I have the following linear system: $$\dot{x} = \sigma - kx$$ $$\dot{y} = -kx+k_1(y_m-y)$$
With $k,k_1,\sigma,y_m\in \mathbb{R}^+$. (That is, all strictly positive constants.)
My textbook claims that these can be simplified into one linear equation, but I did not see how so I decided to attempt to solve the equilibria directly.
Solving for $x$ first: $$0 = \sigma - kx \implies x = \sigma/k$$
And then substituting and solving for $y$: $$0 = -k(\sigma/k)+k_1(y_m-y)$$ $$0 = -\sigma+k_1y_m-k_1y$$ $$k_1y=-\sigma+k_1y_m$$ $$y = -\sigma/k_1 + y_m$$
Did I do this correctly? Is the equilibrium of the system at $$x = \sigma/k, \ y = \sigma/k_1+y_m?$$
I can't see any problem with our OP jeanquilt's solution for the equilibrium values of $x$ and $y$:
$x = -\dfrac{\sigma}{k}, \tag 1$
$y = -\dfrac{\sigma}{k_1} + y_m. \tag 2$
Meanwhile, from
$\dot y = -kx + k_1(y_m - y), \tag 3$
we have
$\ddot y = -k \dot x -k_1 \dot y; \tag 4$
then we may use
$\dot x = \sigma -kx \tag 5$
to obtain
$\ddot y = -k\sigma + k^2x -k_1 \dot y; \tag 6$
from (3) we have
$x = \dfrac{1}{k}(-\dot y + k_1(y_m - y)); \tag 7$
if we substitute this back into (6) we find
$\ddot y = -k\sigma + k(-\dot y + k_1(y_m - y)) - k_1\dot y = -k\sigma - (k + k_1)\dot y - kk_1y + kk_1y_m, \tag 8$
or
$\ddot y + (k + k_1)\dot y +kk_1y + (k\sigma - kk_1y_m) = 0, \tag 9$
an equation in only $y$; the equation for $x$ is already in single-variable form.