem: Three decades of change. ICLARM studies and reviews (Vol. 15, pp. 109–116)

The bifurcation occurring at a = 25,000 in the queen conch example is a saddle node bifurcation, because the transition from two equilibria to zero equilibria is preceded by the appearance of a semistable (or saddle) equilibrium. A more colorful name for this bifurcation is blue sky catastrophe, as two equilibria vanish into the blue sky as a increases past the value 25,000. Another important type of bifurcation is the pitchfork bifurcation illustrated by the next example. A look ahead at Figure 6.45 indicates the source of this name: one equilibrium bifurcates into three to create a pitchfork.

Biological switches

In order to respond appropriately to changes in their environment, living cells receive environmental signals and respond to these signals in variety of ways, including gene expression and cell activity. Similar to your computer, information processing is carried out by a complex network of switches. Unlike your computer, these switches are not made from electronic parts; rather, they consist of interacting genes and the proteins activated by these genes. Here, we examine two forms of biological switches, an autocatalytic gene introduced in Problem 41 of Section 6.5 and a neural switch for memory.

Behind the motions, sensations, and thoughts of every animal lies a vast network of cells, the nervous system. The network comprises billions of cells called neurons. A typical neuron is illustrated in Figure 6.47, although neurons with various shapes and sizes make up the total neural system of any animal.

Figure 6.47 Sketch of a neuron with inset showing details of a synapse

Neurons specialize in carrying “messages” from one part of the body to another through an electrochemical process that typically causes a voltage spike to travel along the membrane of the neural cell. Messages are received by dendrites, which look like tentacles attached to the cell body. The chemical messages pass through these tentacles into the cell body and then out through one main, long axon. The end of this axon then communicates with dendrites of neurons further down the neural chain, thereby passing messages along from one neuron to the next. Messages between two neurons are usually passed in the form of a chemical flux of so-called neurotransmitters. Excitatory neurotransmitters trigger “go” signals that allow messages to be passed to the next neuron in the communication line. Inhibitory neurotransmitters produce “stop” signals that prevent messages from being forwarded. A single neuron “integrates” incoming signals to determine whether or not to pass the information along to other cells. The activity within a single neuron is typically measured by the rate at which it “fires” voltage spikes.

The simplest model of a population of neurons is the Wilson-Cowan model. It assumes that the entire population of neurons fire at the same rate y (units are number of spikes per millisecond) and are of the same type (i.e., release the same type of neurotransmitters).

Level 1 DRILL PROBLEMS

Draw the phase lines requested in Problems 1 to 6.

Sketch bifurcation diagrams for the equations in Problems 7 to 12.

Consider the model

of an autocatalytic gene from Example 4. In Problems 13 to 16, two of the parameters are specified. Sketch a bifurcation diagram with respect to the third parameter.

13. k = 1, c = 2 with a as the bifurcation parameter

14. k = 2, c = 1 with a as the bifurcation parameter

15. a = 10, k = 1 with c as the bifurcation parameter

16. a = 10, c = 1 with k² as the bifurcation parameter

Consider the Wilson-Cowan model for the values of b and c specified in Problems 17 to 22. Sketch the bifurcation diagram with respect to parameter a.

17. b = 5, c = 1

18. b = 4, c = 1

19. b = 8, c = 1

20. b = 4, c = 2

21. b = 8, c = 2

22. b = 12, c = 2

Level 2 APPLIED AND THEORY PROBLEMS

SIS model in Epidemiology

Mathematical epidemiologists often use these symbols: S to denote the number of individuals in a population who are susceptible to a disease, I to denote the number of people infected with the disease, and R to denote the number of individuals who have recovered and are now immune. If no individuals die then the total number of individuals in the population is N = S + I + R. This kind of model is called a SIR model. In the case that all individuals who recover are immediately susceptible, then R = 0 for all time and the model is called an SIS model. Many sexually transmitted infections (STIs)–for example, gonorrhea–do not confer immunity and are best described by SIS models. This is one reason why a single round of antibiotics, even if administered widely on a population basis, will not have a long-term effect in lowering the incidence of STIs.

