How do you solve a coupled ODE when one of the ODE results in a vector of length 3 and the other results in a scalar of length 1?

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L'O.G. on 5 Oct 2023
Edited: Torsten on 5 Oct 2023
For instance, in the following example that I found online, if dz(2) were actually a vector, how would you modify this?
[v z] = ode45(@myode,[0 500],[0 1]);
function dz = myode(v,z)
alpha = 0.001;
C0 = 0.3;
esp = 2;
k = 0.044;
f0 = 2.5;
dz = zeros(2,1);
dz(1) = k*C0/f0*(1-z(1)).*z(2)./(1-esp*z(1));
dz(2) = -alpha*(1+esp*z(1))./(2*z(2));
end

Torsten on 5 Oct 2023
Edited: Torsten on 5 Oct 2023
All solution components have to be aggregated in one big vector z, and also the derivatives have to be supplied in this vector form. E.g. if the unknows were composed of a vector x of length 4 and a vector y of length 7, you had to work with vectors z and dz of length 4 + 7 = 11.
Torsten on 5 Oct 2023
Edited: Torsten on 5 Oct 2023
Let x be a scalar and y a vector of length 2.
Let the equations be
dx/dt = x
dy1/dt = 2*y1
dy2/dt = 3*y2
with initial conditions
x(0) = 1,
y1(0) = 2,
y2(0) = 3.
Then you can set up the problem as
x0 = 1;
y10 = 2;
y20 = 3;
z0 = [x0;[y10;y20]];
tspan = [0 1];
[T,Z] = ode45(@fun,tspan,z0);
X = Z(:,1);
Y = Z(:,2:3);
figure(1)
plot(T,X)
figure(2)
plot(T,Y)
function dzdt = fun(t,z)
x = z(1);
y = z(2:3);
dxdt = x;
dydt = [2*y(1);3*y(2)];
dzdt = [dxdt;dydt];
end
L'O.G. on 5 Oct 2023
Thank you

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