Steady State case for differential equations

I need to write these equations in steady state form (time derivatives set to 0), and then substitute in values for the parameters (e,d,q,f). However, I don't understand how to do this as if I set the derivatives to 0, when solving the equations, the parameters no longer appear?

e x'[t] == (1 - x[t])*x[t] + q y[t] -
x[t] y[t]
d y'[t] == -q y[t] - x[t] y[t] +
2 f z[t]
z'[t] == x[t] - z[t]

Try rule /. _'[t]->0

eqn = {e x'[t] == (1 - x[t])*x[t] + q y[t] - x[t] y[t],d y'[t] == -q y[t] - x[t] y[t] + 2 f z[t],z'[t] == x[t] - z[t]} /. _'[t]->0
(*{0 == (1 - x[t]) x[t] + q y[t] - x[t] y[t], 0 == -q y[t] - x[t] y[t] + 2 f z[t], 0 == x[t] - z[t]}*)

These equations might be solved for {x[t],y[t],z[t]}

The two parameters e,d are irrelevant for the solution!

The equations reduce to algebraic :

{0 == (1 - x)*x + q y - x y,
0 == -q - x y + 2 f z,
0 == x - z}

You can use Solve or NSolve.

• I'd divide both sides by d and e first, to not lose them when you set x'[t] and y'[t] to zero. Mar 16 '20 at 13:00
• @ChrisK How does that make any difference? Once you start solving the equations, you multiply by d and e and they're gone again. Mar 16 '20 at 13:34
• @SjoerdSmit Umm, good point :) Maybe I was thinking ahead to a stability analysis, where d and e would matter. Mar 16 '20 at 14:01