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

Key Concepts

Euler's Method

Euler's method is a basic numerical procedure used to approximate solutions to ordinary differential equations. It involves starting from an initial value and then taking small steps along the x-axis, updating the solution based on the value of the derivative. This method is conceptually straightforward and serves as an introduction to more complex numerical techniques for differential equations.

Initial Value Problems

An initial value problem (IVP) involves finding a function that satisfies a given differential equation and meets a specified value at an initial point. The problem typically provides both the differential equation and an initial condition. IVPs are fundamental in modeling dynamic systems where the state of the system at a particular point is known and future states are determined by the evolution rule given by the differential equation.

Step Size

The step size, often denoted as h or dx, is the interval used in numerical methods like Euler's method to increment the independent variable. It directly affects the approximation's accuracy; smaller step sizes generally yield better approximations but require more computational effort. The choice of step size is crucial as it influences both the local truncation error at each step and the overall global error of the approximation.

Exact Solution

The exact solution of a differential equation refers to a closed-form or analytic expression that satisfies both the differential equation and the given initial condition. Obtaining the exact solution, when possible, allows for a benchmark to assess and compare the accuracy of numerical approximations made using methods like Euler's method.

Accuracy and Error Analysis

Accuracy in numerical methods is measured by how closely the computed approximations match the exact solution. Error analysis involves studying both local truncation error (the error made in a single step due to linear approximation) and global error (the accumulation of these local errors over many steps). Investigating the accuracy provides insights into the reliability of the numerical method being used and helps in fine-tuning parameters like step size for better results.

Transcript

00:01 So for this question, we are given that y prime is equal to x times 1 minus y, y of 1 is equal to 0, and dx is equal to 0 .2.

00:11 So for this question, we're asked to do basically two different things.

00:14 First, we want to use euler's method to calculate the first three approximations to the given initial value problem.

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

00:31 So first thing to remember here is from this section, i recall, is euler's method.

00:42 So first thing that we need to remember is that our xn, that's just going to equal to xn plus dx, and then our yn, the thing that we're finding, is going to equal to yn minus 1 plus dx, and then our yn, the thing that we're finding, is going to equal to yn minus 1, plus f of xn minus 1 y n minus 1 times d x.

01:14 So for our steps here, we're asked to find the first three approximation.

01:17 So we're going to find y1 through 3 using these two formulas.

01:23 And then we'll find the actual solution after that.

01:28 So first things, i'm going to find my x, not, x1, and x2.

01:35 So first, starting with my x -not, that's something that we're already given at first.

01:41 So my x -not here is just in my initial value, and so then therefore that's going to be my y -not.

01:48 So my x -not is just equal to 1.

01:52 My x -1 is going to equal to my x -not plus dx.

01:57 Here we know dx is 0 .2, so it's going to be 1 plus 0 .2, it's just equal to 1 .2.

02:04 And then so when i'm my x2 is just going to equal my x1 plus dx which is 1 .4 so now that we have that we are ready to find our y1 through 3 so starting with y1 well i know that that's just going to equal to y0 plus the function evaluate x0 and y0 so i want to have y0 plus my function here is x times 1 minus y so that this is going to become x0 times 1 minus y0.

02:43 And then the last thing we need is this dx.

02:49 And so plugging in my values that i know, i know again from my initial value that my y0 is just 0.

02:59 My x not is 1 and 1 minus 0.

03:04 And then my dx again is just 0 .2.

03:09 So when i multiply all that together and add, i get that this first one is just point.

03:18 And i'm going to keep on going...

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