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Tammy Todd

Answered question

2021-02-21

Solve the given system of differential equations.
$D x + D y + (D + 1) z = 0$

$D x + y = e^{t}$

$D x + y - 2 z = 50 \sin (2 t)$

krolaniaN

Skilled2021-02-22Added 86 answers

The given system of differential equations is,

D x + D y + (D + 1) z = 0

D x + y = e^{t}

D x + y - 2 z = 50 \sin (2 t)

Putting the value of

D x + y

in the third equation, we get,

D x + y - 2 z = 50 \sin (2 t)

\Rightarrow e^{t} - 2 z = 50 \sin (2 t)

\Rightarrow z = \frac{1}{2} e^{t} - 25 \sin (2 t)

Now differentiating both sides with respect to t, we get,

D z = \frac{1}{2} e^{t} - 50 \cos (2 t)

\Rightarrow (D + 1) z = D z + z = (\frac{1}{2} e^{t} - 50 \cos (2 t)) + \frac{1}{2} e^{t} - 25 \sin (2 t)

\Rightarrow (D + 1) z = e^{t} - 25 \sin (2 t) - 50 \cos (2 t)

Putting

D x = e^{t} - y

and the value of

(D + 1) z

in first equation, we get,

e^{t} - y + D y + e^{t} - 25 \sin 2 t - 50 \cos 2 t = 0

\Rightarrow D y - y = 25 \sin 2 t + 50 \cos 2 t - 2 e^{t}

So the integrating factor of this differential equation is,

I . F = e^{\int (- 1) d t}

= e^{- t}

Hence the solution is,

y e^{- t} = \int (25 \sin 2 t + 50 \cos 2 t - 2 e^{t}) e^{- t} d t

= \int 25 \sin 2 t e^{- t} d t + \int 50 \cos 2 t e^{- t} d t - 2 \int e^{t} e^{- t} d t

= - 10 e^{- t} \cos 2 t - 5 e^{- t} \sin 2 t + 20 e^{- t} \sin 2 t - 10 e^{- t} \cos 2 t - 2 t + C

\Rightarrow y = - 10 \cos 2 t - 5 \sin 2 t + 20 \sin 2 t - 10 \cos 2 t - 2 t e^{t} + C e^{t}

\Rightarrow y = - 20 \cos 2 t + 15 \sin 2 t - 2 t e^{t} + C e^{t}

Now putting this value of

y

in the given second differential equation, we get:

D x = e^{t} - y

= e^{t} - (- 20 \cos 2 t + 15 \sin 2 t - 2 t e^{t} + C e^{t})

= e^{t} + 20 \cos 2 t - 15 \sin 2 t + 2 t e^{t} - C e^{t}

Integrating both sides, we get,

x = \int (e^{t} + 20 \cos 2 t - 15 \sin 2 t + 2 t e^{t} - C e^{t}) d t

= e^{t} + 10 \sin 2 t + \frac{15}{2} \cos 2 t + 2 t e^{t} - 2 e^{t} - C e^{t} + D

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