Where am I going wrong?Where am I going wrong? Where am I going wrong? inverse sine of the sine MATLAB Answers — New Questions

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i want to use ltpda and then plist. But if i use it as ‘ltpda_startup’ , matlab cant recognize it and after that ‘plist’. In 2016 and 2017 version of matlab it worked. How can i fix it in 2023 version or is there any alternative solutions for it?i want to use ltpda and then plist. But if i use it as ‘ltpda_startup’ , matlab cant recognize it and after that ‘plist’. In 2016 and 2017 version of matlab it worked. How can i fix it in 2023 version or is there any alternative solutions for it? i want to use ltpda and then plist. But if i use it as ‘ltpda_startup’ , matlab cant recognize it and after that ‘plist’. In 2016 and 2017 version of matlab it worked. How can i fix it in 2023 version or is there any alternative solutions for it? image processing, ltpda MATLAB Answers — New Questions

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I have a file of color map with known boundary coordinates and color bar scale either horizontal or vertical. How can we transform the color map into contour map, such that the value of contours expressed the same in the color bar values?I have a file of color map with known boundary coordinates and color bar scale either horizontal or vertical. How can we transform the color map into contour map, such that the value of contours expressed the same in the color bar values? I have a file of color map with known boundary coordinates and color bar scale either horizontal or vertical. How can we transform the color map into contour map, such that the value of contours expressed the same in the color bar values? colormap, contour map MATLAB Answers — New Questions

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excelfilepath= ‘D:mirror1.xlsx’;
data =readtable(excelfilepath);
value=data{1,1};

app.EditField.Value = num2str(value);
end
what is wrong with code
it gives parse erorr
thanksexcelfilepath= ‘D:mirror1.xlsx’;
data =readtable(excelfilepath);
value=data{1,1};

app.EditField.Value = num2str(value);
end
what is wrong with code
it gives parse erorr
thanks excelfilepath= ‘D:mirror1.xlsx’;
data =readtable(excelfilepath);
value=data{1,1};

app.EditField.Value = num2str(value);
end
what is wrong with code
it gives parse erorr
thanks app designer, excel MATLAB Answers — New Questions

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% Code for 3x+1 Problem
seed = 10^309; % Starting number
x = seed; % Initialize x with the seed
trajectory = x; % Store the trajectory
tic
while x ~= 1
if mod(x, 2) == 0 % Check if x is even
x = x / 2;
else % If x is odd
x = 3 * x + 1;
end
trajectory = [trajectory, x]; % Append x to trajectory
end
toc
% Plot the trajectory
figure;
plot(trajectory, ‘-o’, ‘LineWidth’, 1, ‘MarkerSize’, 5,’MarkerFaceColor’, ‘r’);
title(‘Plot of the Collatz Sequence for positive Seed 10^310’);
xlabel(‘Step’);
ylabel(‘Value’);
grid on;% Code for 3x+1 Problem
seed = 10^309; % Starting number
x = seed; % Initialize x with the seed
trajectory = x; % Store the trajectory
tic
while x ~= 1
if mod(x, 2) == 0 % Check if x is even
x = x / 2;
else % If x is odd
x = 3 * x + 1;
end
trajectory = [trajectory, x]; % Append x to trajectory
end
toc
% Plot the trajectory
figure;
plot(trajectory, ‘-o’, ‘LineWidth’, 1, ‘MarkerSize’, 5,’MarkerFaceColor’, ‘r’);
title(‘Plot of the Collatz Sequence for positive Seed 10^310’);
xlabel(‘Step’);
ylabel(‘Value’);
grid on; % Code for 3x+1 Problem
seed = 10^309; % Starting number
x = seed; % Initialize x with the seed
trajectory = x; % Store the trajectory
tic
while x ~= 1
if mod(x, 2) == 0 % Check if x is even
x = x / 2;
else % If x is odd
x = 3 * x + 1;
end
trajectory = [trajectory, x]; % Append x to trajectory
end
toc
% Plot the trajectory
figure;
plot(trajectory, ‘-o’, ‘LineWidth’, 1, ‘MarkerSize’, 5,’MarkerFaceColor’, ‘r’);
title(‘Plot of the Collatz Sequence for positive Seed 10^310’);
xlabel(‘Step’);
ylabel(‘Value’);
grid on; collatz conjecture, 3x +1 problem MATLAB Answers — New Questions

