Use the Euler method with d x=0.2 to estimate y(2) if y^'=y / x and y(1)=2 . What is the exact v

Use the Euler method with d x=0.2 to estimate y(2) if y^'=y...

Key Concepts

Euler Method

The Euler method is a simple numerical technique used to approximate solutions of ordinary differential equations (ODEs) with a given initial value. It involves using a fixed step size to progress along the curve by updating the solution using the slope provided by the differential equation at current points, thereby constructing an approximate solution iteratively.

Ordinary Differential Equation (ODE) and Initial Value Problem

An ordinary differential equation (ODE) describes a function’s rate of change with respect to one independent variable. When paired with an initial condition, it forms an initial value problem (IVP), which seeks a solution curve that not only satisfies the differential equation but also passes through a specified point, ensuring the solution is uniquely determined over its domain.

Step Size in Numerical Methods

The step size in numerical methods, such as the Euler method, is the increment by which the independent variable is increased during each iteration. A smaller step size generally leads to a more accurate approximation but requires more computational work, whereas a larger step size might reduce computation but can lead to significant errors in the estimated solution.

Exact Solution using Separation of Variables

The exact solution of certain ODEs can be obtained by the method of separation of variables, where the variables are rearranged so that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the other side. This approach allows for the integration of both sides independently, thereby yielding an analytic expression for the solution that satisfies the initial condition.

Transcript

00:01 So first question it tells us to use the oiler method with dx is equal to 0 .2 to estimate y of 2 if y prime is equal to y over x and y of 1 is equal to 2.

00:12 So that's the first part and the second part is asked what is the exact value of y of 2.

00:18 So there's going to have two parts really says problem.

00:21 So first i'm going to look at what i'm given.

00:22 So i'm given that y prime is equal to y over x.

00:29 I also know that dx is equal to 0 .2 and given the initial condition that y of 1 is equal to 2 and what we want to find is what y of 2 is so we're going to use the oilers method for this so we're starting at y of 1 so x equals 1 we want to get to x equals 2 so first just to recall what the oil and method is the oil method we know that our solution y n that's just going to equal to yn minus 1 plus the function evaluated at xn minus 1, yn minus 1, times delta x.

01:17 And then we also know that xn, well, that's just going to equal xn minus 1 plus dx.

01:30 So we are starting at y of 1 is equal 2.

01:35 So that 1 is our x not, the 2 is our y -not.

01:39 I know for my x -n i'm adding dx to it every single time so i just need to go through my y -ends until i am one step before this two so before my x -ends using this equation now the x of zero first is just equal to one x1 is going to equal to one plus and x -1 is going to equal to one plus delth dx, which is 0 .2.

02:19 So i'm going to 1 .2.

02:22 And i'm just adding points each time.

02:24 So x2 is equal to 1 .4, 1 .6, and then i'm stopping at 1 .8.

02:35 And then now i'm going to find my y's.

02:38 And that last one's going to be my approximate value for y of 2.

02:45 So first, y1, that's just going to equal to y not, plus the function evaluated.

02:53 At y0 x not so y not over x not since our function is y over x times d x so my y not here we're given is two plus we're going to have y0 2 over x not which we're given as 1 times d x which we're given as 0 .2 and so when you evaluate that you should get 2 .4 and so i'm going to do that each time so y2, that's going to be my, now my y1, which is 2 .4, plus y1 over x1, which is 1 .2, times my dx .2.

03:46 And for that, you should get 2 .8, and keep going in this pattern.

03:53 So now y3 is going to be y2, 2 .8 plus 2 .8 over x2, which is 1 .4 times 0 .2, which is 3 .2, which is 3 .2.

04:08 Y4 is equal to y3, which is 3 .2, plus 3 .2 over x3, which is 1 .6, times 0 .2, which you should get 3 .6 for that.

04:25 And then y5 is equal to y4, 3 .6, plus 3 .6 over x4, which is 1 .8, times 0 .2, which is equal to 4.

04:41 So that answers our first part of the question.

04:44 This is what estimates y of 2...

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