Estimate the following solutions using Euler's method with...

Estimate the following solutions using Euler's method with $n=5$ steps over the interval $t=[0,1] .$ If you are able to solve the initial-value problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler’s method. How accurate is Euler’s method? $$ y^{\prime}=y^{2} \ln (x+1), y(0)=1 . \quad \text { Exact solution is } y=-\frac{1}{(x+1)(\ln (x+1)-1)}$$

Estimate the following solutions using Euler's method with $n=5$ steps over the interval $t=[0,1] .$ If you are able to solve the initial-value problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler’s method. How accurate is Euler’s method?
$$ y^{\prime}=y^{2} \ln (x+1), y(0)=1 . \quad \text { Exact solution is } y=-\frac{1}{(x+1)(\ln (x+1)-1)}$$

Key Concepts

Step Size

The step size in a numerical method represents the interval between points in the discretization of the domain over which the method approximates the solution. Its size directly influences the approximation's accuracy; smaller steps typically result in more accurate estimates but require more computational effort.

Exact Solution

An exact solution to an ODE is an analytical expression that satisfies both the differential equation and the initial condition. When available, it serves as a benchmark for evaluating the accuracy of numerical methods, such as Euler’s method.

Error Analysis

This concept involves examining the difference between the numerical approximation and the exact solution. Understanding sources of errors, such as local truncation error and accumulated global error, is crucial for evaluating and improving the efficiency and reliability of numerical methods like Euler's method.

Initial Value Problem (IVP)

An IVP involves finding a function that satisfies a differential equation and meets a specified initial condition. This setup is essential in many applications as it uniquely determines the solution by providing a starting point from which the behavior of the system is developed.

Euler's Method

A numerical technique used to approximate solutions to ordinary differential equations (ODEs) with a given initial value. By iterating small steps from the start point, the method uses the derivative at the current point to estimate the value at the next point, providing an approach to solve problems where an analytic solution is difficult to obtain.

Ordinary Differential Equations (ODEs)

These equations involve functions of a single variable and their derivatives. Solving ODEs is a central topic in calculus and applied mathematics, and they model a broad range of phenomena in science and engineering.

Transcript

00:01 For this problem, i set up this table for calculating different values using oilers method.

00:06 And we know that we're given n is equal to 5 because we want to have five sub intervals.

00:12 And we know that h is going to be equal to our range, which goes from 0 to 1.

00:17 So that'll be 1 over n, which is 5.

00:22 So that's going to be 0 .2.

00:26 And from here we can go ahead and first calculate each of our x of n values.

00:31 So since our step size is 0 .2, that just means that with each increase in n by 1, we're going to increase x of n by 0 .2.

00:42 So what i mean by that is our x of ns are going to be 0 .2, 0 .4, 0 .6, 0 .8, and 1.

00:52 And next i'm going to start by calculating each of my f of x of n minus or not each of them but my first f of x of n minus 1 y of n minus 1 so to calculate for n is equal to 1 we're going to find x of 1 minus 1 so 0 y sub 0 and essentially all we have to do is plug this into our differential equation.

01:21 So that's going to tell us that y prime is equal to y squared, so that'll be 0 squared, or excuse me, 1 squared, times the natural log, i won't write that in preface, times the natural log of x plus 1, so that'll be 0 plus 1.

01:45 And this is going to end up being equal to 0.

01:49 So i'll fill that in right here.

01:51 And from here, we can go ahead and use our formula that we have from oilers ' theorem to find what y sub n is.

01:59 Y sub n is equal to y sub n minus 1 plus h times the f of x sub n minus 1, y sub n minus 1 that we found.

02:10 So this is going to be equal to y sub 0, which we said was 1, plus h, which is 0, which is 0, so that'll be equal to 1.

02:21 And i'll go ahead and do our next step in blue...

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