(d^2 +a^2 )y=tan ax by the method of variation of parameters


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Question Number 47035 by 23kpratik last updated on 04/Nov/18

$(d^{2} + a^{2}) y = t a n a x b y t h e m e t h o d o f v a r i a t i o n o f p a r a m e t e r s$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

$p l s c l a r i f y d m e a n s \frac{d}{d x} i s i t . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Nov/18

	$y = \frac{1}{D^{2} + a^{2}} t a n a x$ $= \frac{1}{D^{2} + a^{2}} \times \frac{2 i s i n a x}{i \times 2 c o s a x}$ $= \frac{1}{D^{2} + a^{2}} \times \frac{e^{i a x} - e^{- i a x}}{i \times (e^{i a x} + e^{- i a x})}$ $e^{i a x} = c o s a x + i s i n a x$ $e^{- i a x} = c o s a x - i s i n a x$ $y = \frac{1}{(D + i a) (D - i a)} \times \frac{1}{i} \times \frac{e^{i 2 a x} - 1}{e^{i 2 a x} + 1}$ $= w a i t . . .$
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