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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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Please answer both with work,please use a paper and picture, thanks ​
Irina-Kira [14]

Answer:

a.  $ \frac{\textbf{17}}{\textbf{4}} $

b.  $ \frac{\textbf{3}}{\textbf{8}} $

Step-by-step explanation:

a. $ \textbf{3} \hspace{1mm} \textbf{+} \hspace{1mm} \textbf{1}\frac{\textbf{1}}{\textbf{4}} $

A mixed fraction of the form $ a\frac{x}{y} = a + \frac{x}{y} $

$ \therefore 3 + 1\frac{1}{4} = 3 + 1 + \frac{1}{4} $

$ = 4 + \frac{1}{4} $

$ = \frac{\textbf{17}}{\textbf{4}} $

b. $ \textbf{2} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{1}\frac{\textbf{5}}{\textbf{8}} $

A mixed fraction of the form $ -c\frac{a}{b} = - c - \frac{a}{b} $

$ \therefore 2 - 1\frac{5}{8} = 2 - 1 - \frac{5}{8} $

$ = 1 - \frac{5}{8} $

$ = \frac{8 - 5}{8} $

$ = \frac{\textbf{3}}{\textbf{8}} $

Hence, the answer.

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Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1
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