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Integration by Partial Fractions

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Question

Find the second order derivatives of $e^{6 x} cos 3 x$

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Solution

Let $y = e^{6 x} cos 3 x$
$\frac{d y}{d x} = \frac{d}{d x} (e^{6 x} . cos 3 x) = cos 3 x . \frac{d}{d x} (e^{6 x}) + e^{6 x} . \frac{d}{d x} (cos 3 x)$
$= cos 3 x . e^{6 x} \frac{d}{d x} (6 x) + e^{6 x} . (- sin 3 x) . \frac{d}{d x} (3 x)$
$= 6 e^{6 x} cos 3 x - 3 e^{6 x} sin 3 x$ ....(1)
$∴$ $\frac{d^{2} x}{d y^{2}} = \frac{d}{d x} (6 e^{6 x} cos 3 x - 3 e^{6 x} sin 3 x) = 6 \frac{d}{d x} (e^{6 x} cos 3 x) - 3 \frac{d}{d x} (e^{6 x} sin 3 x)$
$= 6 [6 e^{6 x} cos 3 x - 3 e^{6 x} sin 3 x] - 3 [sin 3 x \frac{d}{d x} (e^{6 x}) + e^{6 x} \frac{d}{d x} (sin 3 x)$
$= 36 e^{6 x} cos 3 x - 18 e^{6 x} sin 3 x - 3 [sin 3 x . e^{6 x} .6 + e^{6 x} cos 3 x .3]$
$= 27 e^{6 x} cos 3 x - 36 e^{6 x} sin 3 x = 9 e^{6 x} (3 cos 3 x - 4 sin 3 x)$

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