System of 2nd order ODE with Euler.
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Hi,
I had a system of 2 2nd order ODE.
I got to this point :

I need to find the approximate solutions of y2(t).
M1, M2, G, L1,L2 are variables given by the user.

These are the initials conditions which are given by the user also(i guess ?)
Im a bit lost in what should i do. I know how euler works but not with this type of system.
thanks !
4 件のコメント
採用された回答
Jan
2021 年 5 月 27 日
編集済み: Jan
2021 年 5 月 27 日
You got it almost. I've fixed a typo and expanded the Euler method to collect the output as matrix.
% function main
xinit = 0;
xfinal = 3;
h = 0.05;
y0 = [1, 0, 0, 0]; % As many elements as the system has
[x, y] = euler_explicit(@fnc, xinit, xfinal, h, y0);
plot(x, y);
% end
function [x, y] = euler_explicit(f, xinit, xfinal, h, y0)
x = xinit : h : xfinal;
n = length(x);
y = zeros(n, numel(y0));
y(1, :) = y0;
for k = 1:n - 1
y(k + 1, :) = y(k, :) + h * f(x(k), y(k, :));
end
end
function dy = fnc(t,Y)
L1 = 1;
L2 = 2;
M1 = 2;
M2 = 3;
g = 1;
K = 1 / (L1 * L2 * (M1 + M2*sin(Y(1) - Y(2)).^2));
% ^ was missing
Y4 = K*((M1+M2)*g*L1*sin(Y(1))*cos(Y(1)-Y(2)) - (M1+M2)*g*L1*sin(Y(2)) + (M1+M2)*L1^2*sin(Y(1) - Y(2))*Y(3)^2 + M2*L1*L2*sin(Y(1)-Y(2))*cos(Y(1)-Y(2))*Y(4)^2);
Y3 = K*(-(M1+M2)*g*L2*sin(Y(1)) + M2*g*L2*sin(Y(2))*cos(Y(1)-Y(2)) - M2*L1*L2*sin(Y(1) - Y(2))*cos(Y(1)-Y(2))*Y(3)^2 - M2*L2^2*sin(Y(1)-Y(2))*Y(4)^2);
dy = [Y(3), Y(4), Y3, Y4];
end
その他の回答 (1 件)
Torsten
2021 年 5 月 27 日
- y0 must be a 4x1 vector, not a scalar.
- ye = zeros(4,n) instead of ye=zeros(1,n)
- ye(:,1) = y0 instead of ye(1) = y0
- ye(:,i+1) = ye(:,i) + h*f(x(i),ye(:,i)) instead of the expression in your loop
- dy = [Y(3);Y(4);Y3;Y4] instead of the row vector in your code
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