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Dahasolnce [82]
3 years ago
15

Find the general solution of yʹ − 3y = 8e3t + 4sin t .

Mathematics
2 answers:
Arada [10]3 years ago
8 0

Answer:

ye^{-3t} = 8t - \frac{2e^{-3t}}{5}(3sin t + cost)+C'

Step-by-step explanation:

Given differential equation,

y' - 3y = 8e^{3t} + 4sin t

\frac{dy}{dt}-3y = 8e^{3t} + 4sin t

Since, the above equation is of the type of linear differential equation

\frac{dy}{dx}+Py=Q,

In which P = -3, Q = 8e^{3t} + 4sin t,

Thus, the integrating factor,

I.F. = e^{\int (-3) dt}=e^{-3t}

Hence, the solution of the given differential equation would be,

y\times I.F. = \int I.F.\times Q dt

\implies y\times e^{-3t}=\int e^{-3t}\times (8e^{3t} + 4sin t)=\int 8+4e^{-3t} sin tdt

y\times e^{-3t} = \int 8 dt + 4\int e^{-3t} sin tdt

\implies y\times e^{-3t} = 8t + 4\int e^{-3t} sin tdt------(1)

Let,

I=\int e^{-3t} sin tdt-----(2)

Integrating by parts,  ( First term = sin t, second term = e^{-3t} )

I=-\frac{e^{-3t} sint}{3} - \int -\frac{e^{3t} cost}{3} dt+C

I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}-\int -\frac{e^{-3t} sint}{9}dt]+C

I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}+\frac{1}{9}\int e^{-3t} sint dt]+C

I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}+\frac{1}{9} I]+C ( from equation (2))

I+\frac{I}{9}=-\frac{e^{-3t} sint}{3} - \frac{e^{-3t} cost }{9}+C

\frac{10}{9}I=\frac{-3e^{-3t} sint-e^{-3t} cost }{9}+C

I=\frac{1}{10}(-3e^{-3t} sint-e^{-3t} cost)+C' ( where, C' = 9C/10 )

I=-\frac{e^{-3t}}{10}(3sin t + cost)+C'

From equation (1),

y\times e^{-3t} = 8t - \frac{4e^{-3t}}{10}(3sin t + cost)+C'

ye^{-3t} = 8t - \frac{2e^{-3t}}{5}(3sin t + cost)+C'

Blizzard [7]3 years ago
6 0

Answer:

y =C_1e^{3t} +  Ate^{3t}+ Bsint + Ccost

Step-by-step explanation:

{y}^{'} - 3y = 8e^{3t} + 4 sin t

\frac{\mathrm{d} y}{\mathrm{d} t} - 3y = 8e^{3t}+4sint

writing characteristic equation;

( m - 3 ) y = 0

m = 3

y = C_1e^{3t}

for particular solution

y_p = Ate^{3t}+ Bsint + Ccost

hence general solution becomes

y = C.F + P.I

y =C_1e^{3t} +  Ate^{3t}+ Bsint + Ccost

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