Do you have to solve for the functions

Which of the following systems of linear ODE's is equivalent as the following 3rd order differential equation?

$3 sin\: 3t\: \frac {d^{3} x}{d t^{3}} + 2cos \: 3t\: \frac {d^{2} x}{d t^{2}} - wt\: ln\:t \: \frac{dx}{dt} = e^{xt}\: lnx$

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A.

$y'_{1} = y_{2} + y_{3} + y'_{2}$

$y'_{2} = y_{3} - y_{2} - y'_{1}$

$\large y'_{3} = [(e^{y_{1}t} ln \:y_{1}) - (wt ln\: t )\: y_{2}] (1 + \frac{1}{9}cot^{2}\: 3t) - (\frac {2}{3}tan\: 3t) \: y_{3}$

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B.

$y'_{1} = y_{3}$

$y'_{2} = y_{3} - y_{2} - y_{1}$

$\large y'_{3} = \frac {(e^{y_{1}t} ln \:y_{1}) + (wt ln\: t )\: y_{2}}{3 sin \: 3t } - (\frac {2}{3}cot\: 3t) \: y_{3}$

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C.

$y'_{1} = y_{2}$

$y'_{2} = y_{3}$

$\large y'_{3} = csc^{2} 3t\: [\frac {(e^{y_{1}t} ln \:y_{1}) + (wt ln\: t )\: y_{2}}{3 sin \: 3t }] - (\frac {2}{3}tan\: 3t) \: y_{3}$

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D.

$y'_{1} = y_{2}$

$y'_{2} = y_{3}$

$\large y'_{3} = \frac{1}{3} csc\: 3t [(e^{y_{1}t} ln \:y_{1}) + (wt ln\: t )\: y_{2}] - (\frac {2}{3}cot \: 3t) \: y_{3}$

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Note: $f' (t) = \frac {df}{dt}$