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I am trying to add a bulleted list in a Matlab annotation textbox. I’ve tried using LaTex commands but all I get is: Unable to interpret latex string, such as

str = ‘begin{itemized} item First Line item Second Line end{itemized}’;
figure;
ha = annotation(‘textbox’, [0.5 0.5 0.4 0.2], ‘Interpreter’, ‘latex’);
set(ha, ‘String’, str)

This is similar to the way you put a table in an annotated textbox:

<http://www.mathworks.com/matlabcentral/answers/99523-how-do-i-create-a-latex-table-in-a-matlab-text-box Latex Table>I am trying to add a bulleted list in a Matlab annotation textbox. I’ve tried using LaTex commands but all I get is: Unable to interpret latex string, such as

str = ‘begin{itemized} item First Line item Second Line end{itemized}’;
figure;
ha = annotation(‘textbox’, [0.5 0.5 0.4 0.2], ‘Interpreter’, ‘latex’);
set(ha, ‘String’, str)

This is similar to the way you put a table in an annotated textbox:

<http://www.mathworks.com/matlabcentral/answers/99523-how-do-i-create-a-latex-table-in-a-matlab-text-box Latex Table> I am trying to add a bulleted list in a Matlab annotation textbox. I’ve tried using LaTex commands but all I get is: Unable to interpret latex string, such as

str = ‘begin{itemized} item First Line item Second Line end{itemized}’;
figure;
ha = annotation(‘textbox’, [0.5 0.5 0.4 0.2], ‘Interpreter’, ‘latex’);
set(ha, ‘String’, str)

This is similar to the way you put a table in an annotated textbox:

<http://www.mathworks.com/matlabcentral/answers/99523-how-do-i-create-a-latex-table-in-a-matlab-text-box Latex Table> latex, bulleted list, annotation MATLAB Answers — New Questions

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I was working on my code like 10 days ago in 2025a, the code is running and the plot is showing correctly. I decided to run again to take some screenshot of my plot but my plot is no longer displaying correctly. If I print the plot, I will see the correct plot on the print dialog but the original plot window is showing up very bad. Screenshot applied. Please any immediate help will be appreciated as this is my school assignment.

My computer is Linux Ubuntu 22.04

clear all; close all; clc;

%% System Parameters
% Physical parameters of the DC motor
Ra = 1.0; % Armature resistance (Ohm)
La = 0.5; % Armature inductance (H)
J = 0.01; % Moment of inertia (Nms^2/rad)
b = 0.1; % Damping coefficient (Nms)
KT = 0.01; % Torque constant (Nm/A)
Ke = 0.01; % Back EMF constant (Nm/A)
K = Ke; % In SI units, Ke = Ki = K

%% Transfer Function Definition
% Open loop transfer function: θ(s)/Ea(s)
% G(s) = K / [(Js + b)(Las + Ra) + K^2]

% Coefficients calculation
num = K;
den_coeff = [J*La, (J*Ra + b*La), (b*Ra + K^2)];
den = den_coeff;

% Create transfer function
G = tf(num, den);

% Display the transfer function
fprintf(‘Open Loop Transfer Function:n’);
fprintf(‘G(s) = %.4f / (%.4fs^2 + %.4fs + %.4f)n’, num, den(1), den(2), den(3));

%% Performance Specifications
% Desired specifications:
% – Settling time (Ts) < 2 seconds
% – Percentage Overshoot (PO) < 5%
% – Steady State Error (SSE) < 1%

fprintf(‘nDesired Specifications:n’);
fprintf(‘- Settling time (Ts) < 2 secondsn’);
fprintf(‘- Percentage Overshoot (PO) < 5%%n’);
fprintf(‘- Steady State Error (SSE) < 1%%nn’);

%% Step 1: Open Loop Response
fprintf(‘=== STEP 1: OPEN LOOP RESPONSE ===n’);

figure(1);
[y1, t1] = step(G);
plot(t1, y1, ‘b-‘, ‘LineWidth’, 2);
title(‘Open Loop Step Response’);
xlabel(‘Time (seconds)’);
ylabel(‘Angular Position (rad)’);
grid on;

