Respuesta :
Answer:
a) [tex]y(t) = y_{0}e^{4t} + 2[/tex]. It does not have a steady state
b) [tex]y(t) = y_{0}e^{-4t} + 2[/tex]. It has a steady state.
Step-by-step explanation:
a) [tex]y' -4y = -8[/tex]
The first step is finding [tex]y_{n}(t)[/tex]. So:
[tex]y' - 4y = 0[/tex]
We have to find the eigenvalues of this differential equation, which are the roots of this equation:
[tex]r - 4 = 0[/tex]
[tex]r = 4[/tex]
So:
[tex]y_{n}(t) = y_{0}e^{4t}[/tex]
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:
[tex]y_{p}(t) = C[/tex]
So
[tex](y_{p})' -4(y_{p}) = -8[/tex]
[tex](C)' - 4C = -8[/tex]
C is a constant, so (C)' = 0.
[tex]-4C = -8[/tex]
[tex]4C = 8[/tex]
[tex]C = 2[/tex]
The solution in the form is
[tex]y(t) = y_{n}(t) + y_{p}(t)[/tex]
[tex]y(t) = y_{0}e^{4t} + 2[/tex]
b) [tex]y' +4y = 8[/tex]
The first step is finding [tex]y_{n}(t)[/tex]. So:
[tex]y' + 4y = 0[/tex]
We have to find the eigenvalues of this differential equation, which are the roots of this equation:
[tex]r + 4 = [/tex]
[tex]r = -4[/tex]
So:
[tex]y_{n}(t) = y_{0}e^{-4t}[/tex]
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:
[tex]y_{p}(t) = C[/tex]
So
[tex](y_{p})' +4(y_{p}) = 8[/tex]
[tex](C)' + 4C = 8[/tex]
C is a constant, so (C)' = 0.
[tex]4C = 8[/tex]
[tex]C = 2[/tex]
The solution in the form is
[tex]y(t) = y_{n}(t) + y_{p}(t)[/tex]
[tex]y(t) = y_{0}e^{-4t} + 2[/tex]
