Answer:
a)
. It does not have a steady state
b)
. It has a steady state.
Step-by-step explanation:
a) 
The first step is finding
. So:

We have to find the eigenvalues of this differential equation, which are the roots of this equation:


So:

Since this differential equation has a positive eigenvalue, it does not have a steady state.
Now as for the particular solution.
Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

So


C is a constant, so (C)' = 0.



The solution in the form is


b) 
The first step is finding
. So:

We have to find the eigenvalues of this differential equation, which are the roots of this equation:


So:

Since this differential equation does not have a positive eigenvalue, it has a steady state.
Now as for the particular solution.
Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

So


C is a constant, so (C)' = 0.


The solution in the form is