% Calculate performance metrics for open loop
stepinfo_ol = stepinfo(G);
fprintf(‘Open Loop Performance:n’);
fprintf(‘Rise Time (Tr): %.4f secondsn’, stepinfo_ol.RiseTime);
fprintf(‘Settling Time (Ts): %.4f secondsn’, stepinfo_ol.SettlingTime);
fprintf(‘Peak Overshoot: %.2f%%n’, stepinfo_ol.Overshoot);

% Steady state error calculation (for unit step)
% For Type 0 system: ess = 1/(1+Kp) where Kp = lim s->0 G(s)
Kp_ol = dcgain(G);
ess_ol = 1/(1+Kp_ol) * 100; % Convert to percentage
fprintf(‘Steady State Error: %.2f%%n’, ess_ol);
fprintf(‘Comments: Open loop system is unstable or has poor performancenn’);

Bad Plot

Print DialogI was working on my code like 10 days ago in 2025a, the code is running and the plot is showing correctly. I decided to run again to take some screenshot of my plot but my plot is no longer displaying correctly. If I print the plot, I will see the correct plot on the print dialog but the original plot window is showing up very bad. Screenshot applied. Please any immediate help will be appreciated as this is my school assignment.

My computer is Linux Ubuntu 22.04

clear all; close all; clc;

%% System Parameters
% Physical parameters of the DC motor
Ra = 1.0; % Armature resistance (Ohm)
La = 0.5; % Armature inductance (H)
J = 0.01; % Moment of inertia (Nms^2/rad)
b = 0.1; % Damping coefficient (Nms)
KT = 0.01; % Torque constant (Nm/A)
Ke = 0.01; % Back EMF constant (Nm/A)
K = Ke; % In SI units, Ke = Ki = K

%% Transfer Function Definition
% Open loop transfer function: θ(s)/Ea(s)
% G(s) = K / [(Js + b)(Las + Ra) + K^2]

% Coefficients calculation
num = K;
den_coeff = [J*La, (J*Ra + b*La), (b*Ra + K^2)];
den = den_coeff;

% Create transfer function
G = tf(num, den);

% Display the transfer function
fprintf(‘Open Loop Transfer Function:n’);
fprintf(‘G(s) = %.4f / (%.4fs^2 + %.4fs + %.4f)n’, num, den(1), den(2), den(3));

%% Performance Specifications
% Desired specifications:
% – Settling time (Ts) < 2 seconds
% – Percentage Overshoot (PO) < 5%
% – Steady State Error (SSE) < 1%

fprintf(‘nDesired Specifications:n’);
fprintf(‘- Settling time (Ts) < 2 secondsn’);
fprintf(‘- Percentage Overshoot (PO) < 5%%n’);
fprintf(‘- Steady State Error (SSE) < 1%%nn’);

%% Step 1: Open Loop Response
fprintf(‘=== STEP 1: OPEN LOOP RESPONSE ===n’);

figure(1);
[y1, t1] = step(G);
plot(t1, y1, ‘b-‘, ‘LineWidth’, 2);
title(‘Open Loop Step Response’);
xlabel(‘Time (seconds)’);
ylabel(‘Angular Position (rad)’);
grid on;

% Calculate performance metrics for open loop
stepinfo_ol = stepinfo(G);
fprintf(‘Open Loop Performance:n’);
fprintf(‘Rise Time (Tr): %.4f secondsn’, stepinfo_ol.RiseTime);
fprintf(‘Settling Time (Ts): %.4f secondsn’, stepinfo_ol.SettlingTime);
fprintf(‘Peak Overshoot: %.2f%%n’, stepinfo_ol.Overshoot);

% Steady state error calculation (for unit step)
% For Type 0 system: ess = 1/(1+Kp) where Kp = lim s->0 G(s)
Kp_ol = dcgain(G);
ess_ol = 1/(1+Kp_ol) * 100; % Convert to percentage
fprintf(‘Steady State Error: %.2f%%n’, ess_ol);
fprintf(‘Comments: Open loop system is unstable or has poor performancenn’);

Bad Plot

Print Dialog I was working on my code like 10 days ago in 2025a, the code is running and the plot is showing correctly. I decided to run again to take some screenshot of my plot but my plot is no longer displaying correctly. If I print the plot, I will see the correct plot on the print dialog but the original plot window is showing up very bad. Screenshot applied. Please any immediate help will be appreciated as this is my school assignment.

