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Question Number 4925 by sanusihammed last updated on 22/Mar/16

I f y = {(s e c x + t a n x)}^{p} . w h e r e p i s a c o n s t a n t

p r o v e t h a t c o s x d y / d x - p y = 0

Answered by prakash jain last updated on 22/Mar/16

\frac{d y}{d x} = p {(\sec x + \tan x)}^{p - 1} (\sec x \tan x + \sec^{2} x)

\frac{d y}{d x} = p {(\sec x + \tan x)}^{p - 1} \sec x (\sec x + \tan x)

\frac{d y}{d x} = p \sec x {(\sec x + \tan x)}^{p}

\frac{d y}{d x} = p \sec x y

\cos x \frac{d y}{d x} - p y = 0

Answered by Rojaye Shegz last updated on 30/Mar/16

W e s t a r t h e r e;

\int \sec x d x = \log (\sec x + \tan x)

p \int \sec x d x = \log {(\sec x + \tan x)}^{p}

p \int \sec x d x = \log y

p \times \frac{d}{d x} (\int \sec x d x) = \frac{1}{y} \frac{d y}{d x}

p \times y \times \sec x = \frac{d y}{d x}

p y = \cos x \frac{d y}{d x}

∴ \cos x \frac{d y}{d x} - p y = 0