91Ó°ÊÓ

We have seen that the error in approximating the solution to an initial value problem is proportional to \(\Delta t .\) That is, if \(E_{\Delta t}\) is the Euler's method approximation to the solution to an initial value problem at \(\bar{t},\) then $$ y(\bar{t})-E_{\Delta t} \approx K \Delta t $$ for some constant of proportionality \(K .\) In this problem, we will see how to use this fact to improve our estimates, using an idea called accelerated convergence. a. We will create a new approximation by assuming the error is exactly proportional to \(\Delta t,\) according to the formula $$ y(\bar{t})-E_{\Delta t}=K \Delta t . $$ Using our earlier results from the initial value problem \(d y / d t=y\) and \(y(0)=1\) with \(\Delta t=0.2\) and \(\Delta t=0.1,\) we have $$ \begin{array}{l} y(1)-2.4883=0.2 \mathrm{~K} \\ y(1)-2.5937=0.1 \mathrm{~K} \end{array} $$ This is a system of two linear equations in the unknowns \(y(1)\) and \(K .\) Solve this system to find a new approximation for \(y(1)\). (You may remember that the exact value is \(y(1)=\) \(e=2.71828 \ldots . .)\) b. Use the other data, \(E_{0.05}=2.6533\) and \(E_{0.025}=2.6851\) to do similar work as in (a) to obtain another approximation. Which gives the better approximation? Why do you think this is? c. Let's now study the initial value problem $$ \frac{d y}{d t}=t-y, y(0)=0 $$ Approximate \(y(0.3)\) by applying Euler's method to find approximations \(E_{0.1}\) and \(E_{0.05}\). Now use the idea of accelerated convergence to obtain a better approximation. (For the sake of comparison, you want to note that the actual value is \(y(0.3)=0.0408 .\) )

Step by step solution

01

- Set Up the Linear Equations

We are given the equations based on the Euler’s method approximations:\[ y(1) - 2.4883 = 0.2 K \]\[ y(1) - 2.5937 = 0.1 K \]These equations represent a system of two linear equations with unknowns \( y(1) \) and \( K \).

02

- Subtract the Linear Equations

Subtract the second equation from the first to eliminate \( y(1) \):\[ (y(1) - 2.4883) - (y(1) - 2.5937) = 0.2 K - 0.1 K \]This simplifies to:\[ 2.5937 - 2.4883 = 0.1 K \]\[ 0.1054 = 0.1 K \]

03

- Solve for K

Solve for \( K \) by dividing both sides by 0.1:\[ K = \frac{0.1054}{0.1} = 1.054 \]

04

- Substitute Back to Find y(1)

Substitute \( K \) back into one of the original equations to solve for \( y(1) \):\[ y(1) - 2.4883 = 0.2 \times 1.054 \]\[ y(1) - 2.4883 = 0.2108 \]\[ y(1) = 2.6991 \]

05

- Set Up Other Linear Equations

For part (b), we use the following two equations:\[ y(1) - 2.6533 = 0.05 K \]\[ y(1) - 2.6851 = 0.025 K \]

06

- Subtract the Other Linear Equations

Subtract the second equation from the first to eliminate \( y(1) \):\[ (y(1) - 2.6533) - (y(1) - 2.6851) = 0.05 K - 0.025 K \]This simplifies to:\[ 2.6851 - 2.6533 = 0.025 K \]\[ 0.0318 = 0.025 K \]

07

- Solve for K in New Equation

Solve for \( K \) by dividing both sides by 0.025:\[ K = \frac{0.0318}{0.025} = 1.272 \]

08

- Substitute Back to Find y(1)

Substitute \( K \) back into one of the new equations to solve for \( y(1) \):\[ y(1) - 2.6533 = 0.05 \times 1.272 \]\[ y(1) - 2.6533 = 0.0636 \]\[ y(1) = 2.7169 \]

09

- Compare Approximations

The approximations are 2.6991 and 2.7169. Comparing these to the exact value of 2.71828, the second set of approximations with smaller \( \Delta t \) provides a better result, which generally holds because smaller steps in Euler's method reduce error.

10

- Apply Euler's Method for New IVP

For part (c), use Euler's method with \( \Delta t = 0.1 \) and \( \Delta t = 0.05 \) to solve the new initial value problem \( \frac{dy}{dt} = t - y \), \( y(0) = 0 \).

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Key Concepts

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

Initial Value Problem

An initial value problem (IVP) in differential equations involves finding a function that satisfies a given differential equation and an initial condition. The initial condition provides the value of the function at a specific point, typically where the independent variable (e.g., time) equals zero. This foundational concept is crucial in various fields such as physics, engineering, and economics.

For example, in our exercise, the IVP is given by: \( \frac{dy}{dt} = y \) with \( y(0) = 1. \) The task is to find the solution to the differential equation that meets the initial condition specified. Euler's method is one approach to solving IVPs by approximating the solution iteratively, step-by-step.

This method uses a small step size \( \Delta t \) to move from the initial condition to approximate values of the function at subsequent points. Although Euler's method is straightforward, it might introduce errors, which leads us to examine ways to accelerate convergence and improve approximations.

Accelerated Convergence

Accelerated convergence is a method used to improve the accuracy of approximations by leveraging the known behavior of errors. In Euler's method, the error in approximating an initial value problem is proportional to the step size \( \Delta t. \) The error form can be represented as \( y(t) – E_{\Delta t} \approx K \Delta t, \) where \( E_{\Delta t} \) is the Euler's method approximation, and \( K \) is a constant of proportionality.

By using multiple approximations with different step sizes, we can create a system of linear equations. Solving these equations allows us to estimate the constant \( K \) and then achieve a more accurate approximation of the desired value. For instance, given the systems:
\( y(1) - 2.4883 = 0.2K \)
\( y(1) - 2.5937 = 0.1K \)
We solved these equations to find \( y(1) \approx 2.6991 \) and \( K \approx 1.054. \) Repeating this for other datasets and comparing the approximations shows that smaller step sizes generally yield better accuracy due to reduced approximation error.

Approximation Error

Approximation error is an essential concept when using numerical methods to solve differential equations. In Euler's method, the approximation error indicates the difference between the actual value of the solution and the estimated value obtained through iterations using a step size \( \Delta t. \)

Error can be analyzed by considering how it scales with the step size. Generally, the error in Euler's method is directly proportional to \( \Delta t. \) This means that reducing the step size will decrease the error, leading to a more accurate approximation. For example, in the dataset: \( (y(1) - 2.6533 = 0.05K \) and \( y(1) - 2.6851 = 0.025K), \) solving provided an approximation of \( y(1) \approx 2.7169 \), which is closer to the exact value \( e = 2.71828 \).

To minimize the approximation error, smaller step sizes should be used, but this will increase computational resource demands. Therefore, techniques like accelerated convergence help balance accuracy and efficiency by leveraging multiple step sizes to better estimate the solution.