Given the ordinary differential equation
$$ \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \end{bmatrix} = \begin{bmatrix} -x_1+x_2+1 \\ 2x_1-3x_2-2 \\ \end{bmatrix} $$ i have to check the stability of the equilibrium point. Solving the RHS for $0$ gives us the equilibrium $$\hat x=\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and now i want to use the usual eigenvalue stability criterion for linear ODEs, given that our original equation rewrites to
$$ \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & -3 \\ \end{bmatrix}x + \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}, $$ so we basically have a linear ODE which got "shifted" by a constant, so is not of the usual form $\dot x=Ax$, for which we had formulated the eigenvalue stability criterion, but of the form $\dot x = Ax +b$. Checking the eigenvalues of our matrix, we see that all eigenvalues are either negative or semisimple, hence making $\hat x$ a stable equilibrium point.
The question is: How can i justify the use of this stability test?