Chapter 8: Problem 9
Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.
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Chapter 8: Problem 9
Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.
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Explain how the growth rate function determines the solution of a population model.
Use a calculator or computer program to carry out the following steps. a. Approximate the value of \(y(T)\) using Euler's method with the given time step on the interval \([0, T]\). b. Using the exact solution (also given), find the error in the approximation to \(y(T)\) (only at the right endpoint of the time interval). c. Repeating parts (a) and (b) using half the time step used in those calculations, again find an approximation to \(y(T)\). d. Compare the errors in the approximations to \(y(T)\). $$\begin{array}{l}y^{\prime}(t)=t-y, y(0)=4 ; \Delta t=0.2, T=4; \\\y(t)=5 e^{-t}+t-1\end{array}$$
The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. The delivery of a drug (such as an antibiotic) through an intravenous line may be modeled by the differential equation \(m^{\prime}(t)+k m(t)=I,\) where \(m(t)\) is the mass of the drug in the blood at time \(t \geq 0, k\) is a constant that describes the rate at which the drug is absorbed, and \(I\) is the infusion rate. Let \(I=10 \mathrm{mg} / \mathrm{hr}\) and \(k=0.05 \mathrm{hr}^{-1}\). a. Draw the direction field, for \(0 \leq t \leq 100,0 \leq y \leq 600\). b. What is the equilibrium solution? c. For what initial values \(m(0)=A\) are solutions increasing? Decreasing?
Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t).$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{2 t}{t^{2}+1} y(t)=1+3 t^{2}, \quad y(1)=4$$
When an infected person is introduced into a closed and otherwise healthy community, the number of people who contract the disease (in the absence of any intervention) may be modeled by the logistic equation $$ \frac{d P}{d t}=k P\left(1-\frac{P}{A}\right), P(0)=P_{0} $$ where \(k\) is a positive infection rate, \(A\) is the number of people in the community, and \(P_{0}\) is the number of infected people at \(t=0\) The model also assumes no recovery. a. Find the solution of the initial value problem, for \(t \geq 0\), in terms of \(k, A,\) and \(P_{0}\) b. Graph the solution in the case that \(k=0.025, A=300,\) and \(P_{0}=1\) c. For a fixed value of \(k\) and \(A\), describe the long-term behavior of the solutions, for any \(P_{0}\) with \(0 < P_{0} < A\)
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