We are given the nonlinear system:
$\tag 1 x' = y, ~~ y'= -8 \sin x - 2y,$
where $-2\pi \le x \le 2\pi$.
We start off by finding the fixed points of the system. To do this, we need to find those points where $x'$ and $y'$ are simultaneously equal to zero, over the given range in $(1)$.
So, $x' = 0$, when $y = 0$, and $y' = -8 \sin x - 2y = 0$ reduces to (since we must have $y = 0$ from the previous observation):
$y' = -8 \sin x -2(0) = 0 \rightarrow x = -2\pi, -\pi, 0, \pi, 2\pi$.
This gives us a total of five fixed (critical) points as:
$$(-2\pi, 0), (-\pi, 0), (0,0), (\pi, 0), (2\pi, 0).$$
In order to linearize the system, we have to investigate the behavior of the Jacobian of the system at those fixed points. So, the Jacobian matrix is:
$$A = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y}\\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ - 8 \cos x & -2 \end{bmatrix}$$
Next, we need to evaluate the Jacobian matrix $A$ at each of those fixed points.
At $(-2\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
At $(-\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (-\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$
At $(0, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (0) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
At $(\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$$
At $(2\pi, 0)$, we have:
$$A = \begin{bmatrix} 0 & 1 \\ - 8 \cos (2\pi) & -2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$$
Notice, we have two identical sets of matrices given the periodicity of $\cos x$ over our five fixed points.
We are asked to linearize the system near every equilibrium point, and describe the behaviour of the linearized system. So, now we need to determine the behavior of these two matrices by looking at their eigenvalues. So, we get:
$A = \begin{bmatrix} 0 & 1 \\ 8 & -2 \end{bmatrix}$, has eigenvalues $\lambda_1 = -4$ and $\lambda_2 = 2$. These are real eigenvalues with opposite sign, so this is an unstable saddle node.
$A = \begin{bmatrix} 0 & 1 \\ -8 & -2 \end{bmatrix}$, has eigenvalues $\lambda_1 = -1 + \sqrt{7}i$ and $\lambda_2 = -1 - \sqrt{7}i$. These are a complex conjugate pair, with negative real part, so this is a stable spiral point.
Note: The only thing that allows us to use this linearization is that we do not get borderline cases from the fixed points. You had better make sure you are clear on this last statement!
Lastly, we can draw the phase portrait to show the direction field, the five fixed points and many solutions $x(t)$ and $y(t)$ in order to visualize this and to compare to our analyses. Of course, we should see two unstable saddle nodes and three stable spirals from the analyses above.
Make sure to pick out the five fixed points $(x, y)$ and that you see what we derived.
Regards
To visualize the solution, it helps to graph it in the $x_1 x_2$-plane for various $c_1, c_2$.
I will map out most of the details and you can add the rest. Start with the first eigenvalue/vector solution and we have in scalar form:
$$x_1 = c_1 e^t, x_2 =2 c_1 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = 2 x_1$. If $c_1 \gt 0$, we are in quadrant $I$ and if $c_1 \lt 0$, we are in quadrant $III$.
Repeating this for the second eigenvalue/vector, we have:
$$x_1 = c_2 e^t, x_2 =-2 c_2 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = -2 x_1$. If $c_2 \gt 0$, we are in quadrant $IV$ and if $c_2 \lt 0$, we are in quadrant $II$.
The solution is a linear combination of these two as:
$$x(t) = c_1 e^{t} \left( \begin{array}{c} 1\\ 2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c} 1\\ -2\\ \end{array}\right)$$
So, we can plot the eigenvectors as (blue is $n_1$ and yellow is $n_2$):
Now, what happens to the solutions as $t \rightarrow + \infty$? Just draw handfuls of those around the lines in each quadrant (hint - they are asymptotic to the line and move away from the origin as this is an unstable node). This will give you a phase portrait that is an unstable node.
For part $b.$, we have the initial point $(2, 3)$. Since we know we are going to be asymptotic to the eigenvector in quadrant $I$, it starts at ($2, 3)$ and goes to this line.
For part $c.$, we have $x(0) = (2, 3)$, with:
$$x(t) = c_1 e^{t} \left( \begin{array}{c} 1\\ 2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c} 1\\ -2\\ \end{array}\right)$$
We end up with $c_1 = \dfrac 74, c_2 = \dfrac 14$, so have:
$$x_1(t) = \frac{7}{4} e^t+\frac{1}{4} e^{2t}, x_2(t) = \frac{7}{2}e^{t}-\frac{1}{2} e^{2t}$$
A plot of this shows:
One last note, it helps to plot this parametrically and we get:
Best Answer
As the $2$ by $2$ matrix has two real eigenvalues of multiplicity one, it can be diagonalized $$ \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}. $$ Look at diagonalization as a linear coordinate change. In the new coordinates, $(y_1, y_2)$, the ODE system has the form $$ \begin{cases} y'_1 = \lambda_1 y_1 \\ y'_2 = \lambda_2 y_2, \end{cases} $$ so its solutions are given by $$ \tag{1} \begin{cases} y_1(t) = C e^{\lambda_1 t} \\ y_2(t) = D e^{\lambda_2 t}, \end{cases} $$ where $C$ and $D$ are real constants. $C = 0$ (or $D = 0$) correspond to the solutions whose first (or second) coordinate is constantly equal to zero.
Fix, for the moment, $C \ne 0$ and $D \ne 0$. By eliminating $t$ in $(1)$ we obtain $$ \left\lvert \frac{y_1}{C} \right\rvert^{\lambda_2} = \left\lvert \frac{y_2}{D} \right\rvert^{\lambda_1}, $$ hence $$ \lvert y_1 \rvert = E \, \lvert y_2 \rvert^{\lambda_2/\lambda_1}, $$ where $E = \lvert D \rvert \lvert C \rvert^{-\lambda_2 / \lambda_1}$. By symmetry, we can restrict ourselves to the first quadrant: $$ y_2 = E \, y_1^{\lambda_2/\lambda_1}, \quad E > 0. $$ As $\lambda_2 < \lambda_1 < 0$, $\lambda_2/\lambda_1 > 1$. We have thus obtained a family of "generalized parabolas", tangent at the origin to the $y_1$-axis.
Remember that we are still in the new coordinates. Returning to the old coordinates we do some rotating and stretching, and those operations do not change tangencies. In particular, the trajectories (except on the one-dimensional invariant subspaces spanned by eigenvectors) are curves tangent at the origin to an eigenvector corresponding to the larger eigenvalue, as in the picture.
Notice that, as $t \to \infty$, both coordinates of any solution converge to zero, so the arrows on the curves must be directed "toward" the origin.