My computer is Linux Ubuntu 22.04

clear all; close all; clc;

%% System Parameters
% Physical parameters of the DC motor
Ra = 1.0; % Armature resistance (Ohm)
La = 0.5; % Armature inductance (H)
J = 0.01; % Moment of inertia (Nms^2/rad)
b = 0.1; % Damping coefficient (Nms)
KT = 0.01; % Torque constant (Nm/A)
Ke = 0.01; % Back EMF constant (Nm/A)
K = Ke; % In SI units, Ke = Ki = K

%% Transfer Function Definition
% Open loop transfer function: θ(s)/Ea(s)
% G(s) = K / [(Js + b)(Las + Ra) + K^2]

% Coefficients calculation
num = K;
den_coeff = [J*La, (J*Ra + b*La), (b*Ra + K^2)];
den = den_coeff;

% Create transfer function
G = tf(num, den);

% Display the transfer function
fprintf(‘Open Loop Transfer Function:n’);
fprintf(‘G(s) = %.4f / (%.4fs^2 + %.4fs + %.4f)n’, num, den(1), den(2), den(3));

%% Performance Specifications
% Desired specifications:
% – Settling time (Ts) < 2 seconds
% – Percentage Overshoot (PO) < 5%
% – Steady State Error (SSE) < 1%

fprintf(‘nDesired Specifications:n’);
fprintf(‘- Settling time (Ts) < 2 secondsn’);
fprintf(‘- Percentage Overshoot (PO) < 5%%n’);
fprintf(‘- Steady State Error (SSE) < 1%%nn’);

%% Step 1: Open Loop Response
fprintf(‘=== STEP 1: OPEN LOOP RESPONSE ===n’);

figure(1);
[y1, t1] = step(G);
plot(t1, y1, ‘b-‘, ‘LineWidth’, 2);
title(‘Open Loop Step Response’);
xlabel(‘Time (seconds)’);
ylabel(‘Angular Position (rad)’);
grid on;

% Calculate performance metrics for open loop
stepinfo_ol = stepinfo(G);
fprintf(‘Open Loop Performance:n’);
fprintf(‘Rise Time (Tr): %.4f secondsn’, stepinfo_ol.RiseTime);
fprintf(‘Settling Time (Ts): %.4f secondsn’, stepinfo_ol.SettlingTime);
fprintf(‘Peak Overshoot: %.2f%%n’, stepinfo_ol.Overshoot);

% Steady state error calculation (for unit step)
% For Type 0 system: ess = 1/(1+Kp) where Kp = lim s->0 G(s)
Kp_ol = dcgain(G);
ess_ol = 1/(1+Kp_ol) * 100; % Convert to percentage
fprintf(‘Steady State Error: %.2f%%n’, ess_ol);
fprintf(‘Comments: Open loop system is unstable or has poor performancenn’);

Bad Plot

Print Dialog matlab, plotting MATLAB Answers — New Questions

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In absence of a fluid level gauge (it was found in earlier versions of simulink real time instrumentation panels such as in 2018b), how can I modify the linear gauge used in app designer to mimic the functionality of a fluid level gauge? That type of gauge was very useful and for some reason isn’t part of the new instrumenation options.In absence of a fluid level gauge (it was found in earlier versions of simulink real time instrumentation panels such as in 2018b), how can I modify the linear gauge used in app designer to mimic the functionality of a fluid level gauge? That type of gauge was very useful and for some reason isn’t part of the new instrumenation options. In absence of a fluid level gauge (it was found in earlier versions of simulink real time instrumentation panels such as in 2018b), how can I modify the linear gauge used in app designer to mimic the functionality of a fluid level gauge? That type of gauge was very useful and for some reason isn’t part of the new instrumenation options. fluid level gauge, instrumenation MATLAB Answers — New Questions

