13.5: Exercises

13.1. Explaining connections. Explain the connection between Equation (13.2) and the equations Equation \(11.2\) and Equation (12.4).

13.2. Slope fields. Consider the differential equations given below. In each case, draw a slope field, determine the values of \(y\) for which no change takes place - such values are called steady states - and use your slope field to predict what would happen starting from an initial value \(y(0)=1\).

(a) \(\frac{d y}{d t}=-0.5 y\)

(b) \(\frac{d y}{d t}=0.5 y(2-y)\)

(c) \(\frac{d y}{d t}=y(2-y)(3-y)\)

13.3. Drawing slope fields. Draw a slope field for each of the given differential equations:

(a) \(\frac{d y}{d t}=2+3 y\)

(b) \(\frac{d y}{d t}=-y(2-y)\)

(c) \(\frac{d y}{d t}=2-3 y+y^{2}\)

(d) \(\frac{d y}{d t}=-2(3-y)^{2}\)

(e) \(\frac{d y}{d t}=y^{2}-y+1\)

(f) \(\frac{d y}{d t}=y^{3}-y\)

(g) \(\frac{d y}{d t}=\sqrt{y}(y-2)(y-3)^{2}, y \geq 0\)

13.4. Linear or Nonlinear. Identify which of the differential equations in Exercise 13.2 and 13.3 is linear and which nonlinear.

13.5. Using slope fields. For each of the differential equations (a) to \((\mathrm{g})\) in Exercise 13.3, plot \(\frac{d y}{d t}\) as a function of \(y\), draw the motion along the \(y\)-axis, identify the steady state(s) and indicate if the motions are toward or away from the steady state(s).

13.6. Direction field. The direction field shown in the figure below corresponds to which differential equation?

(a) \(\frac{d y}{d t}=r y(y+1)\)

(b) \(\frac{d y}{d t}=r(y-1)(y+1)\)

(c) \(\frac{d y}{d t}=-r(y-1)(y+1)\)

(d) \(\frac{d y}{d t}=r y(y-1)\)

(e) \(\frac{d y}{d t}=-r y(y+1)\)

13.7. Differential equation. Given the differential equation and initial condition

\[\frac{d y}{d t}=y^{2}(y-a), y(0)=2 a \nonumber \]

where \(a>0\) is a constant, the value of the function \(y(t)\) would

(a) approach \(y=0\);

(b) grow larger with time;

(c) approach \(y=a\);

(d) stay the same;

(e) none of the above.

13.8. There’s a hole in the bucket. Water flows into a bucket at constant rate \(I\). There is a hole in the container. Explain the model

\[\frac{d h}{d t}=I-k \sqrt{h} \nonumber \]

Analyze the behavior predicted. What would the height be after a long time? Is this result always valid, or is an additional assumption needed? (hint: recall Example 12.3.)

13.9. Cubical crystal. A crystal grows inside a medium in a cubical shape with side length \(x\) and volume \(V\). The rate of change of the volume is given by

\[\frac{d V}{d t}=k x^{2}\left(V_{0}-V\right) \nonumber \]

where \(k\) and \(V_{0}\) are positive constants.

(a) Rewrite this as a differential equation for \(\frac{d x}{d t}\).

(b) Suppose that the crystal grows from a very small "seed." Show that its growth rate continually decreases.

(c) What happens to the size of the crystal after a very long time?

(d) What is its size (that is, what is either \(x\) or \(V\) ) when it is growing at half its initial rate?

13.10. The Law of Mass Action. The Law of Mass Action in Section \(13.1\) led to the assumption that the rate of a reaction involving two types of molecules (A and B) is proportional to the product of their concentrations, \(k \cdot a \cdot b\).

Explain why the sum of the concentrations, \(k \cdot(a+b)\) would not make for a sensible assumption about the rate of the reaction.

13.11. Biochemical reaction. A biochemical reaction in which a substance \(S\) is both produced and consumed is investigated. The concentration \(c(t)\) of \(S\) changes during the reaction, and is seen to follow the differential equation

\[\frac{d c}{d t}=K_{\max } \frac{c}{k+c}-r c \nonumber \]

where \(K_{\max }, k, r\) are positive constants with certain convenient units. The first term is a concentration-dependent production term and the second term represents consumption of the substance.

(a) What is the maximal rate at which the substance is produced? At what concentration is the production rate \(50 \%\) of this maximal value?

(b) If the production is turned off, the substance decays. How long would it take for the concentration to drop by \(50 \%\) ?

(c) At what concentration does the production rate just balance the consumption rate?

