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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).

Short Answer

Expert verified
Euler's method and linear interpolation were used to approximate solutions of a differential equation and specific values respectively. The step size \(h\) was also calculated to maintain the precision within the required limit.

Step by step solution

01

Prepare the variables for Euler's method

The differential equation is given by \(y^{\prime} = \frac{1}{t^{2}} - \frac{y}{t} - y^{2}\) with the initial value at \(y(1) = -1\) and \(h = 0.05\). Prepare two columns for \(t\) and \(y\). Initialize them with \(t = 1\) and \(y = -1\).
02

Execute Euler's method

Use Euler's method formula \(y_{i+1} = y_i + h*f(t_i, y_i)\) repeatedly where \(f(t_i, y_i) = \frac{1}{t^{2}} - \frac{y}{t} - y^{2}\) until reaching \(t = 2\). After running the process, save the results of \(t\) and \(y\). Compare these values with actual values given by the exact solution \(y(t)=-1 / t\).
03

Perform linear interpolation

Use linear interpolation to approximate the requested values \(y(1.052)\), \(y(1.555)\) and \(y(1.978)\). For this, locate the two points in the list from Step 2 where each value would fall between, and perform interpolation calculation. The standard formula is: \(y(x) = y1 + (x - x1) * ((y2 - y1) / (x2 - x1))\). Compare these interpolated values with the actual ones calculated by the exact solution.
04

Compute h for the required condition

In order to find h that satisfies \(\left|y\left(t_{i}\right)-w_{i}\right| \leq 0.05\), we use the Eq. (5.10) which is given in the breakdown of the problem. Substituting the values from the Euler approximation and the exact solution values calculate h

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Key Concepts

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

Euler's Method
Euler's method is a straightforward technique to approximate solutions to initial-value problems for ordinary differential equations (ODEs). It is especially useful when an exact solution is tough to find. By using small step sizes, Euler's method helps improve accuracy. The basic premise is to step from one known point to the next by using the derivative to estimate the slope.Euler's method works well for initial-value problems like this: knowing the initial condition, we compute subsequent values by following a simple formula. Here, given the differential equation \[ y^{\prime} = \frac{1}{t^{2}} - \frac{y}{t} - y^{2} \]and a starting point at \(t = 1\) where \(y = -1\), we use Euler's formula to progress through the interval with a step size, \(h\), of 0.05:\[ y_{i+1} = y_i + h \cdot f(t_i, y_i) \]Where \(f(t_i, y_i)\) is our function defined as \(\frac{1}{t^{2}} - \frac{y}{t} - y^{2}\). By repeating this step-by-step update, we generate an approximate solution graph as values of \(t\) reach up to 2. It’s crucial to compare these estimated values against the exact solution for validation.
Initial-Value Problem
Initial-value problems set a stage for solving differential equations by specifying a starting point. These problems typically define the value of the function at a particular point, known as the initial condition. For instance, in our exercise, we have:- The differential equation: \( y^{\prime} = \frac{1}{t^{2}} - \frac{y}{t} - y^{2} \)- The initial condition: \( y(1) = -1 \)This setup implies that at \(t = 1\), the value of \(y\) is \(-1\). The goal is to determine how \(y\) changes as \(t\) progresses through the given interval \(1 \leq t \leq 2\).Initial-value problems are foundational in exploring the dynamic behavior described by differential equations. Approaches like Euler’s method become tools to numerically explore this behavior when exact analytical solutions are complicated or impossible to get. Thus, such problems are prevalent in physics, engineering, and any field that models changing systems.
Linear Interpolation
Linear interpolation is a technique used to estimate unknown values that fall within the range of a set of known data points. It's like connecting the dots but much more mathematical! When we perform linear interpolation, we assume that the change between points is linear, which simplifies calculations.In the context of our current problem, after we use Euler's method and calculate a series of approximate values for \(y\), linear interpolation helps us estimate the value of \(y\) at points not explicitly calculated, such as \(y(1.052)\), \(y(1.555)\), and \(y(1.978)\). The formula used is:\[ y(x) = y_1 + (x - x_1) \cdot \left( \frac{y_2 - y_1}{x_2 - x_1} \right) \]Where \((x_1, y_1)\) and \((x_2, y_2)\) are the surrounding known data points. This provides an efficient way to estimate values by drawing a straight line between known points and using that line to infer the unknown values between them, ensuring those values maintain a logical progression.

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Most popular questions from this chapter

Water flows from an inverted conical tank with circular orifice at the rate $$ \frac{d x}{d t}=-0.6 \pi r^{2} \sqrt{2 g} \frac{\sqrt{x}}{A(x)} $$ where \(r\) is the radius of the orifice, \(x\) is the height of the liquid level from the vertex of the cone, and \(A(x)\) is the area of the cross section of the tank \(x\) units above the orifice. Suppose \(r=0.1 \mathrm{ft}\), \(g=32.1 \mathrm{ft} / \mathrm{s}^{2}\), and the tank has an initial water level of \(8 \mathrm{ft}\) and initial volume of \(512(\pi / 3) \mathrm{ft}^{3}\). Use the Runge-Kutta method of order four to find the following. a. The water level after 10 min with \(h=20 \mathrm{~s}\) b. When the tank will be empty, to within \(1 \mathrm{~min}\).

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 $$ y^{\prime}=f(t, y), \quad a \leq t \leq b, \quad y(a)=\alpha. $$

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

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?

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