Use Euler's method to calculate the first three approximations to the given initial value proble

Use Euler's method to calculate the first three...

Key Concepts

Accuracy and Error Analysis

Accuracy and error analysis in numerical methods involves comparing the approximated solution to the exact or analytically determined solution of a differential equation. This comparison helps in assessing the efficiency of the numerical method and understanding the accumulated error over successive steps. In Euler's Method, error analysis is essential for evaluating the balance between step size and the precision of the result.

Step Size

The step size, also known as the increment, is a crucial parameter in numerical methods like Euler's Method. It determines the interval between successive points at which the approximate solution is computed. A smaller step size generally leads to a more accurate approximation of the solution, albeit at the cost of increased computational effort, while a larger step size may reduce accuracy but speed up calculations.

Ordinary Differential Equation

An Ordinary Differential Equation (ODE) describes a relationship between a function and its derivatives. It is fundamental in modeling problems in science and engineering where the rate of change of a quantity is linked to the quantity itself or other variables. Techniques such as Euler's Method are used to approximate the solutions of ODEs when an analytical solution is not easily obtainable.

Initial Value Problem

An Initial Value Problem (IVP) involves finding a function that not only satisfies a differential equation but also meets specified initial conditions. These conditions provide the starting point for the solution, and the differential equation describes how the function changes. Solving an IVP usually requires numerical or analytical methods to determine the function over a desired interval.

Euler's Method

Euler's Method is a first-order numerical procedure used to approximate solutions of ordinary differential equations (ODEs) with a given initial value. It involves taking small steps along the curve defined by the ODE and using the derivative (slope) at the current point to estimate the value of the function at the next point. This method provides a simple and intuitive approach to solving problems where an analytical solution might be difficult or impossible to obtain.

Transcript

00:01 For this problem, we're given that y prime is equal to y square times 1 plus 2x.

00:05 We're given initial condition that y of negative 1 is equal to 1, and our dx is equal to 0 .5.

00:11 So this question is really asking us to do two things.

00:13 First, we need to use euler's method to calculate the first three proxaminations using this initial value problem.

00:21 And then next, we want to calculate the exact solution and look at the accuracy of our approximations.

00:28 So first things that we need to recall is what exactly o 'yler's formula is.

00:34 So first thing to find our x -0, x -1, et cetera, our x -not comes from our initial value that we're given.

00:44 So our x -0 is just our negative 1, and 1 is our y -not.

00:50 To find the rest of the x -0s, or the x -ns, sorry, that is just going to come from a formula that xn is equal to xn minus 1 plus dx and then to find the rest of the y ends that is going to be y n minus 1 plus f of xn minus 1 y n minus 1 d x so first i'm just going to calculate our x's so x1 over it's we ready know x not is equal to negative 1 that's what we're given so next x1 is going to equal to negative 1 plus dx, which here is 0 .5.

01:40 So that's going to equal negative 0 .5.

01:47 And then x2 is going to equal x1 plus that plus 0 .5, which is just 0.

01:59 Okay.

02:00 Now we're going to actually find our approximation.

02:01 So first, y1, that's just going to equal to y0 plus f of x0, y, 0.

02:08 So it's just plugging x0 and y0 if x and y respectively so that's going to be y not squared times 1 plus 2 x not d x well again our y0 already given that's just 1 so it would be 1 plus 1 squared x not is negative 1 and d x is equal to 0 .5 and then when you put that in a calculator you should get that y1 is equal to 0 .5.

02:48 And i're going to continue similarly, just now adding one to these.

02:52 So now we're going to have y1 plus y1 squared times 1 plus 2x1 dx.

03:02 So our y1 we just found to be 0 .5.

03:09 Our x1, we found to be negative 0 .5.

03:14 And then our dx is still 0 .5.

03:18 And then when you multiply all of this out, you again actually get 0 .5...

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