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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
Using Euler's method, the estimate for \( f(2) \) is approximately 3.375.

Step by step solution

01

- Determine step size

Given the interval \text{from } 0 \text{ to } 2 \text{ and } n=4, \text{calculate the step size } h.\[ h = \frac{2 - 0}{4} = 0.5 \]
02

- Initialize variables

Set the initial condition: \( t_0 = 0, y_0 = 5\).
03

- Apply Euler’s method for the first step

Calculate the first step using Euler's method.\[ y_{1} = y_0 + h \cdot f(t_0, y_0) \]\[ f(t_0, y_0) = 2t_0 - y_0 + 1 = 2(0) - 5 + 1 = -4 \]\[ y_{1} = 5 + 0.5 \cdot (-4) = 5 - 2 = 3 \]
04

- Apply Euler’s method for the second step

Calculate the second step.\[ t_1 = 0 + 0.5 = 0.5 \]\[ f(t_1, y_1) = 2t_1 - y_1 + 1 = 2(0.5) - 3 + 1 = 1 - 3 + 1 = -1 \]\[ y_{2} = y_1 + h \cdot f(t_1, y_1) = 3 + 0.5 \cdot (-1) = 3 - 0.5 = 2.5 \]
05

- Apply Euler’s method for the third step

Calculate the third step.\[ t_2 = 0.5 + 0.5 = 1.0 \]\[ f(t_2, y_2) = 2t_2 - y_2 + 1 = 2(1.0) - 2.5 + 1 = 2 + 1 - 2.5 = 0.5 \]\[ y_{3} = y_2 + h \cdot f(t_2, y_2) = 2.5 + 0.5 \cdot 0.5 = 2.5 + 0.25 = 2.75 \]
06

- Apply Euler’s method for the fourth step

Calculate the fourth step.\[ t_3 = 1.0 + 0.5 = 1.5 \]\[ f(t_3, y_3) = 2t_3 - y_3 + 1 = 2(1.5) - 2.75 + 1 = 3 - 2.75 + 1 = 1.25 \]\[ y_{4} = y_3 + h \cdot f(t_3, y_3) = 2.75 + 0.5 \cdot 1.25 = 2.75 + 0.625 = 3.375 \]

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Key Concepts

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

Numerical Methods
Numerical methods are techniques used to approximate solutions for mathematical problems. In many cases, these problems cannot be solved analytically with simple formulas. For example, differential equations often rely on numerical methods to find approximate solutions. Euler's method is one such numerical approach, particularly useful for solving initial value problems. It approximates the solution by taking small steps along the curve, starting from the initial point and using the equation’s slope to find subsequent points. This method is easy to implement and provides a straightforward way to approximate moderately complex equations.
Differential Equations
An initial value problem (IVP) involves finding a function that satisfies a differential equation and meets specified initial conditions. For instance, in the given problem, the initial condition is y(0) = 5, meaning that when the time t is 0, the value of the function y is 5. Initial conditions are crucial because they define the starting point of the solution.
Euler’s method and other numerical methods often require these initial conditions as a starting point to calculate successive values of the function. By iteratively applying the method, we can approximate the function across the interval of interest.
Step Size
The step size \(h\) is a critical parameter in numerical methods like Euler's method. It determines the interval between each point where the function will be approximated. The smaller the step size, the more accurate the approximation, but this comes with increased computational effort. In the given exercise, the interval is from 0 to 2, with \(n = 4\) steps. Therefore, the step size \(h\) is calculated as: \(\frac{2 - 0}{4} = 0.5\)

In practice, choosing the right step size involves a trade-off between accuracy and efficiency. A smaller step size means more points and often higher accuracy but requires more calculations. Conversely, a larger step size is computationally cheaper but may lead to less accurate results. Starting with a moderate step size and refining it as needed based on the problem is usually a good strategy.

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