MatLab will integrate numerically even without numbers

Hello,

I am running a code to solve a system of ODES. In my main code, I make a loop around the y value, and then attempt to solve this equation system with ode23tb (I also tried ode15, but the wrong result is the same):

function Deriv = EDOFun (t,x)
global x0 p W0 y
global A c beta cs T
syms z
Deriv = zeros(2,1);
f=sqrt(2/pi/T^2)*exp(-2*(t-2*T).^2/T^2);
W = W0*(f);

sol = int(x(1)*x(2)*z*exp(-z/x0),z,0,Inf);
Deriv(1) = beta*A*x(2) + c*cs*x(2)*x(1) - (c/y)*x(1);
Deriv(2) = W - A*x(2) - c*cs*(p*x(2)*x(1)) + ((1-p)/x0^2)*sol);

The problem here is that both x(1) and x(2) should be dependent of the z in the integral. If I choose to name x instead of z for it, however, the program will not work. And if I use it this way, MatLab will simply calculate the integral disregarding the fact that x(1) and x(2) should be a solution of this system of equations.

Is there a way MatLab can actually do this calculation, or should I go back to the math in my problem to be able to fix it?

[John D'Errico]

You cannot use a numerical solver, thus ODE23tb, to solve a problem with symbolic variables in it. I suppose you have eliminated z in the integral though. But I would at the very least do some basic mathematics. For example,

sol = int(x(1)*x(2)*z*exp(-z/x0),z,0,Inf);

is nothing more than

x(1)*x(2)*x0^2

as long as x0 is real and non-zero.

What you don't show is how x(1) and x(2) are dependent on z.

[Jessica Dipold]

I did eliminate z in the integral, or I thought I did at least.

Well, x(1) and x(2) are actually the solutions of the ODE, so I do not know how they are dependent of z, which is actually x originally. The original equation system is this one

I will try your suggested math solution first anyways to see how it shifts my answer curves. Thank you for your fast response too!