= Solution
{c}
The second <equilibrium point> equation gives $y=-(x+1)/2$. Substituting into the first gives $x+e^{(x-1)/2}=0$, so $x<0$. Put $x=-a$, with $a>0$. Then
$$
a=e^{-(a+1)/2}
\iff ae^{(a+1)/2}=1,\qquad y=(a-1)/2.
$$
The function $h(a)=ae^{(a+1)/2}$ is strictly increasing on $a\geq0$, since $h'(a)=e^{(a+1)/2}(1+a/2)>0$. It starts at zero and tends to infinity, so it takes the value one exactly once. Moreover $h(1)=e>1$, so $0<a<1$. This proves uniqueness of the stated <equilibrium point>.
At that point $e^{x+y}=a$, and the <Jacobian matrix> is
$$
J=\begin{pmatrix}1+a&a\\1&2\end{pmatrix}.
$$
Its trace is $3+a$ and determinant $2+a$. The characteristic discriminant is $(3+a)^2-4(2+a)=(a+1)^2$, so its two <eigenvalues> are
$$
\boxed{\lambda_1=1,\qquad\lambda_2=2+a.}
$$
They are distinct and positive, with <eigenvectors> $(1,-1)$ and $(a,1)$ respectively. Thus \b[the linearized <critical point> is an <unstable node>]. It is a <hyperbolic equilibrium>, so the original nonlinear <equilibrium point> is also locally repelling; this last observation is stronger than the requested linearized classification.
Back to article page
