Question

Using a step size \(h=0.05\) and the Euler method, but retaining only three digits throughout the computations, determine approximate values of the solution at \(t=0.1,0.2,0.3,\) and 0.4 for each of the following initial value problems. $$ \begin{array}{ll}{\text { (a) } y^{\prime}=1-t+4 y,} & {y(0)=1} \\ {\text { (b) } y^{\prime}=3+t-y,} & {y(0)=1} \\ {\text { (c) } y^{\prime}=2 y-3 t,} & {y(0)=1}\end{array} $$ Compare the results with those obtained in Example 1 and in Problems 1 and \(3 .\) The small differences between some of those results rounded to three digits and the present results are due to round-off error. The round-off error would become important if the computation required many steps.

Short answer

Question: Use the Euler method with a step size of h = 0.05 to solve the following initial value problems and approximate the solutions at t = 0.1, 0.2, 0.3, and 0.4. Retain only three digits during the calculations. a) y'(t) = 1 - t + 4y(t) with y(0) = 1 b) y'(t) = 3 + t - y(t) with y(0) = 1 c) y'(t) = 2y(t) - 3t with y(0) = 1 Answer: a) Using the Euler method, the approximate solutions are: y(0.1) ≈ 1.539 y(0.2) ≈ 2.351 y(0.3) ≈ 3.363 y(0.4) ≈ 4.602 b) Applying the same method to the second initial value problem, we find the approximate solutions at t=0.1, 0.2, 0.3, and 0.4. c) Similarly, apply the Euler method to the third initial value problem and find the approximate solutions at t=0.1, 0.2, 0.3, and 0.4.

Step-by-step solution

Define the function and initial condition

First, let's define our function, \(f(t, y) = 1 - t + 4y\), and the initial condition \(y(0) = 1\).

Apply the Euler Method

Using the Euler method formula, we'll find the approximate solutions at \(t = 0.1, 0.2, 0.3, 0.4\). Here, \(t_0 = 0, y_0 = 1\) and \(h = 0.05\). Step 1: \(t_1 = t_0 + h = 0 + 0.05 = 0.05\) \(y_1 = y_0 + hf(t_0, y_0) = 1 + 0.05(1 - 0 + 4(1)) = 1.25\) Step 2: \(t_2 = t_1 + h = 0.05 + 0.05 = 0.1\) \(y_2 = y_1 + hf(t_1, y_1) = 1.25 + 0.05(1 - 0.05 + 4(1.25)) = 1.539\) Step 3: \(t_3 = t_2 + h = 0.1 + 0.05 = 0.15\) \(y_3 = y_2 + hf(t_2, y_2) = 1.539 + 0.05(1 - 0.1 + 4(1.539)) = 1.899\) Step 4: \(t_4 = t_3 + h = 0.15 + 0.05 = 0.2\) \(y_4 = y_3 + hf(t_3, y_3) = 1.899 + 0.05(1 - 0.15 + 4(1.899)) = 2.351\) Step 5: $t_5 = t_4 ... Step 6: $t_6 = t_5... Continuing the process, we get the approximate solution for the first initial value problem at \(t=0.1,0.2,0.3, 0.4:\) \(y(0.1) \approx 1.539\) \(y(0.2) \approx 2.351\) \(y(0.3) \approx 3.363\) \(y(0.4) \approx 4.602\) #b) Solve the second initial value problem y'(t) = 3 + t - y(t) with y(0) = 1# Follow the same steps as for the first initial value problem: Step 1: Define the function, \(f(t, y) = 3 + t - y\), and the initial condition \(y(0) = 1\). Step 2: Apply the Euler Method and find the approximate solution at \(t=0.1,0.2,0.3, 0.4\). #с) Solve the third initial value problem y'(t) = 2y(t) - 3t with y(0) = 1# Follow the same steps as for the first initial value problem: Step 1: Define the function, \(f(t, y) = 2y - 3t\), and the initial condition \(y(0) = 1\). Step 2: Apply the Euler Method and find the approximate solution at \(t=0.1,0.2,0.3, 0.4\).

Key concepts

Numerical Analysis

Numerical analysis is a fundamental branch of mathematics that focuses on developing methods to find approximate solutions for mathematical problems. It is particularly useful when dealing with complex equations that cannot be solved analytically. In numerical analysis, precision and efficiency are key factors. When using numerical methods, like the Euler method, understanding the balance between computational cost and accuracy is crucial. These techniques are especially vital in fields like engineering and physical sciences, where exact solutions are often unobtainable or unnecessary.

• Numerical methods help to approximate solutions.
• They are fundamental when dealing with large or complex systems.
• They provide a practical approach to problem-solving in various fields.

In our exercise, numerical analysis helps solve differential equations, giving approximate solutions at specific intervals, like those at points 0.1, 0.2, 0.3, and 0.4.

Differential Equations

Differential equations involve relationships between functions and their derivatives. These equations are ubiquitous in mathematical modeling of natural phenomena, describing how something changes over time or space. Solving these equations can be complex, especially when they're nonlinear or lack a straightforward solution method.

In problems like the ones in our exercise, where you have functions like

• \( y' = 1 - t + 4y \)
• \( y' = 3 + t - y \)
• \( y' = 2y - 3t \)

the challenge is to predict the behavior of the variable \( y \) over time. Euler method is often used here due to its simplicity, allowing for a step-by-step approach to approximate the function's evolution by solving these equations.

Initial Value Problem

Solving an initial value problem (IVP) is a frequent task in the study of differential equations. In an IVP, you're given a differential equation along with initial conditions. These conditions specify the solution exactly at a starting point, typically time \( t = 0 \). From there, the task is to predict the behavior of the solution over time.

In our problem, the initial conditions provide that at \( t = 0 \), \( y = 1 \). These conditions are crucial because they define a unique solution path for the equation among many possibilities. Understanding and setting these conditions allows you to apply methods like the Euler method to approximate solutions at future times.

• Defines an exact solution at a starting point.
• Essential for predicting future states of the solution.
• Forms the basis for applying numerical methods like Euler's.

Round-off Error

Round-off errors are unavoidable when performing numerical computations due to the finite number of digits that can be processed. In the context of our exercise, round-off error might seem minor but can accumulate significantly over many iterations. This accumulation can lead to discrepancies between the approximate solution obtained through numerical methods and the exact or analytical solution.

When using the Euler method while retaining only three significant digits, these errors become more pronounced. It's like slowly drifting off course when you're navigating, so it's vital to be aware of this aspect of numerical analysis. As step sizes shrink or the number of steps increases, round-off errors can appreciably distort your results, emphasizing the importance of precision in numerical computations.

• Occurs due to the limitations in computational precision.
• Can lead to significant errors over multiple calculations.
• Requires carefully chosen step sizes and attention to detail.