Lorentz equation using the Euler method

I would like to plot the Lorentz equation using the Euler method.(If possible don't use ODE)

Please give me how to plot it.

for loop seems difficult and i don't understand.

I think the following is completely different

p=10 %初期値
r=28;
b=8/3;
x(1)=10;
y(1)=12;
z(1)=25;
dy(1)=p*y(1)+p*y(2);
dy(2)=-y(1)*y(3)+r*y(1)-y(2);
dy(3)=y(1)*y(2)-b*y(3);
for i=1:length(t)-1;
　　 y(i+1)=y(i)+v(i)*h
end
plot

[darova]

Can you show original equations?

[sz]

lorenz equation

Is this ok?

Please take care of me.

%  lorenz equation
dx/dt=-px+py
dy/dt=-xz+rx-y
dz/dt=xy-bz      p=10,　r=28,　b=8/3
x(0)=10, y(0)=12, z(0)=25

euler

y(t+h)=y(t)+h y′(t)

Answer 1

[darova]

Try this solution

[A Name]

I have the same question and I am stuck. I need to use the euler foward time stepping and this is the code i have so far:

could you please help I am not very confident on matlab

%First define sigma=s, rho=r and beta=b for this question

s=10
r=28;
b=8/3;
%  lorenz equation
dx/dt=-sx+sy
dy/dt=-xz+rx-y
dz/dt=xy-bz

%Choose initail conditions
x(0)=10, y(0)=12, z(0)=25

y(1,1) = 10;
y(1,2) = 12;
y(1,3) = 25;
for i=1:length(t)-1;
dy(i,1) = s*y(i,1)+s*y(i,2);
dy(i,2) = -y(i,1)*y(i,3)+r*y(i,1)-y(i,2);
dy(i,3) = y(i,1)*y(i,2)-b*y(i,3);
 y(i+1,:)=y(i,:)+dy(i,:)*h(i)
end
plot(t,y)

[Kian Halim]

Did you ever get this working?