Let's look at your function symbolically.
xo = 400;
yo = 7000;
x = 260;
y = 6954;
a = 0.5; b =0.4;
K = b*xo-a*yo;
% t= 0:0.5:50
syms t
y = K./(exp((K*t)-(K*(log(b*xo/yo)/(-(K)))))-a);
vpa(y, 5)
That exponential term is the exponential of a very large (in magnitude) negative number. It very quickly underflows to 0.
format longg
fun = @(t) exp(-3340*t-3.7785);
t = logspace(-9, 0, 10).';
results = table(t, fun(t), 'VariableNames', ["t", "fun(t)"])
If we evaluated that at a symbolic value of 50, what's the value?
vpa(fun(sym(50)), 10)
When you're talking about numbers that small you're not quite at the probability of a monkey typing Hamlet first try, but maybe one of Shakespeare's shorter plays? For most other intents and purposes, that value is 0.
