Question

Consider the initial-value problem $$ y^{\prime}=3 x^{2}-y^{2}+1, \quad y(0)=2 $$ Use Euler's method with a step size of \(0.1\) to generate a table of values for the solution for values of \(x\) between 0 and \(1 .\)

Short answer

Generate values using Euler's method from \( x = 0 \) to \( x = 1 \) by calculating successive approximations.

Step-by-step solution

Understand the Problem

We are given a differential equation \( y' = 3x^2 - y^2 + 1 \) with an initial condition \( y(0) = 2 \). We are asked to solve this using Euler's method, which is a numerical method for solving ordinary differential equations with a given initial value.

Review Euler's Method Formula

Euler's method uses the formula: \[ y_{n+1} = y_n + h f(x_n, y_n) \] where \( f(x_n, y_n) \) is the right-hand side of the differential equation and \( h \) is the step size. For this problem, \( f(x, y) = 3x^2 - y^2 + 1 \) and \( h = 0.1 \).

Initialize Values

Start with the initial values from the problem: \( x_0 = 0 \) and \( y_0 = 2 \).

Apply Euler's Method for Each Step

For each step, calculate the next values of \( y \) using Euler's method formula. Start at \( x_0 = 0 \) and move in steps of 0.1 till \( x = 1 \).- **When \( x_0 = 0 \):** Compute \( y_1 = y_0 + 0.1 \times (3 \times 0^2 - 2^2 + 1) = 2 + 0.1 \times (-3) = 1.7 \), thus \( y(0.1) = 1.7 \).- **When \( x_1 = 0.1 \):** Compute \( y_2 = y_1 + 0.1 \times (3 \times 0.1^2 - 1.7^2 + 1) = 1.7 + 0.1 \times (-1.49) = 1.551 \), thus \( y(0.2) = 1.551 \).- **Repeat similarly up to \( x = 1 \).**

Complete the Table of Results

Continue applying Euler's method step by step until you reach \( x = 1 \).\**Results: (approximations)** - \( y(0) = 2 \) - \( y(0.1) = 1.7 \) - \( y(0.2) = 1.551 \) - Continue this process to fill in the table for all values up to \( x = 1 \).

Conclude

The table of values shows the approximate solution at each step from \( x = 0 \) to \( x = 1 \) using a step size of 0.1. This method gives an approximate visualization of how \( y \) changes with respect to \( x \).

Key concepts

Understanding Initial Value Problems

An initial value problem involves solving a differential equation along with a specified value at a given point, called the 'initial condition'. In this exercise, the problem is defined by the ordinary differential equation (ODE) \( y' = 3x^2 - y^2 + 1 \) with an initial condition given as \( y(0) = 2 \). This informs us of both the starting point of the problem and a specific value of \( y \) when \( x \) is zero. The essence is to find how \( y \) changes when \( x \) moves from its initial point, using the information provided by the differential equation.

The Role of Numerical Methods

Numerical methods are techniques used to approximate solutions for problems that may not have an exact analytical solution or where an exact solution is difficult to find. Euler's method is among the simplest numerical methods for approximating solutions to initial value problems. It provides a step-by-step approach to finding the approximate solution at discrete points, using initial conditions and a combination of differential equations. This allows us to visualize the behavior of a function under consideration without resorting to complex mathematics.

Diving into Ordinary Differential Equations

Ordinary differential equations like \( y' = 3x^2 - y^2 + 1 \) describe relationships between a function and its derivatives. An ordinary differential equation may be of any order, but for elementary cases like this one, we deal with first-order ODEs, where the highest derivative is of the first order. Solving ODEs involves finding functions that satisfy the equation - a process that Euler's method helps to approximate through a series of calculated steps. This process is particularly useful when the equation is non-linear or simple algebraic manipulation doesn't suffice.

Significance of Step Size in Euler's Method

The step size \( h \) is a crucial component in Euler's Method, defining the distance between consecutive \( x \)-values at which approximations of \( y \) are made. In this exercise, using a step size of 0.1 means calculating new \( y \) values in increments of 0.1 along the \( x \)-axis. A smaller step size usually leads to more accurate solutions since it accounts for more points along the curve, albeit requiring more computations. Conversely, a larger step size might introduce greater errors, though reducing calculation effort. Therefore, balancing these aspects is key in numerical analysis.