Trouble plotting Second Order Equation

I'm trying to plot the numerical solution of the second order equation: y'' + sin(y) =0, with the intital values of y(0) being 0.1, 0.7, 1.5, and 3.0, and y'(0) =0. However, the following code is giving me an incorrect graph, and I'm not sure how to resolve this.

rhs = @(t, y) [y(2); -sin(y(1))];
[xa, ya] = ode45(rhs, [0 10], [0.1 0]);
figure, hold on
for y0 = 0:0
for yp0 = 0.1:0.7:1.5:3.0
[tfor, yfor] = ode45(rhs, [0 6.28], [y0 yp0]);
[tbak, ybak] = ode45(rhs, [0 6.28], [y0 yp0]);
plot(tfor, yfor(:,1))
plot(tbak, ybak(:,1))
end
end

Answer 1

[Sam Chak]

Hi @Nina

It appears that you tried to simulate the undamped pendulum-like ODE with multiple initial angles.

If I'm correct, then you can fix your code like the following:

% undamped pendulum
odefcn = @(t,y) [y(2); -sin(y(1))];

% settings
y10   = [0.1 0.7 1.5 3.0];      % initial angle (in radian)
y20   = 0;                      % initial angular rate
tspan = [0 20];

% solving the ODE in loops
for j = 1:length(y10)
    [t, y] = ode45(odefcn, tspan, [y10(j) y20]);
    plot(t, y(:,1))
    hold on
end
hold off

% display graph properties
grid on
legend(strcat('y_{1}(0) = ', num2str(y10')), 'location', 'Best')
xlabel({'$t$'}, 'Interpreter', 'latex')
ylabel({'$y_{1}(t)$'}, 'Interpreter', 'latex')
title({'Solutions of $\ddot{y} = - \sin(y)$, with multiple initial conditions'}, 'Interpreter', 'latex')