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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

In the previous exercise, all infected individuals remained in the population to spread the disease. A more realistic proposal is to introduce a third variable \(z(t)\) to represent the number of individuals who are removed from the affected population at a given time \(t\) by isolation, recovery and consequent immunity, or death. This quite naturally complicates the problem, but it can be shown (see [Ba2]) that an approximate solution can be given in the form $$ x(t)=x(0) e^{-\left(k_{1} / k_{2}\right) z(t)} \quad \text { and } \quad y(t)=m-x(t)-z(t) $$ where \(k_{1}\) is the infective rate, \(k_{2}\) is the removal rate, and \(z(t)\) is determined from the differential equation $$ z^{\prime}(t)=k_{2}\left(m-z(t)-x(0) e^{-\left(k_{1} / k_{2}\right) z(t)}\right) $$ The authors are not aware of any technique for solving this problem directly, so a numerical procedure must be applied. Find an approximation to \(z(30), y(30)\), and \(x(30)\), assuming that \(m=100,000\), \(x(0)=99,000, k_{1}=2 \times 10^{-6}\), and \(k_{2}=10^{-4}\).

Use the Modified Euler method to approximate the solutions to each of the following initial-value problems, and 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 \(y(t)=t /(1+\ln t)\). 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 \(h=0.2 ;\) actual solution \(y(t)=\) \(-3+2\left(1+e^{-2 t}\right)^{-1}\) d. \(\quad y^{\prime}=-5 y+5 t^{2}+2 t, \quad 0 \leq t \leq 1, \quad y(0)=\frac{1}{3}\), with \(h=0.1\); actual solution \(y(t)=t^{2}+\frac{1}{3} e^{-5 t}\).

Use the Runge-Kutta for Systems Algorithm to approximate the solutions of the following higherorder differential equations, and compare the results to the actual solutions. a. \(\quad y^{\prime \prime}-3 y^{\prime}+2 y=6 e^{-t}, \quad 0 \leq t \leq 1, \quad y(0)=y^{\prime}(0)=2\), with \(h=0.1\); actual solution \(y(t)=2 e^{2 t}-e^{t}+e^{-t}\). b. \(\quad t^{2} y^{\prime \prime}+t y^{\prime}-4 y=-3 t, \quad 1 \leq t \leq 3, \quad y(1)=4, \quad y^{\prime}(1)=3\), with \(h=0.2\); actual solution \(y(t)=2 t^{2}+t+t^{-2}\). c. \(\quad y^{\prime \prime \prime}+y^{\prime \prime}-4 y^{\prime}-4 y=0, \quad 0 \leq t \leq 2, \quad y(0)=3, \quad y^{\prime}(0)=-1, \quad y^{\prime \prime}(0)=9\), with \(h=0.2\); actual solution \(y(t)=e^{-t}+e^{2 t}+e^{-2 t}\). d. \(\quad t^{3} y^{\prime \prime \prime}+t^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=8 t^{3}-2, \quad 1 \leq t \leq 2, \quad y(1)=2, \quad y^{\prime}(1)=8, \quad y^{\prime \prime}(1)=6\), with \(h=0.1 ; \quad\) actual solution \(y(t)=2 t-t^{-1}+t^{2}+t^{3}-1\).

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 \(h=0.1\)

In Exercise 9 we considered the problem of predicting the population in a predator-prey model. Another problem of this type is concerned with two species competing for the same food supply. If the numbers of species alive at time \(t\) are denoted by \(x_{1}(t)\) and \(x_{2}(t)\), it is often assumed that, although the birth rate of each of the species is simply proportional to the number of species alive at that time, the death rate of each species depends on the population of both species. We will assume that the population of a particular pair of species is described by the equations $$ \frac{d x_{1}(t)}{d t}=x_{1}(t)\left[4-0.0003 x_{1}(t)-0.0004 x_{2}(t)\right] \quad \text { and } \quad \frac{d x_{2}(t)}{d t}=x_{2}(t)\left[2-0.0002 x_{1}(t)-0.0001 x_{2}(t)\right]. $$ If it is known that the initial population of each species is 10,000 , find the solution to this system for \(0 \leq t \leq 4\). Is there a stable solution to this population model? If so, for what values of \(x_{1}\) and \(x_{2}\) is the solution stable?

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