An arbitrary signal $v(t)$ pass through the following system,
$w'(t) + 5 w(t) = v'''(t) + 320v''(t) + 40 v' (t) + 40v(t)$
How to determine the coefficient of the following differential equation, where the input signal is $w(t)$ and the output signal is $z(t)$, So that the output $z(t) = v(t)$ (after long time when there is no transient exist any longer)?
$z'''(t) + d_2 z''(t) + d_1 z'(t) + d_0 z(t) = e_3 w'''(t) + e_2 w''(t) + e_1 w'(t) + e_0 w(t)$
You have $a(D)w=b(D)v$ where $D$ is the differentiation operator and $a,b$ are polynomials so that $a(D)$ is a linear differential operator.
To undo this in $d(D)z=e(D)w$ so that $z=v$ plus transient terms the easiest way is to have $e(D)$ be a multiple of $a(D)$, and then $d(D)$ the same multiple of $b(D)$. As the orders of $d(D)$ and $b(D)$ already match, the multiple is by a constant that can be chosen to be $1$, $e(D)=a(D)$, $d(D)=b(D)$. Then $b(D)(z-v)=0$ and you need to check that all roots of $b(z)=0$ have negative real part, using Hurwitz or Lienard-Chipart.