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The plot of the rather complicated function B(t,d) seems correct (see figure 1).
The plot of its definite integral from 0 to t (in the case d=0) is incorrect (see figure 2). It should have constant value 1 on the interval [1,2], but the plot indicates a constant value of 0 on that interval.
What is going on? Did I misuse the symbolic math functions?
PS Both plots look correct for integers d>1.

syms t k d;
cutoff(t) = piecewise(t < 0, 0, t >= 0, t);
parity(k) = 1-2*mod(k,2);
f(t,k,d) = 1/factorial(d)*parity(k)*nchoosek(d+1, k)*cutoff(t-k)^d;
B(t,d) = symsum(f,k,0,floor(t));
IB(t,d) = int(B(t,d),0,t);

bad_d=0;

figure(1)
fplot(B(t,bad_d), [-1,2], ‘LineWidth’, 2);

figure(2)
fplot(IB(t,bad_d), [-1,2], ‘LineWidth’, 2);The plot of the rather complicated function B(t,d) seems correct (see figure 1).
The plot of its definite integral from 0 to t (in the case d=0) is incorrect (see figure 2). It should have constant value 1 on the interval [1,2], but the plot indicates a constant value of 0 on that interval.
What is going on? Did I misuse the symbolic math functions?
PS Both plots look correct for integers d>1.

syms t k d;
cutoff(t) = piecewise(t < 0, 0, t >= 0, t);
parity(k) = 1-2*mod(k,2);
f(t,k,d) = 1/factorial(d)*parity(k)*nchoosek(d+1, k)*cutoff(t-k)^d;
B(t,d) = symsum(f,k,0,floor(t));
IB(t,d) = int(B(t,d),0,t);

bad_d=0;

figure(1)
fplot(B(t,bad_d), [-1,2], ‘LineWidth’, 2);

figure(2)
fplot(IB(t,bad_d), [-1,2], ‘LineWidth’, 2); The plot of the rather complicated function B(t,d) seems correct (see figure 1).
The plot of its definite integral from 0 to t (in the case d=0) is incorrect (see figure 2). It should have constant value 1 on the interval [1,2], but the plot indicates a constant value of 0 on that interval.
What is going on? Did I misuse the symbolic math functions?
PS Both plots look correct for integers d>1.

syms t k d;
cutoff(t) = piecewise(t < 0, 0, t >= 0, t);
parity(k) = 1-2*mod(k,2);
f(t,k,d) = 1/factorial(d)*parity(k)*nchoosek(d+1, k)*cutoff(t-k)^d;
B(t,d) = symsum(f,k,0,floor(t));
IB(t,d) = int(B(t,d),0,t);

bad_d=0;

figure(1)
fplot(B(t,bad_d), [-1,2], ‘LineWidth’, 2);

figure(2)
fplot(IB(t,bad_d), [-1,2], ‘LineWidth’, 2); symbolic math, definite integral, plot MATLAB Answers — New Questions

​

Hi all,
I want to test the setup to use kron to solve the 2D poisson equation with dirichlet boundary condition on two sides (left and bottom) and non-zero neumann boundary conditions on the other sides. I know the theoretical solution which is and thus the forcing function is . I have the code herein:
%grid declaration
N = 100;
x = linspace(0,1,N);
dx = x(2)-x(1);
y = x;
[X,Y] = meshgrid(x,y);

%analytical solution and forcing function
u_an = 8*X.^2.*Y.^2;
f = 16*(X.^2+Y.^2);

%creation of the matrix (complete matrix including end points)
e = ones(N,1);
T = spdiags([e -2*e e],[-1 0 1],N,N);
T(1,1:2) = [dx^2 0]; %dirichlet matrix (first order approximation)
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
fullT = full(T);
I = speye(N);
L = kron(T,I)+kron(I,T);
L = L/dx^2;
fullL = full(L);

%rhs including boundary conditions and forcing function
b = zeros(N,N);
b(2:end-1,2:end-1) = f(2:end-1,2:end-1);
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary

%solution of the matrix equation and reshaping
b1 = reshape(b,N^2,1);
u = Lb1;
u = u;
u = reshape(u,N,N);

