= 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> is
$$
\dot z=Mz+r(z),\qquad\frac{\|r(z)\|}{\|z\|}\longrightarrow0.
$$
The <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> $v_1,v_2$,
$$
\boxed{z(t)=c_1e^{\lambda_1t}v_1+c_2e^{\lambda_2t}v_2.}
$$
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 is
$$
e^{at}(p\cos bt-q\sin bt),\qquad e^{at}(p\sin bt+q\cos bt).
$$
For a repeated <eigenvalue> $\lambda$, there are two cases:
* If two independent <eigenvectors> exist, $M=\lambda I$ and $\boxed{z(t)=e^{\lambda t}(c_1v_1+c_2v_2)}$.
* Otherwise choose a <Jordan chain> $(M-\lambda I)v=0$, $(M-\lambda I)w=v$. Then $\boxed{z(t)=e^{\lambda t}\bigl[c_1v+c_2(w+tv)\bigr]}$.
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.
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