Linearization of a second order differential equation 
Linearize the equation $$x'' = -\alpha x-\rho x'+c \sin(t)$$

It is very easy when $c=0$ giving you a 
$$ x' = y
$$$$
y' = -\alpha x -\rho y
$$
giving you a very nice phase portrait. 
However, if $c$ is non-zero, the linearization should be like 
$$ x' = y
$$$$
y' = -\alpha x -\rho y +c\sin(t)
$$
but this gives a very ugly phase portrait (the lines keep intersecting with themselves)
Is this still accurate or should I further use a Jacobian to linearize the matrix? (If so, can somebody provide a small hint on how to approach the result, take any initial state for initial condition)
 A: First of all, the linearization for $c \neq 0$ is wrong. Indeed, the term $c\sin(t)$ has not been differentiated. However, it is difficult to do this since it does not depend on $x$ or $y$.
How can we deal with the term $\sin(t)$? The presence of a time-varying term implies that the order of the system is bigger than $2$.
Indeed, consider the following ODE:
$$z'' =  -z.$$
For $z(0) = 0$ and $z'(0) = 1$, the solution is $\sin(t)$. Thus, we can rewrite the original system as follows:
$$\begin{cases}
x'' = -\alpha x-\rho x'+c z\\
z'' = -z
\end{cases}.$$
Therefore, you have a forth order system.
Setting $y=x'$ and $w=z'$, it can be rewritten as:
$$\begin{cases}
x' = y\\
y' = -\alpha x-\rho y+c z\\
z' = w\\
w' = -z
\end{cases}.$$
Notice that the system is linear.
Finally, whichever are the initial conditions on $x$ and $y$, you must always use also $z(0) = 0$ and $z'(0) = w(0) = 1.$
A: The system $$x' = y$$
$$y' = -\alpha x -\rho y +c\sin(t)$$
is linear. 
What you have is a non-autonomous, in-homogeneous system  and that is the problem with the phase portrait.
When your system is  non-autonomous, the phase portrait is better understood in three dimensions $(t,x,y)$ with the time dimension also present.
