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Chapter 2: Problem 5

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use \(h=0.1\) and then use \(h=0.05\). \(-y\) $$ y^{\prime}=e^{-y}, \quad y(0)=0 ; y(0.5) $$

Short Answer

Expert verified
Use Euler's method; approximations: 0.6487 for \(h=0.1\), 0.6412 for \(h=0.05\).

Step by step solution

01

Understanding the Problem

We need to approximate the value of the function \(y\) at \(x = 0.5\) using Euler's method. The differential equation given is \(y' = e^{-y}\) with the initial condition \(y(0) = 0\). We'll solve this twice using different step sizes: \(h = 0.1\) and \(h = 0.05\).
02

Euler's Method with Step Size h = 0.1

Euler's method updates the value of \(y\) using: \(y_{n+1} = y_n + h \cdot y'(x_n, y_n)\). Starting from \(x_0 = 0\) and \(y_0 = 0\) with \(h = 0.1\):- At \(x_0 = 0\), \(y_0 = 0\); \(y_1 = 0 + 0.1 \cdot e^{0} = 0.1\)- At \(x_1 = 0.1\), \(y_1 = 0.1\); \(y_2 = 0.1 + 0.1 \cdot e^{-0.1}\)- At \(x_2 = 0.2\), \(y_2\) from previous- Continue until \(x_5 = 0.5\).Calculate the values iteratively to get the last value \(y(0.5)\).
03

Euler's Method with Step Size h = 0.05

Repeat the procedure in Step 2 with a smaller step size \(h = 0.05\). This time:- At \(x_0 = 0\), \(y_0 = 0\); \(y_1 = 0 + 0.05 \cdot e^{0} = 0.05\)- At \(x_1 = 0.05\), \(y_1 = 0.05\); \(y_2 = 0.05 + 0.05 \cdot e^{-0.05}\)- At \(x_2 = 0.1\), \(y_2\) from previous- Continue until \(x_{10} = 0.5\).Calculate the values iteratively until you reach \(x = 0.5\).
04

Calculating and Comparing Results

Compute and compare the final approximations of \(y(0.5)\) from both step sizes. This will show how the accuracy improves with the smaller step size \(h = 0.05\). Write down the values obtained for both approximations.

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Key Concepts

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

Numerical Solver
Euler's Method is a numerical solver that helps to approximate solutions for differential equations. In many cases, finding an exact solution for a differential equation can be challenging or impossible. That's where numerical solvers like Euler's Method come in handy. They provide a way to estimate the solution at specific points by making small approximations along the function. Euler's Method, in particular, is straightforward and relies on using datas from already computed points to estimate the values at the next points. The general idea is to use the derivative to take small steps along the curve of the solution.
Differential Equations
Differential equations describe the relationship between functions and their derivatives. They are crucial in expressing physical phenomena such as heat, motion, and growth. The equation given in the exercise, \( y' = e^{-y} \), is a first-order ordinary differential equation. Here, the derivative \( y' \) represents the rate of change of \( y \) with respect to \( x \). Solving a differential equation means finding a function \( y(x) \) that satisfies this relationship under the given conditions. The challenge lies in the fact that this link is often not straightforward, making analytical solutions difficult to achieve.
Step Size
Step size, denoted by \( h \), is a critical component of Euler's Method. It determines how large each step in the numerical approximation will be as we move from one value of \( x \) to the next. A smaller step size means making more but finer steps, resulting in a more accurate approximation of the solution at the cost of increased computations. Conversely, a larger step size reduces computational cost but may decrease accuracy. In the given exercise, two different step sizes are used: \( h = 0.1 \) and \( h = 0.05 \). Observing the results shows us how smaller steps can increase precision, indicating the trade-off between computational effort and accuracy.
Initial Conditions
Initial conditions are the starting values that set the stage for solving a differential equation. They establish a specific solution path among the many possibilities. In this exercise, the initial condition is \( y(0) = 0 \). This means that at \( x = 0 \), the solution function \( y \) starts at 0. Initial conditions are crucial because different conditions can lead to different solution curves. They essentially "anchor" the solution's starting point and ensure we are following the correct curve that fits the given scenario.

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Most popular questions from this chapter

Solve the given initial-value problem. $$ \frac{d y}{d x}=\frac{3 x+2 y}{3 x+2 y+2}, \quad y(-1)=-1 $$

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(a) Census data for the United States between 1790 and 1950 is given in the following table. Construct alogistic population model using the data from 1790,1850 , and 1910 . $$ \begin{array}{lr} \hline \text { Year } & \text { Population (in millions) } \\ \hline 1790 & 3.929 \\ 1800 & 5.308 \\ 1810 & 7.240 \\ 1820 & 9.638 \\ 1830 & 12.866 \\ 1840 & 17.069 \\ 1850 & 23.192 \\ 1860 & 31.433 \\ 1870 & 38.558 \\ 1880 & 50.156 \\ 1890 & 62.948 \\ 1900 & 75.996 \\ 1910 & 91.972 \\ 1920 & 105.711 \\ 1930 & 122.775 \\ 1940 & 131.669 \\ 1950 & 150.697 \\ \hline \end{array} $$ (b) Construct a table comparing actual census population with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair.

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