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Tcecarenko [31]
3 years ago
10

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

Mathematics
1 answer:
WINSTONCH [101]3 years ago
6 0

Answer:

a) y(t) = y_{0}e^{4t} + 2. It does not have a steady state

b) y(t) = y_{0}e^{-4t} + 2. It has a steady state.

Step-by-step explanation:

a) y' -4y = -8

The first step is finding y_{n}(t). So:

y' - 4y = 0

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

r - 4 = 0

r = 4

So:

y_{n}(t) = y_{0}e^{4t}

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

So

(y_{p})' -4(y_{p}) = -8

(C)' - 4C = -8

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

-4C = -8

4C = 8

C = 2

The solution in the form is

y(t) = y_{n}(t) + y_{p}(t)

y(t) = y_{0}e^{4t} + 2

b) y' +4y = 8

The first step is finding y_{n}(t). So:

y' + 4y = 0

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

r + 4 =

r = -4

So:

y_{n}(t) = y_{0}e^{-4t}

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

So

(y_{p})' +4(y_{p}) = 8

(C)' + 4C = 8

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

4C = 8

C = 2

The solution in the form is

y(t) = y_{n}(t) + y_{p}(t)

y(t) = y_{0}e^{-4t} + 2

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