91Ó°ÊÓ

Solve the following stiff initial-value problems using Euler's method, and compare the results with the actual solution. a. \(\quad y^{\prime}=-9 y, \quad 0 \leq t \leq 1, \quad y(0)=e\), with \(h=0.1\); actual solution \(y(t)=e^{1-9 t}\). b. \(\quad y^{\prime}=-20\left(y-t^{2}\right)+2 t, \quad 0 \leq t \leq 1, \quad y(0)=\frac{1}{3}\), with \(h=0.1 ;\) actual solution \(y(t)=t^{2}+\frac{1}{3} e^{-20 t}\). c. \(\quad y^{\prime}=-20 y+20 \sin t+\cos t, \quad 0 \leq t \leq 2, \quad y(0)=1\), with \(h=0.25\); actual solution \(y(t)=\sin t+e^{-20 t}\) d. \(\quad y^{\prime}=50 / y-50 y, \quad 0 \leq t \leq 1, \quad y(0)=\sqrt{2}\), with \(h=0.1 ;\) actual solution \(y(t)=\left(1+e^{-100 t}\right)^{1 / 2}\).

Use the Extrapolation Algorithm with tolerance \(T O L=10^{-4}, \operatorname{hmax}=0.25\), and hmin \(=0.05\) to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. \(\quad y^{\prime}=t e^{3 t}-2 y, \quad 0 \leq t \leq 1, \quad y(0)=0 ;\) actual solution \(y(t)=\frac{1}{5} t e^{3 t}-\frac{1}{25} e^{3 t}+\frac{1}{25} e^{-2 t}\). b. \(\quad y^{\prime}=1+(t-y)^{2}, \quad 2 \leq t \leq 3, \quad y(2)=1 ;\) actual solution \(y(t)=t+1 /(1-t)\). c. \(\quad y^{\prime}=1+y / t, \quad 1 \leq t \leq 2, \quad y(1)=2 ;\) actual solution \(y(t)=t \ln t+2 t\). d. \(\quad y^{\prime}=\cos 2 t+\sin 3 t, \quad 0 \leq t \leq 1, \quad y(0)=1 ;\) actual solution \(y(t)=\frac{1}{2} \sin 2 t-\frac{1}{3} \cos 3 t+\frac{4}{3}\).

Use all the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use exact starting values, and compare the results to the actual values. a. \(\quad y^{\prime}=t e^{3 t}-2 y, \quad 0 \leq t \leq 1, \quad y(0)=0\), with \(h=0.2\); actual solution \(y(t)=\frac{1}{5} t e^{3 t}-\frac{1}{25} e^{3 t}+\) \(\frac{1}{25} e^{-2 t}\) b. \(\quad y^{\prime}=1+(t-y)^{2}, \quad 2 \leq t \leq 3, \quad y(2)=1\), with \(h=0.2\); actual solution \(y(t)=t+\frac{1}{1-t}\). c. \(\quad y^{\prime}=1+y / t, \quad 1 \leq t \leq 2, \quad y(1)=2\), with \(h=0.2\); actual solution \(y(t)=t \ln t+2 t\). d. \(\quad y^{\prime}=\cos 2 t+\sin 3 t, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.2 ;\) actual solution \(y(t)=\) \(\frac{1}{2} \sin 2 t-\frac{1}{3} \cos 3 t+\frac{4}{3} .\)

Use each of the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use starting values obtained from the Runge-Kutta method of order four. Compare the results to the actual values. a. \(\quad y^{\prime}=y / t-(y / t)^{2}, \quad 1 \leq t \leq 2, \quad y(1)=1\), with \(h=0.1 ;\) actual solution \(y(t)=\frac{t}{1+\ln t}\). b. \(\quad y^{\prime}=1+y / t+(y / t)^{2}, \quad 1 \leq t \leq 3, \quad y(1)=0\), with \(h=0.2\); actual solution \(y(t)=t \tan (\ln t)\). c. \(\quad y^{\prime}=-(y+1)(y+3), \quad 0 \leq t \leq 2, \quad y(0)=-2\), with \(h=0.1 ;\) actual solution \(y(t)=\) \(-3+2 /\left(1+e^{-2 t}\right)\) d. \(\quad y^{\prime}=-5 y+5 t^{2}+2 t, \quad 0 \leq t \leq 1, \quad y(0)=1 / 3\), with \(h=0.1\); actual solution \(y(t)=t^{2}+\frac{1}{3} e^{-5 t}\).

