I have a system of a mass action type.
$\require{AMScd}$ \begin{CD} X @>{\text{$k$}}>> 2X, 2X @>{\text{$k\cdot v(t)$}}>> ∅ \end{CD}
When $v(t) = v_0(const)$, what is the stable state of x, ${x^s}^t$?
I am thinking
$\frac{dx}{dt}$ = $kX - k*v(t) 2X = 0$.
Therefore, when $v(t) = v_0(const)$, the steady state of x = $\frac{k}{k}$ = 1.
Is it correct? I would appreciate any help!