Let us assume that a susceptible individual encounters and gets infected by infected individuals at a rate proportional to the densities of the product of susceptible and infected individuals in the population. Call this proportionality constant b ≥ 0. The constant b is known as the transmission rate in the epidemiological literature. Also assume that individuals infected with the disease recover from the disease at a constant rate r ≥ 0. Under these assumptions, we obtain this SIS model:

Since I + S = N, we know S = N − I, so

Rearranging terms yields

In Problems 23 to 26 sketch a bifurcation diagram with respect to b for r = 1 and with respect to r for b = 1. Discuss under what conditions the disease persists in a population confined to living in a group of indicated size.

23. N = 1,000 (boarding school)

24. N = 10,000 (army camp)

25. N = 100,000 (isolated town)

26. N = 1,000,000 (isolated city)

Habitat destruction

Consider a population living in a patchy environment. Let y be the fraction of patches occupied by the species of interest. Let c ≥ 0 denote the colonization rate (i.e., the rate at which individuals from one patch colonize an empty patch), d ≥ 0 the rate at which individuals clear out of a patch, and 0 ≤ D ≤ 1 the fraction of patches destroyed by humans. Then we get the following model (developed by Harvard biologists, Richard Levins and David Culver*):

Sketch the bifurcation diagram for this differential equation for the information given in Problems 27 to 32.

27. Assume that D = 0. Sketch bifurcation diagrams for d when c = 1 and for c when d = 0. Under what conditions does the population persist?

28. Assume that D = 0. Sketch bifurcation diagrams for d when c = 2 and for c when d = 1. Under what conditions does the population persist?

29. Assume that D = 0.5. Sketch bifurcation diagrams for d when c = 1 and for c when d = 0.5. Under what conditions does the population persist?

30. Assume that D = 0.5. Sketch bifurcation diagrams for d when c = 2 and for c when d = 2. Under what conditions does the population persist?

31. Assume that c = 3/2. Sketch bifurcation diagrams for D when d = 1/2 and for d when D = 0. Under what conditions does the population persist?

32. Assume that d = 2. Sketch bifurcation diagrams for D when c = 4 and for c when D = 1/2. Under what conditions does the population persist?

Lotka-Volterra predation

In the 1920s two mathematicians, Vito Volterra (1860–1940) and Alfred Lotka (1880–1949) considered models of the density of a prey species, denoted here by the variable x, predated by a species at a density denoted here by the variable y. They wrote two differential equations, one for the prey and one for the predator, in which the prey equation included the predator density and the predator equation included the prey density. In Chapter 8 we will explore how to analyze a system of two interdependent differential equations. Here, however, with the methods covered in this chapter, we can analyze the behavior of the prey equation or the predator equation where the density of the other species appears as a parameter. The general form of the prey equation is

where g(y) is a per capita growth rate of the prey species and h(x) is the rate at which each unit of predator is able to extract prey. Note it is assumed that both g(0) = 0 and h(0) = 0. The general form of the predator equation is

where h(x) is the prey extraction rate per predator appearing in the prey equation, 0 < b < 1 is the efficiency with which predators can convert a unit of consumed prey into their own biomass (ingestion, digestion, metabolism, etc.), and f(y) is the rate at which predators die when they have no prey species to feed upon.

In Problems 33 to 35 sketch the bifurcation diagram for the specified growth and extraction functions in the prey equation in which the density y of the predators is regarded as a parameter in the prey equation, and in Problems 36 to 37 sketch the bifurcation diagram for the specified extraction and mortality functions in the predator equation in which the density x of the prey species is regarded as a parameter.

33. In one form of the Lotka-Volterra model, g(x) is a decreasing linear function and h(x) is proportional to x. Consistent with these assumptions, let g(x) = 0.5 and h(x) = x. Determine under what conditions the population persists.

34. In another form of the Lotka-Volterra model, g(x) is constant and h(x) is an increasing and saturating function of prey density x. Consistent with these assumptions, let g(x) = 0.5 and h(x) = . Under what conditions is the prey population extinction bound?

