 Solution

Translate the equilibrium to the origin by $z=(x-a,y-b{)}^T$. If the vector field is continuously differentiable, its first-order Taylor expansion isThe Jacobian matrix $M$ gives the linearization $\dot{z}=Mz$, whose general solution is $z(t)=e^{tM}z(0)$ using the matrix exponential. Its eigenvalues determine the exponential factors, while its eigenvectors and, when necessary, generalized eigenvectors determine their directions.

For distinct eigenvalues ${\lambda }_1,{\lambda }_2$ and corresponding eigenvectors $v_1,v_2$,For a real matrix with nonreal conjugate eigenvalues, choose conjugate coefficients to obtain real solutions. Equivalently, if $\lambda =a+ib$ and $v=p+iq$, a real pair of independent solutions isFor a repeated eigenvalue $\lambda$, there are two cases:

The factor $t$ in the second case is essential. These are exact solutions of the linearization; the remainder describes the error for the nonlinear differential equation. When no eigenvalue has zero real part, the hyperbolic equilibrium theory justifies the corresponding local qualitative classification. Zero real parts require additional nonlinear analysis.