runge kutta for 2 order ODE

how can I solve this equation

with runge kutta method

Answer 1

[Hiro Yoshino]

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

Answer 2

[Sam Chak]

Hi @Rio Bratasena

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)