How to build a plot differential equations

Hi, I'm solving a differential equation and with the given parameters my graph should show a circle, but it's displaying a void, I don't understand why, please help

M8()

Значения параметров: n = 0,b=0, k = 1, a = 1, h = 0, p = 1
Решение задачи Коши для уравнения
     solver: 'ode45'
    extdata: [1×1 struct]
          x: [0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1]
          y: [2×11 double]
      stats: [1×1 struct]
      idata: [1×1 struct]

Elapsed time is 0.017687 seconds.

function M8
  n = 0;b=0; k = 1; a = 1; h = 0; p = 1; y(1)=0; y(2)=1; x=2;
  f=@(t,y)[y(2);y(2)+2.*n.*(1+a.*y(1).^3).*y(1)+k.^2.*(y(1)+b.*y(1).^3)-h.*sin(p.*x)];
  clc
  disp('Значения параметров: n = 0,b=0, k = 1, a = 1, h = 0, p = 1')
  disp('Решение задачи Коши для уравнения')
  z= ode45(f,[0,1],[0,0]);
  disp(z);
  tic, [~,y] = ode45(f,[0,1],[0,0]);toc
  plot(y(:,1),y(:,2))
  grid on
end

Answer 1

[Sam Chak]

Hi @Maria

Please check your second state equation. It does not produce a circle in the phase portrait. However, I will show you an example that produces a perfect circular trajectory.

M8

function M8
    n = 0; b = 0; k = 1; a = 1; h = 0; p = 1; x = 2;
    f = @(t, y) [y(2); 
               - y(1)];
    % f = @(t, y) [y(2); 
    %              y(2) + 2*n*(1 + a*y(1)^3)*y(1) + k^2*(y(1) + b*y(1)^3) - h*sin(p*x)];
    % disp('Значения параметров: n = 0,b=0, k = 1, a = 1, h = 0, p = 1')
    % disp('Решение задачи Коши для уравнения')
    tspan  = [0, 100];    % 
    y0     = [0, 1];    % initial values: y1(0) = 0, y2(0) = 1
    [t, y] = ode45(f, tspan, y0);
    % disp(z);
    % tic, [~,y] = ode45(f,[0,1],[0,0]);toc
    plot(y(:,1), y(:,2))
    grid on
    axis equal
    xlabel y_1
    ylabel y_2
end

Answer 2

[Walter Roberson]

  n = 0;b=0; k = 1; a = 1; h = 0; p = 1; y(1)=0; y(2)=1; x=2;
  f=@(t,y)[y(2);y(2)+2.*n.*(1+a.*y(1).^3).*y(1)+k.^2.*(y(1)+b.*y(1).^3)-h.*sin(p.*x)];

Your h is 0, so h.*sin(p.*x) is going to be 0, so the second entry in f is effectively

  y(2)+2.*n.*(1+a.*y(1).^3).*y(1)+k.^2.*(y(1)+b.*y(1).^3)

If you let y(1) and y(2) both be 0, then that evaluates to 0.

Therefore with your h value and with those boundary conditions, your ode is constant 0, 0