%visualisation and comparison with the analytical solution
figure(1)
plot3(X,Y,u,’*k’)
hold on
plot3(X,y,u_an,’or’)
hold off
The code isn’t quite accurate but what is bizzare is that if you replace
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary
with
T(end,end-1:end) = [1 -1]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2/dx; %right boundary
b(end,:) = -16*x.^2/dx; %top boundary
it generates the correct result. Could someone please help me understand if I’m setting up the Kronecker product correctly or not and what changes in the two implementations?Hi all,
I want to test the setup to use kron to solve the 2D poisson equation with dirichlet boundary condition on two sides (left and bottom) and non-zero neumann boundary conditions on the other sides. I know the theoretical solution which is and thus the forcing function is . I have the code herein:
%grid declaration
N = 100;
x = linspace(0,1,N);
dx = x(2)-x(1);
y = x;
[X,Y] = meshgrid(x,y);

%analytical solution and forcing function
u_an = 8*X.^2.*Y.^2;
f = 16*(X.^2+Y.^2);

%creation of the matrix (complete matrix including end points)
e = ones(N,1);
T = spdiags([e -2*e e],[-1 0 1],N,N);
T(1,1:2) = [dx^2 0]; %dirichlet matrix (first order approximation)
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
fullT = full(T);
I = speye(N);
L = kron(T,I)+kron(I,T);
L = L/dx^2;
fullL = full(L);

%rhs including boundary conditions and forcing function
b = zeros(N,N);
b(2:end-1,2:end-1) = f(2:end-1,2:end-1);
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary

%solution of the matrix equation and reshaping
b1 = reshape(b,N^2,1);
u = Lb1;
u = u;
u = reshape(u,N,N);

%visualisation and comparison with the analytical solution
figure(1)
plot3(X,Y,u,’*k’)
hold on
plot3(X,y,u_an,’or’)
hold off
The code isn’t quite accurate but what is bizzare is that if you replace
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary
with
T(end,end-1:end) = [1 -1]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2/dx; %right boundary
b(end,:) = -16*x.^2/dx; %top boundary
it generates the correct result. Could someone please help me understand if I’m setting up the Kronecker product correctly or not and what changes in the two implementations? Hi all,
I want to test the setup to use kron to solve the 2D poisson equation with dirichlet boundary condition on two sides (left and bottom) and non-zero neumann boundary conditions on the other sides. I know the theoretical solution which is and thus the forcing function is . I have the code herein:
%grid declaration
N = 100;
x = linspace(0,1,N);
dx = x(2)-x(1);
y = x;
[X,Y] = meshgrid(x,y);

%analytical solution and forcing function
u_an = 8*X.^2.*Y.^2;
f = 16*(X.^2+Y.^2);

%creation of the matrix (complete matrix including end points)
e = ones(N,1);
T = spdiags([e -2*e e],[-1 0 1],N,N);
T(1,1:2) = [dx^2 0]; %dirichlet matrix (first order approximation)
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
fullT = full(T);
I = speye(N);
L = kron(T,I)+kron(I,T);
L = L/dx^2;
fullL = full(L);

%rhs including boundary conditions and forcing function
b = zeros(N,N);
b(2:end-1,2:end-1) = f(2:end-1,2:end-1);
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary

%solution of the matrix equation and reshaping
b1 = reshape(b,N^2,1);
u = Lb1;
u = u;
u = reshape(u,N,N);

%visualisation and comparison with the analytical solution
figure(1)
plot3(X,Y,u,’*k’)
hold on
plot3(X,y,u_an,’or’)
hold off
The code isn’t quite accurate but what is bizzare is that if you replace
T(end,end-1:end) = [dx -dx]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2; %right boundary
b(end,:) = -16*x.^2; %top boundary
with
T(end,end-1:end) = [1 -1]; %neumann matrix (first order approximation)
and
b(:,end) = -16*y’.^2/dx; %right boundary
b(end,:) = -16*x.^2/dx; %top boundary
it generates the correct result. Could someone please help me understand if I’m setting up the Kronecker product correctly or not and what changes in the two implementations? kron, finite difference, mldivide MATLAB Answers — New Questions

​