35. Assume g(x) = 0.5 and h(x) = . Under what conditions is the prey population driven to extinction by predation?

36. Assume b = 0.2 and f(y) = 1. Under what conditions does the predator population persist when h(x) = x?

37. Assume b = 0.2 and f(y) = y. Under what conditions does the predator population persist when h(x) = x?

38. A self-regulatory genetic network. Smolen and colleagues* investigated a model of a single transcription factor, TF-A, that activates its own transcription TF-A by first forming a homodimer, which then activates transcription by binding to enhancers (TF-REs). A rapid equilibrium is assumed between monomeric and dimeric TF-A. The transcription rate saturates with TF-A dimer concentration to a maximal rate a, which is proportional to TF-A phosphorylation. Responses to stimuli are modeled by varying the degree of TF-A phosphorylation. A basal synthesis rate d is present, as well as a first-order process for degradation, −cy. If y denotes the concentration of TF-A then the model is given by

Assume that b = 1, c = 1, and d = 0.1. Sketch a bifurcation diagram over the region 1 ≤ a ≤ 3 and 0 ≤ y ≤ 3. Discuss when you expect to see two stable equilibria.

39. Evolution of cooperation, part 3. Problem 31 in Section 6.5 investigated how individuals who interact frequently and respond to the strategy of their opponents can promote the evolution of cooperation. If opponents interact on average n times and cooperation gives a benefit B to the opponent and a cost C to the cooperator, then the payoff matrix for the tit-for-tat and defect strategies are as shown here:

1. Write a replicator equation for this game.
2. Assume B = 4 and C = 3 and sketch a bifurcation diagram with respect to the parameter n.
3. Discuss the implications for the evolution of cooperation.

40. Suppose the growth rate of a whale population at density N (individuals per million square kilometers of ocean), harvested at a rate h, is given by

where the units of t are years.

1. Sketch a bifurcation diagram with respect to the parameter h as it varies over the interval [0, 8].
2. If h = v N, then sketch a bifurcation diagram with respect to the parameter v as it varies over the interval [0, 0.12].

1. Determine for what values of a, b, and c, y(t) = e^t + at² + bt + c is a solution to t² + = y.
2. Find the general solution to the differential equation = y tan t.
3. In Ludwig von Bertalanffy's classic paper on modeling the growth of individuals, he modeled the growth of a species of guppy (Lebistes reticulatus) using the differential equation

   

   where L(t) is the length of the guppy in millimeters in week t. The data and the best-fitting version of the model are shown in Figure 6.50. Given that the length at one week is 12.5 mm and at two weeks is 15.6 mm, and that the limiting length is 26.1, find L(t).

   

   Figure 6.50 Growth of the guppy species Lebistes reticulatus with the best-fitting von Bertalanffy growth curve.

4. Let = f(y) where the graph of f(y) is as shown here:

   

   Sketch the phase line for this differential equation and sketch solutions of this differential equation with initial conditions y(0) = −0.1 and y = 0.1.

5. Find the implicit solution to the initial value problem lem y = e^t+2y sin t where y(0) = 0.
6. Cooperation can be observed in natural populations. To better understand under what conditions it may evolve, imagine that each time two opponents meet they interact two times. During these interactions, individuals play one of two strategies: defect always or tit-for-tat, in which case an individual initially cooperates but switches to defecting if the opponent defected. If the benefit of interacting with a cooperating individual is B and the cost of cooperating is seven, then the payoff matrix is given by

   

   1. Write a replicator equation for this game where y is the frequency of individuals playing the tit-for-tat strategy.
   2. Linearize the equation at the y = 1 equilibrium and determine for what values of B this equilibrium is stable.
   3. Discuss the implications for the evolution of cooperation.
7. Sketch the bifurcation diagram for the differential equation = ry+y³ with respect to the parameter r.
8. Estimate a solution to the differential equation = 2t(t² − y) with y(0) = 3 over the interval 0 ≤ t ≤ 2 using Euler's method with step size h = 0.4.
9. The radioactive substance gallium-67 (symbol⁶⁷Ga) used in the diagnosis of malignant tumors has a half-life of 46.6 hours. If we start with 100 mg of⁶⁷ Ga, what percentage is lost between the 30th and 35th hours? Is this the same as the percentage lost over any other five-hour period?
10. Sketch the solution to the initial value problem = f(t, y) with y(0) = 0.25 for the slope field shown here:

