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)

First of all, the linearization for $c \
eq 0$ 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 $z'(0) = w(0) = 1.$