/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Approximate the value of e by lo... [FREE SOLUTION] | 91Ó°ÊÓ

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Approximate the value of e by looking at the initial value problem \(y^{\prime}=y\) with \(y(0)=1\) and approximating \(y(1)\) using Euler's method with a step size of \(0.2 .\)

Short Answer

Expert verified
Using Euler's method, the value of \(e\) is approximately 2.48832.

Step by step solution

01

Understanding Euler's Method

Euler's method is a numerical technique to approximate the solutions of initial value problems of the form \(y' = f(x, y)\). Starting from an initial condition \((x_0, y_0)\), we use a step size \(h\) to iteratively compute \(y_{n+1} = y_n + h \cdot f(x_n, y_n)\) and \(x_{n+1} = x_n + h\).
02

Initial Setup

For the problem \(y' = y\) with \(y(0) = 1\), we have the initial condition \((x_0, y_0) = (0, 1)\). Our function is \(f(x, y) = y\), and we are given a step size \(h = 0.2\).
03

First Iteration

Using Euler's method, we compute \(y_1\) as follows: \(y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.2 \cdot 1 = 1.2\). The new point is \((x_1, y_1) = (0.2, 1.2)\).
04

Second Iteration

For the next step, we calculate \(y_2 = y_1 + h \cdot f(x_1, y_1) = 1.2 + 0.2 \cdot 1.2 = 1.44\). The new point becomes \((x_2, y_2) = (0.4, 1.44)\).
05

Third Iteration

Next, we find \(y_3 = y_2 + h \cdot f(x_2, y_2) = 1.44 + 0.2 \cdot 1.44 = 1.728\). Thus, the new point is \((x_3, y_3) = (0.6, 1.728)\).
06

Fourth Iteration

Now calculate \(y_4 = y_3 + h \cdot f(x_3, y_3) = 1.728 + 0.2 \cdot 1.728 = 2.0736\). Therefore, the new point is \((x_4, y_4) = (0.8, 2.0736)\).
07

Fifth (Final) Iteration

Finally, compute \(y_5 = y_4 + h \cdot f(x_4, y_4) = 2.0736 + 0.2 \cdot 2.0736 = 2.48832\). The final point is \((x_5, y_5) = (1.0, 2.48832)\).

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Key Concepts

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

Numerical Approximation
When working with mathematical problems that do not have straightforward analytical solutions, numerical methods become our best allies. Numerical approximation techniques allow us to estimate values that we cannot compute exactly. Techniques like Euler's method, which we use in this exercise, form a foundation in numerical analysis.
The main idea of numerical approximation is that instead of solving the problem directly, we create a sequence of simpler approximations. These approximations use a step-by-step approach to get closer to the true solution. In Euler's method, the step size, denoted by \( h \), defines the intervals at which we estimate our function's value.
  • Smaller step sizes generally lead to more accurate results.
  • The balance between computation effort and accuracy is crucial.
  • Euler's method provides a simple way to approximate differential equations, but understanding its limitations is key.
Through practice and adjustments in step size, we refine our approximations to achieve the desired accuracy.
Initial Value Problems
Initial value problems (IVPs) set the stage for many real-world scenarios that involve differential equations. These problems specify the value of the unknown function at a particular point, commonly denoted as the initial condition.
In our example, the initial value problem is defined by the differential equation \( y' = y \) with the specific initial condition \( y(0) = 1 \). Essentially, we are tasked with finding a function \( y(x) \) such that it satisfies the equation and begins with the given initial value.
  • IVPs are central in modeling physical systems like population growth or cooling of substances.
  • They allow incorporation of specific conditions to ensure a unique solution exists.
  • Navigating from the known initial condition through numerical approximations like Euler's method helps fulfill IVP requirements.
The beauty of IVPs is how they bring theoretical mathematics closer to practical, applicable solutions.
Differential Equations
Differential equations are mathematical tools used to describe how things change. They relate functions and their derivatives, providing insights into processes that involve rates of change.
Differential equations can be found in various forms, from simple linear equations like \( y' = y \) in our exercise, to complex, non-linear systems depicting intricate phenomena. Solving these involves finding a function that satisfies the equation given the derivatives and any initial conditions.
  • They model the behavior and dynamics of the world — from physics to finance.
  • Understanding the type of differential equation helps choose the right solving technique.
  • Numerical methods, like Euler's method, make it feasible to handle equations without simple closed-form solutions.
With differential equations, we capture the essence of continuous change, essential in many scientific fields to explain and predict behavior.

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Most popular questions from this chapter

Take \(y^{\prime}=f(x, y), y(0)=0,\) where \(f(x, y)>1\) for all \(x\) and \(y\). If the solution exists for all \(x,\) can you say what happens to \(y(x)\) as \(x\) goes to positive infinity? Explain.

Sketch slope field for \(y^{\prime}=x^{2}\).

Sketch a phase diagram for different possibilities. Note that these possibilities are \(A>B\), or \(A=B\), or \(A\) and \(B\) both complex (i.e. no real solutions). Hint: Fix some simple \(k\) and \(M\) and then vary \(h\). For example, let \(M=8\) and \(k=0.1\). When \(h=1\), then \(A\) and \(B\) are distinct and positive. The slope field we get is in Figure 1.13 on the next page. As long as the population starts above \(B\), which is approximately 1.55 million, then the population will not die out. It will in fact tend towards \(A \approx 6.45\) million. If ever some catastrophe happens and the population drops below \(B\), humans will die out, and the fast food restaurant serving them will go out of business. When \(h=1.6\), then \(A=B=4\). There is only one critical point and it is unstable. When the population starts above 4 million it will tend towards 4 million. If it ever drops below 4 million, humans will die out on the planet. This scenario is not one that we (as the human fast food proprietor) want to be in. A small perturbation of the equilibrium state and we are out of business. There is no room for error. See Figure \(1.14 .\) Finally if we are harvesting at 2 million humans per year, there are no critical points. The population will always plummet towards zero, no matter how well stocked the planet starts. See Figure 1.15 .

Suppose \(y^{\prime}=f(x, y)\) is such that \(f(x, 1)=0\) for every \(x, f\) is continuous and \(\frac{\partial f}{\partial y}\) exists and is continuous for every \(x\) and \(y\). a) Guess a solution given the initial condition \(y(0)=1\). b) Can graphs of two solutions of the equation for different initial conditions ever intersect? c) Given \(y(0)=0,\) what can you say about the solution. In particular, can \(y(x)>1\) for any \(x\) ? Can \(y(x)=1\) for any \(x\) ? Why or why not?

Solve \(y^{\prime}=x / y\).

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