    

11. A population of animals on Catalina Island is limited by the amount of food available. Suppose it is shown that there were 1,800 individuals present in 1980 and 2,000 in 1986, and studies also suggest that 5,000 individuals can be supported by the conditions present on the island. Use the logistic model = rN(1 − N/K) to predict the size of the population in the year 2000.
12. In the 1960s, research scientist Anna Laird for the first time successfully used the Gompertz model to fit data of growth of tumors. Recall the Gompertz model is given by

    

    where y is the size of the tumor, t is measured in hours, K is the theoretical limiting size of the tumor (never achieved), and α is a positive constant. For one of the rat data sets used by Laird, the growth of the rat's tumor was modeled using the parameter estimates α = 0.02, y(0) = 0.4 grams, and K = 9,600 grams. Find the size of the tumor after t hours.

13. A lake has a volume of 6 billion ft³, and its initial pollutant content is 0.22%. A river whose waters contain only 0.06% pollutants flows into the lake at the rate of 350 million ft³/day, and another river flows out of the lake also carrying 350 million ft³/day. Assume that the water in the two rivers and the lake is always well mixed. How long does it take for the pollutant content to be reduced to 0.15%?
14. The velocity v of a skydiver falling to the ground is governed by the equation = mg − kv², where m is the mass of the skydiver, g is acceleration due to gravity, and k > 0 is a constant proportional to air resistance. Plot the phase line for this equation and find the terminal velocity v(t).
15. Consider the differential equation = ay − y². Verify that y = 0 is an equilibrium and use linearization to determine for what a values 0 is stable or unstable equilibrium.
16. A refined model of harvesting a logistically growing population is given by

    

    where N is the population abundance in thousands of individuals, r is the population's intrinsic rate of growth in the absence of harvesting, K is the carrying capacity of the population in the absence of harvesting, H is the harvesting effort, and a > 0 is the half saturation constant for harvesting. Assume r = 0.05, K = 100, and a = 1.

    1. Sketch a bifurcation diagram for N ≥ 0 and H ≥ 0.
    2. What is the maximal harvesting effort H that is sustainable?
17. Spruce budworm is a serious insect pest in eastern Canada. In years with large outbreaks, these insects can defoliate and kill a major portion of balsam firs in a forest within four years. Mathematical biologist Donald Ludwig and colleagues developed an elegant model of spruce budworm and forest dynamics. For the budworm dynamics, they assumed that the population exhibited logistic growth and experienced bird predation. Bird predation was modeled using a type III functional response. Hence, the dynamics are given by

    

    Assume K = 15, b = a = 1.

    1. Sketch a bifurcation diagram with respect to r. Hint: Solve for r in terms of N and use technology to graph the resulting curve.
    2. Experimental observation suggests that for a young forest r < 0.2. As the forest matures, r slowly increases to 1. Discuss what increasing r slowly implies about the abundance of spruce budworms.
18. A patient requires a concentration of 2 mg per liter of a drug in his bloodstream. The clearance rate of the drug is 0.2 per hour (i.e., a half-life of ≈ 3.5 hours). The patient has five liters of blood and initially has no drug in his bloodstream.
    1. Find the infusion rate required to get a concentration of 2 mg per liter in two hours.
    2. Find the infusion rate required to maintain a concentration of 2 mg per liter in the long term.
19. Consider

    

    1. Sketch the slope field for this differential equation.
    2. Use Euler's method with Δt = 0.5 to approximate the solution satisfying y(0) = 0 over the interval 0 ≤ t ≤ 1.5.
    3. Sketch the numerical solution over your slope field and briefly discuss whether the approximation is reasonable.
20. Consider the initial value problem

    

    1. Find the solution to the initial value problem.
    2. Find y(t) for the answer you found in part b.