Let $ (\bar{x},\bar{y},\bar{z}) $ be an equilibrium of autonomous first-order differential equation: \begin{equation} \begin{cases} \dot{x}= f_1(x,y,z),\\ \dot{y}= f_2(x,y,z), \\ \dot{z}= f_3(x,y,z). \end{cases} \end{equation} Is it possible to say that $(\bar{x},\bar{y},\bar{z})$ is unique equilibrium point of the following system: \begin{equation} \begin{cases} \dot{x}= f_1(x,y,z)-(x-\bar{x}),\\ \dot{y}= f_2(x,y,z)-(y-\bar{y}), \\ \dot{z}= f_3(x,y,z)-(z-\bar{z}). \end{cases} \end{equation}
2026-02-23 11:20:02.1771845602
Equilibrium of autonomous first-order differential equation:
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A point $x^*$ is called an equilibrium (stationary point/critical point) of a system of differential equations :
$$\vec{x}' = \vec{f}(x_1,\dots,x_n)$$
if $x^* \in \mathbb R^n$ is a vector, such :
$$\vec{f}(x^*_1,\dots,x^*_n)=0$$
For your specific example, letting the expressions :
$$\vec{x}' =\vec{F}(x_1,x_2,x_3)=\vec{f}(x_1,x_2,x_3) - (\vec{x}-{\vec{x^*}})$$
we have that :
$$\vec{F}(x_1,x_2,x_3)=\vec{f}(x^*_1,x^*_2,x^*_3)-(\vec{x^*}-\vec{x^*})=0$$
which means that $\vec{x^*}$ is indeed an equilibrium of the alternative system of differential equations that you mentioned.