runge kutta for 2 order ODE

조회 수: 3(최근 30일)
Rio Bratasena
Rio Bratasena 2022년 11월 3일
답변: Sam Chak 2022년 11월 3일
how can I solve this equation
with runge kutta method

답변(2개)

Hiro Yoshino
Hiro Yoshino 2022년 11월 3일
Let us assume that and re-write the problem as follows:
Let's transform it so it can be used in an ode setting:
These are used in the function (vdp1) as follows:
solve this the time interval of [0 10] with initial values of [0.2 0] for x_1, x_2 respectively.
[t,x] = ode45(@vdp1,[0 10],[0.2 0])
t = 89×1
0 0.0001 0.0003 0.0004 0.0005 0.0011 0.0018 0.0024 0.0030 0.0062
x = 89×2
0.2000 0 0.2000 -0.0001 0.2000 -0.0001 0.2000 -0.0002 0.2000 -0.0002 0.2000 -0.0005 0.2000 -0.0007 0.2000 -0.0010 0.2000 -0.0012 0.2000 -0.0024
plot(t,x(:,1),'-o',t,x(:,2),'-o')
title('Solution with ODE45');
xlabel('Time t');
ylabel('Solution x');
legend('x_1','x_2');
function dxdt = vdp1(t,x)
% ODE
dxdt = [x(2); -2*(x(2)+x(1))];
end

Sam Chak
Sam Chak 2022년 11월 3일
Assuming that the symbol is the Dirac impulse. please check if the following responses are expected when you run it with the Runge-Kutta solver.
A = [0 1;
-2 -2];
B = [0; 0.2];
C = eye(2);
D = [0; 0];
sys = ss(A, B, C, D);
impulse(sys)

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