Use the Centered in Space scheme and approximate the...

Use the Centered in Space scheme and approximate the solution to the partial differential equation: $u_{xx} + u_{yy} = 0$, $0 \le x \le 2$, $0 \le y \le 1$ subject to the boundary conditions: $u(0, y) = 1 - y$, $u(2, y) = 3y$, $0 \le y \le 1$ and $u(x, 0) = x$, $u(x, 1) = x^2$, $0 \le x \le 2$ Page 1 of 2 1. Use $h = k = \frac{1}{2}$ and set up the matrix system to be solved.

Transcript

00:01 Hello everyone we are going to solve a question and this question we are given boundary value problem boundary value problem the problem is given us d square y divided by d x square plus dy divided by d x plus x multiplied with y is zero and here we are given conditions y -dice 0 is equals to 1.

00:27 Y -dash 2 is given as y at 2.

00:31 So from here we let that let d square y divided by d x squared is represented by y double derivative and d -y divided by d x is represented as y derivative y -dash.

00:46 So from here we put this values in the equation.

00:50 So our equation becomes our equation becomes our equation becomes.

00:56 Y double derivative plus y dash plus xy is equals to 0 and here we have 4 y -dash is equal to 1 and y -dash 2 is equals to y of 2 so by using the central difference method here we have by using central difference method h is equals to 1 so from here we have the system of linear equations for the computation of values for y equals to 0 here we have to find values of y at 0 y at 1 and y at 2 so from here we solve for x equals to 1 firstly we are solving we are solving 4 x equals to 1 so by putting x equals to 1 here we have double derivative at 1 plus y single derivative at 1 plus x multiplied with y at 1 equals to 0 so from here we say that this is the value as y 2 dash minus y 1 .1 dash divided by 1 plus y 1 plus x here we have y 1 equals to 0 so from here we have y 2 dash minus y 1 dash plus y 1 plus x y1 equals to 0 and this get cancelled so that means where x equals to 1 here we have x equals to 1 so by putting the x equals to 1 y2 dash plus y 1 is equal to 0 so this is our equation number 1 this is our equation number 1 now we are solving for x equals to 1 here we are again solving for x equals to 1 so from here we can say that we are solving for x value equals to 1 so here we have y 1 double derivative plus y 1 derivative plus y 1 derivative plus y equals to 0 so from here we can say that y 2 derivative minus y 0 derivative divided by 2 plus y 1 derivative plus y 1 equals to 0 here by this we have by taking lcm we have y 2 derivative minus y 0 derivative plus 2 times y 1 derivative plus 2 y 1 equals to 0 so from here we have that y 2 derivative is equal to y 2 and y 0 0 .0 derivative by putting the values we have this is y 2 minus 1 plus twice of y 1 derivative and here we have plus twice of y 1 equals to 0 now putting the values here we have y2 plus twice of y1 derivative or here we have plus by putting the value of y 1 derivative here we have twice of y 2 minus y not divided by 2 and minus 1 equals to 0 so from here we have a equation that twice of y 2 plus twice of y 1 minus y 0 equals to 1 this is our equation 2.

04:30 Now we solve these equation for x equals to 0.

04:34 So here we have solving for x equals to 0...

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