A(:) = B, the number of elements in A and B must be the same. error, what am I doing wrong?

figure(1); clf

%Time Step
Ts = 0.001; % time step
t=0:Ts:200 ; 

%  perameters

a1 = 1.25;
a2 = 1;
y = 0.001;
z= 0;
gain = 1.8;
x=0;

for k=2:length(t)

    dxdt = y;
    d2xdt2 = z;
    d3xdt3 = (gain*x-a1*y-a2*z-gain*(x.^3));

       x(k) = x(k-1)+dxdt*Ts;
       y(k) = y(k-1)+d2xdt2*Ts;
       z(k) = z(k-1)+d3xdt3*Ts;

end

figure(1); clf
plot (x,y,'b-'); hold on

Answer 1

[Roger Stafford]

The problem lies with the lines

   x(k) = x(k-1)+dxdt*Ts;
   y(k) = y(k-1)+d2xdt2*Ts;
   z(k) = z(k-1)+d3xdt3*Ts;

On the second pass through your for-loop, the variables 'dxdt', 'd2xdt2', and 'd3xdt3' become two-element arrays. However, you are then attempting to enter these into one-variable values of x(k), y(k), and z(k). They won't fit, hence the error message. You need to reconsider what it is you wish to enter into x(k), y(k), and z(k).

Answer 2

[Walter Roberson]

x=0;

That initializes x(1) = 0

for k=2:length(t)
  ...
  x(k) = x(k-1)+dxdt*Ts;

so when k = 2 (the first iteration), x(2) is going to be assigned. So after that x will be a vector rather than a scalar.

       d3xdt3 = (gain*x-a1*y-a2*z-gain*(x.^3));

That uses all of x, so from k = 3 and later, d3xdt3 will be a vector because x became a vector by the end of k = 2

         z(k) = z(k-1)+d3xdt3*Ts;

d3xdt3 is a vector, so the right hand side is a vector, but you are trying to assign the vector into the single location z(k)