Answer:
a)
. It does not have a steady state
b)
. It has a steady state.
Step-by-step explanation:
a) ![y' -4y = -8](https://tex.z-dn.net/?f=y%27%20-4y%20%3D%20-8)
The first step is finding
. So:
![y' - 4y = 0](https://tex.z-dn.net/?f=y%27%20-%204y%20%3D%200)
We have to find the eigenvalues of this differential equation, which are the roots of this equation:
![r - 4 = 0](https://tex.z-dn.net/?f=r%20-%204%20%3D%200)
![r = 4](https://tex.z-dn.net/?f=r%20%3D%204)
So:
![y_{n}(t) = y_{0}e^{4t}](https://tex.z-dn.net/?f=y_%7Bn%7D%28t%29%20%3D%20y_%7B0%7De%5E%7B4t%7D)
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:
![y_{p}(t) = C](https://tex.z-dn.net/?f=y_%7Bp%7D%28t%29%20%3D%20C)
So
![(y_{p})' -4(y_{p}) = -8](https://tex.z-dn.net/?f=%28y_%7Bp%7D%29%27%20-4%28y_%7Bp%7D%29%20%3D%20-8)
![(C)' - 4C = -8](https://tex.z-dn.net/?f=%28C%29%27%20-%204C%20%3D%20-8)
C is a constant, so (C)' = 0.
![-4C = -8](https://tex.z-dn.net/?f=-4C%20%3D%20-8)
![4C = 8](https://tex.z-dn.net/?f=4C%20%3D%208)
![C = 2](https://tex.z-dn.net/?f=C%20%3D%202)
The solution in the form is
![y(t) = y_{n}(t) + y_{p}(t)](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_%7Bn%7D%28t%29%20%2B%20y_%7Bp%7D%28t%29)
![y(t) = y_{0}e^{4t} + 2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_%7B0%7De%5E%7B4t%7D%20%2B%202)
b) ![y' +4y = 8](https://tex.z-dn.net/?f=y%27%20%2B4y%20%3D%208)
The first step is finding
. So:
![y' + 4y = 0](https://tex.z-dn.net/?f=y%27%20%2B%204y%20%3D%200)
We have to find the eigenvalues of this differential equation, which are the roots of this equation:
![r + 4 =](https://tex.z-dn.net/?f=r%20%2B%204%20%3D%20)
![r = -4](https://tex.z-dn.net/?f=r%20%3D%20-4)
So:
![y_{n}(t) = y_{0}e^{-4t}](https://tex.z-dn.net/?f=y_%7Bn%7D%28t%29%20%3D%20y_%7B0%7De%5E%7B-4t%7D)
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:
![y_{p}(t) = C](https://tex.z-dn.net/?f=y_%7Bp%7D%28t%29%20%3D%20C)
So
![(y_{p})' +4(y_{p}) = 8](https://tex.z-dn.net/?f=%28y_%7Bp%7D%29%27%20%2B4%28y_%7Bp%7D%29%20%3D%208)
![(C)' + 4C = 8](https://tex.z-dn.net/?f=%28C%29%27%20%2B%204C%20%3D%208)
C is a constant, so (C)' = 0.
![4C = 8](https://tex.z-dn.net/?f=4C%20%3D%208)
![C = 2](https://tex.z-dn.net/?f=C%20%3D%202)
The solution in the form is
![y(t) = y_{n}(t) + y_{p}(t)](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_%7Bn%7D%28t%29%20%2B%20y_%7Bp%7D%28t%29)
![y(t) = y_{0}e^{-4t} + 2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_%7B0%7De%5E%7B-4t%7D%20%2B%202)