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An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the three-decimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. \(y^{\prime}=x-y, y(0)=1 ; y(x)=2 e^{-x}+x-1\)

Short Answer

Expert verified
At \(x=0.5\), Euler's method with \(h=0.1\) approximates \(y(0.5) \) closer to the exact solution than \(h=0.25\). The exact value is approximately 0.713.

Step by step solution

01

Understanding Euler's Method

Euler's Method is a numerical technique used to approximate solutions to ordinary differential equations (ODEs) with a given initial value. The general formula is: \( y_{n+1} = y_n + h \, f(x_n, y_n) \), where \( h \) is the step size and \( f(x_n, y_n) \) is the derivative. For this exercise, \( f(x, y) = x - y \).
02

Apply Euler's Method with h=0.25

Starting with \( y(0) = 1 \), apply Euler's method with \( h = 0.25 \) over the interval \( [0, 0.5] \).- **At \( x_0 = 0, y_0 = 1 \):** \( y_1 = y_0 + h \cdot (x_0 - y_0) = 1 + 0.25 \cdot (0 - 1) = 0.75 \)- **At \( x_1 = 0.25, y_1 = 0.75 \):** \( y_2 = y_1 + h \cdot (x_1 - y_1) = 0.75 + 0.25 \cdot (0.25 - 0.75) = 0.625 \)So, at \( x = 0.5 \), the approximation is \( y(0.5) = 0.625 \).
03

Apply Euler's Method with h=0.1

Now, apply Euler's method with \( h = 0.1 \) over the interval \( [0, 0.5] \):- **At \( x_0 = 0, y_0 = 1 \):** \( y_1 = y_0 + h \cdot (x_0 - y_0) = 1 + 0.1 \cdot (0 - 1) = 0.9 \)- **At \( x_1 = 0.1, y_1 = 0.9 \):** \( y_2 = y_1 + h \cdot (x_1 - y_1) = 0.9 + 0.1 \cdot (0.1 - 0.9) = 0.82 \)- **At \( x_2 = 0.2, y_2 = 0.82 \):** \( y_3 = y_2 + h \cdot (x_2 - y_2) = 0.82 + 0.1 \cdot (0.2 - 0.82) = 0.758 \)- **At \( x_3 = 0.3, y_3 = 0.758 \):** \( y_4 = y_3 + h \cdot (x_3 - y_3) = 0.758 + 0.1 \cdot (0.3 - 0.758) = 0.7122 \)- **At \( x_4 = 0.4, y_4 = 0.7122 \):** \( y_5 = y_4 + h \cdot (x_4 - y_4) = 0.7122 + 0.1 \cdot (0.4 - 0.7122) = 0.68198 \)Approximately, at \( x = 0.5 \), \( y(0.5) = 0.68198 \).
04

Calculate the Exact Solution at x=0.5

Substitute \( x = 0.5 \) into the given exact solution \( y(x) = 2e^{-x} + x - 1 \):\[y(0.5) = 2e^{-0.5} + 0.5 - 1 = 2 \cdot e^{-0.5} - 0.5\]Using a calculator, \( 2 \cdot e^{-0.5} \approx 1.2131 \), thus:\[y(0.5) \approx 1.2131 - 0.5 = 0.7131\]
05

Compare Approximations with Exact Solution

Now, compare the approximations from Euler's method with the exact solution:- Euler's method with \( h=0.25: y(0.5) = 0.625 \)- Euler's method with \( h=0.1: y(0.5) = 0.682 \)- Exact solution: \( y(0.5) = 0.7131 \)Notice that smaller step size gives a closer approximation to the exact value.

