I need to solve this first-order system ODE using Matlab - matlab

Question

I have to solve this first-order system ODEs using Matlab.
y' + y - z - u = 0.
z' - y + z - u = 0.
u' - y - z - u = 0.
y(0)=1, z(0)=0, u(0)= 0
The analytical solution of above system of first-order ODEs is:
y = (1/3)e^–x + (1/2)e^–2x + (1/6)e^2x
z = (1/3)e^–x – (1/2)e^–2x + (1/6)e^2x
u = (1/3)e^2x – (1/3)e^–x
I have this code to solve it and plot the numerical solution I obtain using the ODE solvers and the analytical solution I have in the statement.
----------
clear
%options= odeset('RelTol',1e-5);
options= odeset('RelTol',1e-5,'AbsTol',1e-7);
[t23,y23]= ode23('functionB',[0 1],[1 0 0],options);
[t23s,y23s]= ode23s('functionB',[0 1],[1 0 0],options);
figure
ya =((1/3)*exp(-t23) + (1/2)*exp(-2*t23) + (1/6)*exp(2*t23));
za =((1/3)*exp(-t23) - (1/2)*exp(-2*t23) + (1/6)*exp(2*t23));
ua =((1/3)*exp(2*t23) + (1/3)*exp(-t23));
plot(t23,ya,t23,za,t23,ua);
title('\bf{Analytical solutions }')
figure
plot(t23,y23(:,1),'m-.',t23,y23(:,2),'g:', t23,y23(:,3),r);
title('\bf{Numerical solutions using} \it{ode23}')
figure
plot(t23s,y23s(:,1),'m-.',t23s,y23s(:,2),'g:', t23s,y23s(:,3),r);
title('\bf{Numerical solutions using} \it{ode23s}')
%legend('ya','ode23','ode23s',0)
%text(3.4,-1.7,'ya = -2sin(t) - cos(2t) ')
%title('\bf{Analytical and numerical solutions using} \it{ode23, ode23s}')
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And also I have the function functionB:
-------
function dy= functionB(t,y)
%-y+z+u
%y-z+u
%y+z+u
dy =[??????];
-------
I don't know how to write the solutions in the code of the function to use it in my code.
If someone can help me with this I will be so grateful.
Thanks in advance.

Answer 1

The MATLAB documentation has some good examples about this.
You must rearrange your ODE like this:
y' = -y + z + u = 0
z' = y + z + u = 0
u' = y + z + u = 0
You then substitude y by y(1), z by y(2) and u by y(3) and now you can write your function like this:
function dy = functionB(t,y)
dy = zeros(3,1);
dy(1) = -y(1)+y(2)+y(3);
dy(2) = y(1)-y(2)+y(3);
dy(3) = y(1)+y(2)+y(3);
end