Question

I know the solution is 0.2, but it says incorrect for my quiz. I think there is a problem when writing log(x). Can someone help me?

The code provided solves the boundary value problem dạy %= r- cos(x), y(1) = 1, y(5) = 2 , on the interval 1<x<5 using a d.x2 centred approximation of the derivative term and N=100 nodes. 1 4 x Matlab code for the solution of Module 2 xleft = 1; xright = 5; N = 100; x = linspace(xleft,xright,N); x = x'; $this just turns x into a column vector dx = (xright-xleft)/(N-1); %If theres N nodes, theres N-1 separations. yleft = 1; yright = 2; 8 9 11 12 M here is the matrix which when multiplied by the vector y approximates y''(x) numerically for all the INTERNAL nodes. M = (diag(-2*ones (N,1),0) + diag(ones (N-1,1),-1) + diag(ones (N-1,1),1))/dx^2; 14 15 %updating the matrix to treat the boundaries differently. M(1,:) = (1, zeros (1,N-1)]; M(end, :) = (zeros(1,N-1), 1]; 16 17 18 19 20 21 %The RHS vector (dont forget to sort out the boundaries separately). b = x.^2.*cos(x); b(1) = yleft; blend) = yright; 23 24 solve y = M\b; Modify this code to solve the following problem and answer the associated question. dºg_log(x) Y'(0.5) = 0.2, y(1.5) = () on 0.5<x<1.5 with 20 d.x2 х nodes. What is the maximum of the computed y vector on this domain, accurate to 3 decimal places (do not round)? Hint: First plot your solution to see if the boundaries look right. Answer: 1 Check

Answer 1

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Note:

Used and modified the code provided in the question.

Code to copy:

   % change start and end points for x
   xleft = 0.5;
   xright = 1.5;
   % Number of nodes = 20
   N = 20;
   % get 20 equally spaced values for x
   x = linspace(xleft, xright, N);
   x = x'; % this just turns x into a column vector
   % size of steps
   dx = (xright-xleft)/(N-1);
   % start and end for y
   yleft = 0.2;
   yright = 0;
   % M M (remains same)
   M = (diag(-2*ones(N, 1),0) + diag(ones(N-1, 1), -1) + diag(ones(N-1, 1), 1))/dx^2;
   % update M to treat boundaries separately
   M(1,:) = [1, zeros(1, N-1)];
   M(end,:) = [zeros(1, N-1), 1];
   % the RHS vector
   b = x.*sin(x);
   b(1) = yleft;b(end) = yright;
   % solve
   y = M\b;
   % minimum value for y
   maxy = max(y);
   % print the value upto 3 decimal places
   fprintf('Maximum of computed y vector = %.3f', maxy);

Code Screenshot:

сл во оноо ооо л во он $ change start and end points for x xleft = 0.5; xright = 1.5; & Number of nodes = 20 N = 20; § get 20 equally spaced values for x x = linspace (xleft, xright, N); x = x'; & this just turns x into a column vector f size of steps dx = (xright-xleft) / (N-1); * start and end for y yleft = 0.2; yright = 0; $ MM (remains same) M = (diag (-2*ones (N, 1),0) + diag (ones (N-1, 1), -1) + diag (ones (N-1, 1), 1))/dx^2; & update M to treat boundaries separately M(1,:) = [1, zeros (1, N-1)]; M(end, :) = [zeros (1, N-1), 1]; & the RHS vector b = x.*sin(x); b(1) = yleft;b (end) = yright; $ solve y = M\b; & minimum value for y maxy = max(y); & print the value upto 3 decimal places fprintf ("Maximum of computed y vector = $.3', maxy);

Output Screenshot:

Maximum of computed y vector = 0.200