The linear finite difference method to approximate the...

Name: the linear finite difference method to approximate the solution to the following boundary value problems is as follows a y 3y 2y2x3 0x1 y02 y11 use h01 55124
Uploaded: 2022-06-30T23:38:14-08:00
Duration: 19 min 20 s
Channel: Oswaldo Jiménez
Description: the linear finite difference method to approximate the solution to the following boundary value problems is as follows a y 3y 2y2x3 0x1 y02 y11 use h01 55124

The linear Finite-Difference method to approximate the solution to the following boundary-value problems is as follows: (a). y′′ =−3y′ +2y+2x+3, 0<=x<=1, y(0)=2, y(1)=1; use h=0.1.

Transcript

00:01 Consider the following system of linear equations, given here this picture.

00:09 We want to rewrite the above system of linear equations in the matrix form.

00:15 That's part a.

00:17 In part b, we want to use the initial vector x0 equal, a vector with eight components all equal to 0 .1, and apply the jacoby method with two times of the above linear system to 10.

00:33 Approximate value.

00:35 So we obtain an approximate solution to the linear system, and we're going to apply the method twice.

00:42 That is two iterations.

00:47 And first we write the system in a matrix form that is a x equal b.

01:01 A is a coefficient matrix.

01:03 X is the unknown vector which is composed of eight unknown variables, x1, x2, x3, x4, x6, x7 and x8, and right -hand side vector b, which is composed of the independent terms written to the right of the equations.

01:31 And so, matrix a is, okay, for this again, so we do something like this, because we have eight rows and eight columns for a.

02:02 So we see here the first row.

02:07 You have one thing that is good from this way of writing the linear systems is to have the variable order and tabulated in a very good range so that we can see easily what the coefficients are.

02:29 For example, in the first row, 4 is a coefficient of x1, 1 is the coefficient as x2, negative 1, the coefficient of x3 and all other coefficients are zero because we don't see the variables over here in the first equation.

02:46 So we have 4 -1 -901 and 5 zeros in the first row.

02:53 So it's 4 -193, 4 -1 -191, sorry, 1 -191.

03:21 Then in the second and then 5 zeros.

03:24 Three four six seven and eight okay the second rule will be here so we have in order coefficients are one six negative two one negative one and three zeros so we get one six negative two one three zeros that's correct now the third equation over here we see that x1 is not there, so its coefficient is 0.

04:17 So we start with 0 for x1.

04:20 Then we have 1, 5.

04:27 X4 is a 0.

04:31 Then negative 1, 1.

04:37 And to 0s.

04:40 Now we go to the fourth row here.

04:46 Again, we have 0 for x1, and 2 for x2...