Use Taylor's method of order two to approximate the solution for each of the following initial-value problems. a. \(\quad y^{\prime}=y / t-(y / t)^{2}, \quad 1 \leq t \leq 1.2, \quad y(1)=1\), with \(h=0.1\) b. \(\quad y^{\prime}=\sin t+e^{-t}, \quad 0 \leq t \leq 1, \quad y(0)=0\), with \(h=0.5\) c. \(\quad y^{\prime}=\left(y^{2}+y\right) / t, \quad 1 \leq t \leq 3, \quad y(1)=-2\), with \(h=0.5\) d. \(\quad y^{\prime}=-t y+4 t y^{-1}, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.25\)

Given the multistep method $$ w_{i+1}=-\frac{3}{2} w_{i}+3 w_{i-1}-\frac{1}{2} w_{i-2}+3 h f\left(t_{i}, w_{i}\right), \quad \text { for } i=2, \ldots, N-1 $$ with starting values \(w_{0}, w_{1}, w_{2}\) a. Find the local truncation error. b. Comment on consistency, stability, and convergence.

Use Taylor's method of order two to approximate the solution for each of the following initial-value problems. a. \(\quad y^{\prime}=\frac{2-2 t y}{t^{2}+1}, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.1\) b. \(\quad y^{\prime}=\frac{y^{2}}{1+t}, \quad 1 \leq t \leq 2, \quad y(1)=-(\ln 2)^{-1}\), with \(h=0.1\) c. \(\quad y^{\prime}=\left(y^{2}+y\right) / t, \quad 1 \leq t \leq 3, \quad y(1)=-2\), with \(h=0.2\) d. \(\quad y^{\prime}=-t y+4 t / y, \quad 0 \leq t \leq 1, \quad y(0)=1\), with \(h=0.1\)

The study of mathematical models for predicting the population dynamics of competing species has its origin in independent works published in the early part of the 20th century by A. J. Lotka and V. Volterra (see, for example, [Lo1], [Lo2], and [Vo]). Consider the problem of predicting the population of two species, one of which is a predator, whose population at time \(t\) is \(x_{2}(t)\), feeding on the other, which is the prey, whose population is \(x_{1}(t)\). We will assume that the prey always has an adequate food supply and that its birth rate at any time is proportional to the number of prey alive at that time; that is, birth rate (prey) is \(k_{1} x_{1}(t)\). The death rate of the prey depends on both the number of prey and predators alive at that time. For simplicity, we assume death rate (prey) \(=k_{2} x_{1}(t) x_{2}(t)\). The birth rate of the predator, on the other hand, depends on its food supply, \(x_{1}(t)\), as well as on the number of predators available for reproduction purposes. For this reason, we assume that the birth rate (predator) is \(k_{3} x_{1}(t) x_{2}(t)\). The death rate of the predator will be taken as simply proportional to the number of predators alive at the time; that is, death rate (predator) \(=k_{4} x_{2}(t)\). Since \(x_{1}^{\prime}(t)\) and \(x_{2}^{\prime}(t)\) represent the change in the prey and predator populations, respectively, with respect to time, the problem is expressed by the system of nonlinear differential equations $$ x_{1}^{\prime}(t)=k_{1} x_{1}(t)-k_{2} x_{1}(t) x_{2}(t) \quad \text { and } x_{2}^{\prime}(t)=k_{3} x_{1}(t) x_{2}(t)-k_{4} x_{2}(t) $$ Solve this system for \(0 \leq t \leq 4\), assuming that the initial population of the prey is 1000 and of the predators is 500 and that the constants are \(k_{1}=3, k_{2}=0.002, k_{3}=0.0006\), and \(k_{4}=0.5\). Sketch a graph of the solutions to this problem, plotting both populations with time, and describe the physical phenomena represented. Is there a stable solution to this population model? If so, for what values \(x_{1}\) and \(x_{2}\) is the solution stable?

Given the initial-value problem $$ y^{\prime}=\frac{1}{t^{2}}-\frac{y}{t}-y^{2}, \quad 1 \leq t \leq 2, \quad y(1)=-1 $$ with exact solution \(y(t)=-1 / t\) : a. Use Euler's method with \(h=0.05\) to approximate the solution, and compare it with the actual values of \(y\). b. Use the answers generated in part (a) and linear interpolation to approximate the following values of \(y\), and compare them to the actual values. i. \(y(1.052)\) ii. \(y(1.555)\) iii. \(\quad y(1.978)\) c. Compute the value of \(h\) necessary for \(\left|y\left(t_{i}\right)-w_{i}\right| \leq 0.05\) using Eq. (5.10).

a. Derive the Adams-Bashforth Two-Step method by using the Lagrange form of the interpolating polynomial. b. Derive the Adams-Bashforth Four-Step method by using Newton's backward- difference form of the interpolating polynomial.