Canonical Form and DE If we have the differential equation $x'' = x - \cos(x')$, then 
In part a) Compute the corresponding non-linear 2D system and and its (unique) equilibrium
part b) compute the linearized system at the equilibrium and classify its equilibrium.
so in part a, I let y = x' and x'' = y' so I get then:
F(x,y) = (x',y') = (y, x-cos(y))
and the equilibrium is y = 0 and back-substituting, I get x = 1, so the unique equilibrium is (x,y) = (1,0)
in part b, I get the Jacobian matrix of F(x,y) and evaluate it at (1,0) and get the eigenvalues and realize that the system is NOT hyperbolic. 
Now, how can we do part c, d, and e?
c) What can we tell about the non-linear system from the linearized one at the equi-
librium; draw an approximative phase-portrait of the non-linear system around the equilibrium
d) computing the Jordan form of the linearized system matrix and and the linear map
that puts it in Jordan canonical form


*

*e) Is this system a gradient system, a hamiltonian system, or none of these? If so,
what is the hamiltonian/potential function then?


So, for part a), I introduced $y' = x$ and $y'' = x''$, 
I need help especially with part c), d), and e). I did parts a) and b).
b) I get hyperbolic equilibria after computing the Jacobinan matrix and evaluating it at (1,0). The jacobian matrix I get is:
(0, 1  as the first row
(1, (x-cos(y))*sin(y)) as the second row
and evaluating at (1,0) I get hyperbolic equilibriia
 A: We are given the second-order nonlinear system:
$$\tag 1 x'' = x - \cos x'$$ 
Part a
We are asked to compute corresponding non-linear 2D system and and its (unique) equilibrium.
So, lets write this as two equations (linearize it sort of speak). Let $x' = y$, so $x'' = y'$ and substituting this back into $(1)$ yields the system:
$$x' = f(x, y) =  y$$
$$y' = g(x, y) = x -\cos y$$
The equilibrium points are given by the points that simultaneously satisfy $x' = 0$ and $y' = 0$, thus:
$$x'  = y = 0, \text{and}$$ 
$$y' =  x - \cos y = 0$$
We know $y = 0$, so $x - \cos 0 = 0 \rightarrow x = 1$. Thus, we have one critical point at $(x, y) = (1, 0)$.
Part b
Compute the linearized system at the equilibrium and classify its equilibrium.
Here we would calculate the Jacobian of the system and then classify that matrix for each critical point. The Jacobian is given by:
$$\displaystyle J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix}0 & 1\\1 & \sin y\end{bmatrix}$$
Evaluating $J$ at the critical point $(1, 0)$, yields the matrix:
$$A = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$
The eigenvalues for this matrix are given by:
$$|A - \lambda I| = 0 \rightarrow (\lambda-1) (\lambda+1) = 0 \rightarrow \lambda_1 = -1, \lambda_2 = 1.$$
Since these eigenvalues are real and opposite sign, we classify this critical point as a saddle node. Generally speaking, we know that this is a robust case. Also note that we could have several critical points and would need to classify EACH one using this process!
Part c
What can we tell about the non-linear system from the linearized one at the equilibrium; draw an approximative phase-portrait of the non-linear system around the equilibrium?
Lets draw a direction field and plot solutions in a phase portrait for this system. Of course, we expect to see an unstable saddle - node at the critical point $(1,0)$, so look for that in the diagram below.
 
Part d
Compute the Jordan form of the linearized system matrix and and the linear map that puts it in Jordan canonical form.
To find the Jordan Normal Form, we need to find the eigenvalues and eigenvectors for the system.
We already found the two eigenvalues and the corresponding eigenvectors are given by:
$$\lambda_1 = -1, v_1 = (-1, 1)$$
$$\lambda_2 = 1, v_2 = (1, 1)$$
So, to write the Jordan Normal Form, using the eigenvalues and eigenvectors, we form:
$$A = PSP^{-1} = \begin{bmatrix}-1 & 1\\1 & 1 \end{bmatrix} \cdot \begin{bmatrix}-1 & 0\\0 & 1 \end{bmatrix} \cdot \begin{bmatrix}-\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{bmatrix}$$
