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Question Number 115000 by mnjuly1970 last updated on 22/Sep/20

$. . . . n i c e m a t h . . .$ $i f y = {(c o s (2 x))}^{- \frac{1}{2}} t h e n$ $p r o v e :: y + y^{^{″}} = 3 y^{5}$ $. . . m . n . j u l y . 1970 . . .$
	Answered by Dwaipayan Shikari last updated on 22/Sep/20

	$y = \frac{1}{\sqrt{c o s 2 x}}$ $y^{'} = {c o s}^{- \frac{3}{2}} 2 x . s i n 2 x$ $y^{″} = 3 {c o s}^{- \frac{5}{2}} 2 x . {s i n}^{2} 2 x + 2 {c o s}^{- \frac{1}{2}} 2 x$ $y + y^{″} = 3 ({c o s}^{- \frac{5}{2}} 2 x (1 - {c o s}^{2} 2 x) + {c o s}^{- \frac{1}{2}} 2 x)$ $y + y^{″} = 3 {c o s}^{- \frac{5}{2}} 2 x - 3 {c o s}^{- \frac{1}{2}} 2 x + 3 {c o s}^{- \frac{1}{2}} 2 x$ $y + y^{″} = 3 y^{5}$
	Commented by mnjuly1970 last updated on 22/Sep/20

	$v e r y n i c e v e r y n i c e . . . t h a n k$ $y o u s i r . . .$
	Commented by Canovas last updated on 29/Sep/20

	$C o r r e c t$
	Answered by $@y@m last updated on 23/Sep/20
	$y=(1/( (√(cos 2x)))) y^2 =sec2 x ...(1) 2yy′=2sec2xtan 2x yy′=sec2xtan 2x yy′=y^2 tan2 x y′=ytan 2x ...(2) y′′=2ysec^2 2x+tan2 x.y′ y′′=2y.y^4 +tan 2x.ytan 2x {using(1) &(2) y′′=2y^5 +ytan^2 2x y′′=2y^5 +y(sec^2 2x−1) y′′=2y^5 +y(y^4 −1) y+y′′=3y^5$
	$y = \frac{1}{\sqrt{\cos 2 x}}$ $y^{2} = \sec 2 x . . . (1)$ $2 {y y}^{'} = 2 \sec 2 x \tan 2 x$ ${y y}^{'} = \sec 2 x \tan 2 x$ ${y y}^{'} = y^{2} \tan 2 x$ $y^{'} = y \tan 2 x . . . (2)$ $y^{″} = 2 y \sec^{2} 2 x + \tan 2 x . y^{'}$ $y^{″} = 2 y . y^{4} + \tan 2 x . y \tan 2 x {u s i n g (1) & (2)$ $y^{″} = 2 y^{5} + y \tan^{2} 2 x$ $y^{″} = 2 y^{5} + y (\sec^{2} 2 x - 1)$ $y^{″} = 2 y^{5} + y (y^{4} - 1)$ $y + y^{″} = 3 y^{5}$
	Commented by mnjuly1970 last updated on 23/Sep/20

	$v e r y n i c e a w e s o m s o l u t i o n$
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