I am reading into Khasminskii's Stochastic Stability of Differential Equations. At the start of Chapter 1.5, we look at the stability of solutions to the equation $$ \frac{dx}{dt} = G(x, t, \xi(t,\omega)), $$ where $\xi$ is a stochastic process.
Khasminskii writes:
Following the usual procedure of introducing new variables, equal to the deviations of the corresponding coordinates of the "perturbed" motion from their "unperturbed" values, we see that we only need to consider the stability of the solution $x(t) \equiv 0$ of the equation above in which $G$ satisfies the condition $G(0,t,\xi(t,\omega)) = 0$.
Perhaps this is due to my lack of knowledge in standard ODE techniques, but I am having difficulty understanding why this is true. Is there some insight that would explain this passage? More specifically, if I had a solution $y(t)$ which is somewhere non-zero to the above equation, how could this be 'transformed' to the trivial solution $x(t) = 0$?
Yes, this is not very clear and I would have put it in a different way.
But what he wants to say is that, in general, if we have an equilibrium trajectory $x(\cdot)\equiv x^*$, you can use the change of variables $\tilde x:=x-x^*$. Then, we can look at the dynamical system that governs $\tilde x$. Now, the equilibrium trajectory is $\tilde x(\cdot)\equiv 0$.
Note that, in that case, the function $G$ may be different and this should have made it clear in the main text. The $G$ in (1.62) may not be the same as in (1.61), except in special cases.
For instance, if $\dot{x}=-x^3+a^3$, $a>0$, has $a$ as equilibrium point. Letting $\tilde x=x-a$ yields $\dot{\tilde x}=-\tilde{x}(\tilde{x}^2-3a\tilde x+3a^2)$.
However, you can safely ignore that passage in the book, as it is not important. You can just assume that the zero trajectory is an equilibrium of the system.