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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 \(y(t)=\) \(\frac{1}{2} \sin 2 t-\frac{1}{3} \cos 3 t+\frac{4}{3} .\)

Short Answer

Expert verified
The solutions obtained using various orders of Adams-Bashforth methods will be closer to the actual solutions as the order increases. However, the error between the approximate and actual solution can be quantified using methods such as absolute error and relative error.

Step by step solution

01

Solve using First Order Adams-Bashforth method

The first-order Adams-Bashforth method is the same as Euler's method which updates the current value by adding the product of step size and derivative at the current value. Calculate the approximated solution for each problem using this method and compare with the actual solution.
02

Solve using Second Order Adams-Bashforth method

The second-order Adams-Bashforth method considers both the current and previous value. Use the formula \(y_{n+1} = y_{n} + h/2 * (3*f(t_{n},y_{n}) - f(t_{n-1},y_{n-1}))\) to calculate the approximated solution for each problem and compare with the actual solution.
03

Solve using Third Order Adams-Bashforth method

The third-order Adams-Bashforth method considers three previous points. Use the formula \(y_{n+1} = y_{n} + h/12 * (23*f(t_{n},y_{n}) - 16*f(t_{n-1},y_{n-1}) + 5*f(t_{n-2},y_{n-2}))\) to calculate the approximated solution for each problem and compare with the actual solution.
04

Solve using Fourth Order Adams-Bashforth method

The fourth-order Adams-Bashforth method considers four previous points. Use the formula \(y_{n+1} = y_{n} + h/24 * (55*f(t_{n},y_{n}) - 59*f(t_{n-1},y_{n-1}) + 37*f(t_{n-2},y_{n-2}) - 9*f(t_{n-3},y_{n-3}))\) to calculate the approximated solution for each problem and compare with the actual solution.

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Key Concepts

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

Numerical Analysis
Numerical analysis is a branch of mathematics that deals with the development and application of algorithms to approximate the solutions to mathematical problems. This area of study is particularly important when exact solutions are not readily available or are difficult to obtain. In particular, numerical analysis comes into play when working with differential equations which model real-world phenomena, like population growth or heat distribution.

Numerical methods allow us to approximate these solutions using computational techniques. Techniques such as the Adams-Bashforth methods, which are used for approximating the solutions to initial-value problems, are classic examples of numerical analysis in action. They are part of a broader group known as multistep methods, requiring multiple previous points for calculating the future value of a function. By using different orders of the Adams-Bashforth method, we effectively employ the historical data points in varying complexities to improve our approximate solution's accuracy.
Initial-Value Problems
Initial-value problems are a specific type of differential equation where the solution is sought given the initial condition of the function. They are often formulated as 'given this starting point, where will the system be at some point in the future?'. These problems are widespread in physics and engineering – for example, predicting the velocity of a falling object over time given its initial velocity.

In the context of our exercise, the initial-value problems are defined by a differential equation alongside an initial condition, such as \( y(0)=0 \) or \( y(2)=1 \). The challenge is to approximate the function \( y(t) \) over a range of values for \( t \), moving forward from the initial value. The Adams-Bashforth methods approach these problems by developing a formulation that uses past computed values to predict future values, thus constructing the solution incrementally.
Approximation of Solutions
The approximation of solutions to differential equations is a necessity when an analytical or exact solution is infeasible. In our exercise, the Adams-Bashforth methods provide a framework for developing successive approximations of the solution using previously calculated values. As we advance from the first-order to the fourth-order Adams-Bashforth method, we rely on an increasing number of previously calculated values. Each step incorporates more information to refine the approximation of the solution.

By comparing the approximated values with the actual solution provided in the exercises, one can evaluate the accuracy of different orders of the Adams-Bashforth methods. This comparison is crucial as it illustrates the trade-off between computational complexity and the precision of the solution. Lower-order methods are simpler but less accurate, while higher-order methods, which consider more points, generally provide a better approximation at the cost of increased complexity.

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Most popular questions from this chapter

Show that the Midpoint method and the Modified Euler method give the same approximations to the initial-value problem $$ y^{\prime}=-y+t+1, \quad 0 \leq t \leq 1, \quad y(0)=1 $$ for any choice of \(h\). Why is this true?

Discuss consistency, stability, and convergence for the Implicit Trapezoidal method $$ w_{i+1}=w_{i}+\frac{h}{2}\left(f\left(t_{i+1}, w_{i+1}\right)+f\left(t_{i}, w_{i}\right)\right), \quad \text { for } i=0,1, \ldots, N-1, $$ with \(w_{0}=\alpha\) applied to the differential equation $$ y^{\prime}=f(t, y), \quad a \leq t \leq b, \quad y(a)=\alpha. $$

In a book entitled Looking at History Through Mathematics, Rashevsky [Ra], pp. 103-110, considers a model for a problem involving the production of nonconformists in society. Suppose that a society has a population of \(x(t)\) individuals at time \(t\), in years, and that all nonconformists who mate with other nonconformists have offspring who are also nonconformists, while a fixed proportion \(r\) of all other offspring are also nonconformist. If the birth and death rates for all individuals are assumed to be the constants \(b\) and \(d\), respectively, and if conformists and nonconformists mate at random, the problem can be expressed by the differential equations $$ \frac{d x(t)}{d t}=(b-d) x(t) \quad \text { and } \quad \frac{d x_{n}(t)}{d t}=(b-d) x_{n}(t)+r b\left(x(t)-x_{n}(t)\right) $$ where \(x_{n}(t)\) denotes the number of nonconformists in the population at time \(t\). a. Suppose the variable \(p(t)=x_{n}(t) / x(t)\) is introduced to represent the proportion of nonconformists in the society at time \(t\). Show that these equations can be combined and simplified to the single differential equation $$ \frac{d p(t)}{d t}=r b(1-p(t)) $$ b. Assuming that \(p(0)=0.01, b=0.02, d=0.015\), and \(r=0.1\), approximate the solution \(p(t)\) from \(t=0\) to \(t=50\) when the step size is \(h=1\) year. c. Solve the differential equation for \(p(t)\) exactly, and compare your result in part (b) when \(t=50\) with the exact value at that time.

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?

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 \(h=0.1\)

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