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Chapter 10: Problem 7

Use Euler's method with \(n=4\) to approximate the solution \(f(t)\) to \(y^{\prime}=2 t-y+1, y(0)=5\) for \(0 \leq t \leq 2 .\) Estimate \(f(2)\).

Short Answer

Expert verified
Approximation of \(f(2)\) is 3.375.

Step by step solution

01

Define Initial Parameters

Identify the initial condition and the number of steps. Here, the initial condition is given as \(y(0) = 5\) and the number of steps \(n = 4\). The interval is from \(t=0\) to \(t=2\).
02

Calculate Step Size

Calculate the step size \(h\) by dividing the interval length by the number of steps: \[ h = \frac{2-0}{4} = 0.5 \]
03

Euler's Method Formula

Recall Euler's method formula for updating the value of \(y\): \[ y_{k+1} = y_k + h f(t_k, y_k) \ f(t_k, y_k) = 2 t_k - y_k + 1 \]
04

Iteration 1

For \(k=0\), use \(t_0 = 0\) and \(y_0 = 5\):\[ y_1 = y_0 + h (2 t_0 - y_0 + 1) \ y_1 = 5 + 0.5 (2(0) - 5 + 1) = 5 + 0.5 (-4) = 5 - 2 = 3 \]
05

Iteration 2

For \(k=1\), use \(t_1 = 0.5\) and \(y_1 = 3\):\[ y_2 = y_1 + h (2 t_1 - y_1 + 1) \ y_2 = 3 + 0.5 (2(0.5) - 3 + 1) = 3 + 0.5 (1 - 3 + 1) = 3 + 0.5 (-1) = 3 - 0.5 = 2.5 \]
06

Iteration 3

For \(k=2\), use \(t_2 = 1\) and \(y_2 = 2.5\):\[ y_3 = y_2 + h (2 t_2 - y_2 + 1) \ y_3 = 2.5 + 0.5 (2(1) - 2.5 + 1) = 2.5 + 0.5 (2 - 2.5 + 1) = 2.5 + 0.5 (0.5) = 2.5 + 0.25 = 2.75 \]
07

Iteration 4

For \(k=3\), use \(t_3 = 1.5\) and \(y_3 = 2.75\):\[ y_4 = y_3 + h (2 t_3 - y_3 + 1) \ y_4 = 2.75 + 0.5 (2(1.5) - 2.75 + 1) = 2.75 + 0.5 (3 - 2.75 + 1) = 2.75 + 0.5 (1.25) = 2.75 + 0.625 = 3.375 \]
08

Final Answer

After 4 iterations with Euler's method, the approximation of \(f(2)\) is \(y_4 = 3.375\).

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Key Concepts

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

Numerical Analysis
Numerical analysis involves algorithms for solving mathematical problems using numerical approximations rather than exact analytical solutions. This approach is essential for solving problems that cannot be addressed with straightforward analytical methods.
Euler's method, a key topic in numerical analysis, deals with finding approximate solutions to differential equations. This method is particularly useful when exact solutions are not feasible.
The essence of numerical analysis lies in its ability to provide practical solutions for complex problems through iterative approximations.
Differential Equations
Differential equations describe the relationship between a function and its derivatives, representing how a quantity changes over time. These equations are fundamental in modeling real-world phenomena like population growth, heat transfer, and motion.
In the given problem, the differential equation is \( y^{\text{'} }=2 t-y+1 \). This first-order differential equation involves a function \( y \) and its derivative with respect to \( t \). Solving such equations can reveal the behavior of changing systems.
Using Euler鈥檚 method, we approximate solutions of differential equations through discrete steps rather than continuous functions preserving the relationship depicted by derivatives.
Initial Value Problem
An initial value problem specifies the value of the solution at a particular point, serving as a starting point for solving the differential equation. For our exercise, the initial value provided is \( y(0) = 5 \).
Step-by-step, the Euler method determines subsequent values based solely on this initial condition and the defined interval. Understanding the initial condition is crucial as it's the foundation for the entire approximation process.
This initial value problem employs iterative calculations to move from the given starting point (\

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Most popular questions from this chapter

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