Problem 1 - Thomas Calculus

Question

In Exercises \(1-6,\) use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$ y^{\prime}=1-\frac{y}{x}, \quad y(2)=-1, \quad d x=0.5 $$

Step-by-Step Solution

Answer

Euler's approximations are: 1. \( x = 2.5, y \approx -0.25 \) 2. \( x = 3.0, y \approx 0.3 \) Exact solutions differ slightly.

Step 1: Understand the Problem Setup

We are given a differential equation \( y' = 1 - \frac{y}{x} \) with the initial condition \( y(2) = -1 \). We have to use Euler's method to find three approximations with a step size of \( dx = 0.5 \). Additionally, we need to find the exact solution for comparison.

Step 2: Euler's Method Formula

Euler's method involves using the formula \( y_{n+1} = y_n + f(x_n, y_n) \cdot dx \) to approximate the solution of the differential equation. Here, \( f(x, y) = 1 - \frac{y}{x} \) and the step size \( dx = 0.5 \).

Step 3: First Approximation with Euler's Method

Start with the initial condition \( x_0 = 2 \) and \( y_0 = -1 \). Calculate the next approximation:\[\begin{align*}y_1 &= y_0 + f(x_0, y_0) \cdot dx \&= -1 + \left(1 - \frac{-1}{2}\right) \cdot 0.5 \&= -1 + 1.5 \cdot 0.5 \&= -0.25.\end{align*}\]

Step 4: Second Approximation with Euler's Method

Now use \( x_1 = 2.5 \) and \( y_1 = -0.25 \) to find the next approximation:\[\begin{align*}y_2 &= y_1 + f(x_1, y_1) \cdot dx \&= -0.25 + \left(1 - \frac{-0.25}{2.5}\right) \cdot 0.5 \&= -0.25 + 1.1 \cdot 0.5 \&= 0.3.\end{align*}\]

Step 5: Third Approximation with Euler's Method

Use \( x_2 = 3.0 \) and \( y_2 = 0.3 \) to find the next approximation:\[\begin{align*}y_3 &= y_2 + f(x_2, y_2) \cdot dx \&= 0.3 + \left(1 - \frac{0.3}{3}\right) \cdot 0.5 \&= 0.3 + 0.9 \cdot 0.5 \&= 0.75.\end{align*}\]

Step 6: Calculate the Exact Solution

To find the exact solution, solve the differential equation \( y' = 1 - \frac{y}{x} \).\[\begin{align*}\text{Rearrange to } &\frac{dy}{dx} + \frac{y}{x} = 1. \\text{This is a } &\text{linear differential equation which can be solved by using the integrating factor } \mu(x) = e^{\int \frac{1}{x} dx} = x. \\text{Multiply both sides by } & x: x\frac{dy}{dx} + y = x. \\text{Rewrite as } & \frac{d}{dx}(xy) = x. \\text{Integrate: } & xy = \int x \, dx = \frac{x^2}{2} + C. \\text{Solve for } y: & y = \frac{x}{2} + \frac{C}{x}. \end{align*}\]

Step 7: Find the Constant C Using Initial Conditions

Using the initial condition \( y(2) = -1 \), substitute in to find \( C \):\[-1 = \frac{2}{2} + \frac{C}{2} \-1 = 1 + \frac{C}{2} \-2 = \frac{C}{2} \C = -4.\]So, the exact solution is \( y = \frac{x}{2} - \frac{4}{x} \).

Step 8: Compare Euler's Approximations with the Exact Solution

Calculate the exact values at each corresponding \( x \ (x = 2.5, 3.0) \) using the exact solution \( y = \frac{x}{2} - \frac{4}{x} \):- At \( x = 2.5 \):\[y = \frac{2.5}{2} - \frac{4}{2.5} = 1.25 - 1.6 = -0.35.\]- At \( x = 3.0 \):\[y = \frac{3}{2} - \frac{4}{3} = 1.5 - 1.33 \approx 0.17.\]Euler’s method approximations are: \( y_1 = -0.25 \), \( y_2 = 0.3 \), and \( y_3 = 0.75 \). These vary from the exact solutions due to the nature of the numerical approximation.

Key Concepts

Numerical Approximations

Numerical approximations are essential in mathematics, especially when exact solutions are difficult or impossible to find. Euler's Method is one of the simplest and most widely used techniques for numerical approximations. It estimates the value of a function at a certain point using a given initial value and step size. By advancing in small increments and using the derivative to predict the new function values, we approximate the solution to differential equations.
By applying Euler's Method, we start at an initial point and take steps according to the derivative's slope. This method is particularly beneficial in solving problems where analytical solutions require complex calculations or are not feasible. Though simple, Euler’s Method can lead to inaccuracies due to its reliance on straight-line approximations. Yet, it provides a foundational approach to understanding numerical solutions, opening up pathways to more advanced methods such as the Runge-Kutta or Adams-Bashforth methods.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with a specific point where the solution is known. It sets the initial condition of the function, which Euler’s Method then uses as a starting point. Solving an IVP involves finding a function that satisfies both the differential equation and the initial condition.
In our example, the IVP is given as \(y'(x) = 1 - \frac{y}{x}\), with an initial condition \(y(2) = -1\). This means we are looking for a function \(y(x)\) where, when the derivative is substituted back into the equation, the initial condition holds true. The precision of solutions to IVPs is crucial for predicting future behavior of various systems in science and engineering.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They model various phenomena in physics, engineering, and other sciences, such as motion, heat, or waves. Solving a differential equation provides insights into the dynamics of the system it describes.
There are various types of differential equations: ordinary differential equations (ODEs), like the one in this exercise, involve functions of a single variable and their derivatives. Solving them can be done analytically or numerically, depending on the complexity. Euler’s Method provides a quick, albeit rough, numerical solution.
Analytically solving differential equations often involves techniques like separation of variables, integration factors, or transformation methods. Each method caters to specific forms or types but aims to find a function that will satisfy the given differential equation across its domain.