Is there a one-line code for this?
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% I would like to go from A = [1 1;2 2;3 3;4 4;5 5]
% to A = [1 1;2 2;3 3;4 4;5 5;1 2;2 3;3 4;4 5;5 6];
A = [1 1;2 2;3 3;4 4;5 5];
% Multi-line approach
B = A;
B(:,2) = B(:,2) + 1;
A = [A;B];
clear B
% Is there a way to do this in one line of code? I have to do similar operations multiple times,
% and would like to know if there is a way to do so. Otherwise, I will settle for a function.
% Note that the actual matrices are 1,000,000+ x 30 in size.
% Thank you
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% Define the differential equation dy/dx = f(x, y)
% Example: dy/dx = (y^2-x^2)/(y^2+x^2) f = @(x, y) x + y;
% Input initial and final conditions
x0 = input('Enter the initial value of x0: ');
y0 = input('Enter the initial value of y0: ');
xg = input('Enter the final value of xg: ');
h = input('Enter the value of step size h: ');
% Number of steps
n = (xg - x0) / h;
% Runge-Kutta 2nd order iteration
for i = 1:n
k1 = h * f(x0, y0);
k2 = h * f(x0 + h, y0 + k1);
k = (k1 + k2) / 2;
yg = y0 + k; % Update y
x0 = x0 + h; % Update x y0=yg;
end
% Display final result
fprintf('The final value of y at x = %f is y = %f\n', x0, y0);
Is there a way to do entire code in one line
2 个评论
Stephen23
2025-11-20
"Is there a way to do entire code in one line"
x0 = input('Enter the initial value of x0: '); y0 = input('Enter the initial value of y0: '); xg = input('Enter the final value of xg: '); h = input('Enter the value of step size h: '); n = (xg - x0) / h; for i = 1:n; k1 = h * f(x0, y0); k2 = h * f(x0 + h, y0 + k1); k = (k1 + k2) / 2; yg = y0 + k; x0 = x0 + h; y0=yg; end; fprintf('The final value of y at x = %f is y = %f\n', x0, y0);
Dyuman Joshi
2025-11-20
Lol.
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