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(a) To obtain the system, introduce $y = x'$ as a new variable. Then $x' = y$ and $$y' = x'' = - 3 x' - (k - {\rm e}^{-x^2}) x = - 3 y - (k - {\rm e}^{-x^2}) x$$.

(b) For $x' = 0$ we need $y = 0$, and then $$y' = - (k - {\rm e}^{-x^2}) x$$. But $$k > 1 \ge {\rm e}^{-x^2}$$ for all $x$, so the only way to have $y' = 0$ is $x = 0$. The Jacobian matrix at $(0,0)$ is $$\pmatrix{ 0 & 1\cr 1-k & -3\cr}$$, so the linearized system is

$$u' = v, \qquad v' = (1-k) u - 3 v$$

(c) The characteristic polynomial is $r^2 + 3 r - 1 + k$, and its roots are $(-3 \pm \sqrt{13-4k})/2$. If the eigenvalues are real, they are both negative since $k > 1$. If the eigenvalues are complex, their real part is negative. Thus in either case, the critical point is stable.

(d) When $k=3$ the characteristic polynomial is $r^2 + 3 r + 2 = (r+1)(r+2)$ so the eigenvalues are $-1$ and $-2$, and the critical point is a node. The eigenvectors are $$\pmatrix{1\cr -2\cr}$$ for $-2$ (``fast'') and $$\pmatrix{1\cr -1\cr}$$ for $-1$ (``slow''). The trajectories are shown in the first plot below. A solution can cross $x = 0$ at most once. This corresponds to an overdamped spring-mass system.

When $k=4$ the characteristic polynomial is $r^2 + 3 r + 3$. The eigenvalues are complex: $(-3 \pm \sqrt{3} i)/2$. Thus the critical point is a spiral. At $(0,1)$ the velocity is $$\pmatrix{1\cr -3\cr}$$, so the direction is clockwise. The trajectories are sketched in the second plot below. This is not an exact plot for our system, because in the actual spiral the distance to the origin decreases by so much in each revolution that you could not see a whole turn of the spiral in one picture. Solutions will oscillate, crossing $x = 0$ infinitely many times. This corresponds to an underdamped spring-mass system.