13.12. Logistic growth with proportional harvesting. Consider a fish population of density \(N(t)\) growing at rate \(g(N)\), with harvesting, so that the population satisfies the differential equation

\[\frac{d N}{d t}=g(N)-h(N) . \nonumber \]

Now assume that the growth rate is logistic, so \(g(N)=r N \frac{(K-N)}{K}\) where \(r, K>0\) are constant. Assume that the rate of harvesting is proportional to the population size, so that

\[h(N)=q E N \nonumber \]

where \(E\), the effort of the fishermen, and \(q\), the catchability of this type of fish, are positive constants.

Use qualitative methods discussed in this chapter to analyze the behavior of this equation. Under what conditions does this lead to a sustainable fishery?

13.13. Logistic growth with constant number harvesting. Consider the same fish population as in Exercise 12, but this time assume that the rate of harvesting is fixed, regardless of the population size, so that

\[h(N)=H \nonumber \]

where \(H\) is a constant number of fish being caught and removed per unit time. Analyze this revised model and compare it to the previous results.

13.14. Scaling time in the logistic equation. Consider the scaled logistic equation (13.3). Recall that \(r\) has units of \(1 /\) time, so \(1 / r\) is a quantity with units of time. Now consider scaling the time variable in (13.3) by defining \(t=s / r\). Then \(s\) carries no units ( \(s\) is "dimensionless").

Substitute this expression for \(t\) in (13.3) and find the differential equation so obtained (for \(d y / d s\) ).

13.15. Euler’s method applied to logistic growth. Consider the logistic differential equation

\[\frac{d y}{d t}=r y(1-y) . \nonumber \]

Let \(r=1\). Use Euler’s method to find a solution to this differential equation starting with \(y(0)=0.5\), and step size \(h=0.2\). Find the values of \(y\) up to time \(t=1.0\).

13.16. Spread of infection. In the model for the spread of a disease, we used the fact that the total population is constant \((S(t)+I(t)=N=\) constant \()\) to eliminate \(S(t)\) and analyze a differential equation for \(I(t)\) on its own.

Carry out a similar analysis, but eliminate \(I(t)\). Then analyze the differential equation you get for \(S(t)\) to find its steady states and behavior, practicing the qualitative analysis discussed in this chapter.

13.17. Vaccination strategy. When an individual is vaccinated, he or she is "removed" from the susceptible population, effectively reducing the size of the population that can participate in the disease transmission. For example, if a fraction \(\phi\) of the population is vaccinated, then only the remaining \((1-\phi) N\) individuals can be either susceptible or infected, so \(S(t)+I(t)=(1-\phi) N\). When smallpox was an endemic disease, it had a basic reproductive number of \(R_{0}=7\).

What fraction of the population would have had to be vaccinated to eradicate this disease?

13.18. Social media. Sally Sweetstone has invented a new social media App called HeadSpace, which instantly matches compatible mates according to their changing tastes and styles. Users hear about the App from one another by word of mouth and sign up for an account. The account expires randomly, with a half-life of 1 month. Suppose \(y_{1}(t)\) are the number of individuals who are not subscribers and \(y_{2}(t)\) are the number of are subscribers at time \(t\). The following model has been suggested for the evolving subscriber population

\[\begin{aligned} & \frac{d y_{1}}{d t}=b y_{2}-a y_{1} y_{2}, \\ & \frac{d y_{2}}{d t}=a y_{1} y_{2}-b y_{2} . \end{aligned} \nonumber \]

(a) Explain the terms in the equation. What is the value of the constant \(b\) ?

(b) Show that the total population \(P=y_{1}(t)+y_{2}(t)\) is constant.

Note: this is a conservation statement.

(c) Use the conservation statement to eliminate \(y_{1}\). Then analyze the differential equation you obtain for \(y_{2}\).

(d) Use your model to determine whether this newly launched social media will be successful or whether it will go extinct.

13.19. A bimolecular reaction. Two molecules of \(A\) can react to form a new chemical, \(B\). The reaction is reversible so that \(B\) also continually decays back into 2 molecules of \(A\). The differential equation model proposed for this system is

\[\begin{aligned} & \frac{d a}{d t}=-\mu a^{2}+2 \beta b \\ & \frac{d b}{d t}=\frac{\mu}{2} a^{2}-\beta b, \end{aligned} \nonumber \]

where \(a(t), b(t)>0\) are the concentrations of the two chemicals.

(a) Explain the factor 2 that appears in the differential equations and the conservation statement. Show that the total mass \(M=\) \(a(t)+2 b(t)\) is constant.

(b) Use the techniques in this chapter to investigate what happens in this chemical reaction, to find any steady states, and to explain the behavior of the system