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Key Concepts

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

Numerical Approximation
Numerical approximation is a powerful approach used to find approximate solutions to mathematical problems that are difficult or impossible to solve exactly. In the context of differential equations, it allows us to estimate the behavior of functions when the exact solution is unknown or hard to compute. Euler's Method is a fundamental tool in numerical approximation. This method offers a simple yet effective means to approximate the solution of an ordinary differential equation (ODE) given an initial value.
- **Numerical Methods like Euler's Method:** These methods iterate over small steps to project the behavior of the function over a given interval. - **Practical Application:** Useful in engineering, physics, and many other fields where modeling the behavior of dynamic systems is essential. - **Iterative Process:** It involves guesswork within defined bounds to converge upon a solution, which improves with more iterations or smaller steps.
Initial Value Problem
An initial value problem is a type of differential equation accompanied by a value at a starting point, known as the initial condition. Solving these problems involves finding a function that not only satisfies the differential equation but also matches the supplied initial condition.
- **Given Information:** For example, in this exercise, we have an equation: \( y' = x - y \) with an initial condition: \( y(0) = 1 \). - **Solving Strategy:** Start from the initial value, use Euler's Method to project future values.
- **Real-Life Examples:** These problems are pivotal in simulating real-world processes such as population dynamics, heat distribution, or motion under forces.
Step Size in Differential Equations
The step size, denoted as \( h \), is crucial in numerical methods as it defines the interval between calculations when estimating values of a differential equation. Smaller step sizes generally yield more accurate results because they allow the approximation to track the curve of the actual solution more closely. However, they also require more computations.
- **Effect on Accuracy and Efficiency:** - A small step size (e.g., \( h = 0.1 \)) results in better approximations but at the cost of increased computational effort. - A larger step size (e.g., \( h = 0.25 \)) will involve fewer calculations but may lead to less accurate approximations.- **Optimal Balancing:** Finding the right step size often involves a trade-off between computational expense and desired precision.
Exact Solution Comparison
Comparing numerical approximations to an exact solution helps validate the efficiency and accuracy of numerical methods like Euler's. In this exercise, comparing the approximations obtained using two different step sizes to the exact solution provides insight into how the choice of \( h \) affects results.
- **Comparison Results:** - Step size of \( h = 0.25 \) gave an approximation of \( y(0.5) = 0.625 \). - A smaller step size of \( h = 0.1 \) resulted in a more accurate approximation of \( y(0.5) = 0.682 \). - The exact solution, calculated as \( y(0.5) \approx 0.7131 \), shows that the smaller step size prediction is closer to the actual value.- **Significance of Comparison:** By analyzing the discrepancy, one can understand the trade-offs involved in choosing step sizes for numerical methods. This evaluation is crucial in applications where precision is vital.

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Most popular questions from this chapter

A motorboat starts from rest (initial velocity \(v(0)=v_{0}=\) 0). Its motor provides a constant acceleration of \(4 \mathrm{ft} / \mathrm{s}^{2}\) but water resistance causes a deceleration of \(v^{2} / 400 \mathrm{ft} / \mathrm{s}^{2}\) Find \(v\) when \(t=10 \mathrm{~s}\), and also find the limiting velocity as \(t \rightarrow+\infty\) (that is, the maximum possible speed of the boat).

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the Runge-Kutta method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.25 .\) Construct a table showing five-decimal-place values of the approximate solution and actual solution at the points \(x=0.25\) and \(0.5\). $$ y^{\prime}=x-y, y(0)=1 ; y(x)=2 e^{-x}+x-1 $$

(a) Show that if a projectile is launched straight upward from the surface of the earth with initial velocity \(v_{0}\) less than escape velocity \(\sqrt{2 G M} / R\), then the maximum distance from the center of the earth attained by the projectile is $$ r_{\mathrm{max}}=\frac{2 G M R}{2 G M-R v_{0}^{2}} $$ where \(M\) and \(R\) are the mass and radius of the earth, respectively. (b) With what initial velocity \(v_{0}\) must such a projectile be launched to yield a maximum altitude of 100 kilometers above the surface of the earth? (c) Find the maximum distance from the center of the earth, expressed in terms of earth radii, attained by a projectile launched from the surface of the earth with \(90 \%\) of escape velocity.

Use either the exact solution or a computer-generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $$ \frac{d x}{d t}=1-x^{2}, x(0)=3 $$

Consider a population \(P(t)\) satisfying the logistic equation \(d P / d t=a P-b P^{2}\), where \(B=a P\) is the time rate at which births occur and \(D=b P^{2}\) is the rate at which deaths occur. If the initial population is \(P(0)=P_{0}\), and \(B_{0}\) births per month and \(D_{0}\) deaths per month are occurring at time \(t=0\), show that the limiting population is \(M=B_{0} P_{0} / D_{0}\)

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