Question

Find the solution of $y{\sec }^{2}x+(y+7)\tan x\frac{dy}{dx}=0$

• A. $y^7\tan x=k{e}^{-y}$
• B. $y^5\tan x=k{e}^{-y}$
• C. $y^5\tan x=k{e}^y$
• D. $y^7\tan x=k{e}^y$

Answer 1

The correct option is A $y^7\tan x=k{e}^{-y}$

$y{\sec }^{2}x+(y+7)\tan x\frac{dy}{dx}=0$
$\frac{1}{{\cos }^{2}x.\tan x}=-\left(\frac{y+7}{y}\right)\frac{dy}{dx}$
$\frac{2}{\sin 2x}=-(1+\frac{7}{y})\frac{dy}{dx}$
$2cosec2xdx=-(1+\frac{7}{y})dy$
Integrating,
$2\int cosec2xdx=-\int (1+\frac{7}{y})dy$
$2.\frac{\log |tanx|}{2}=-y-7\log y+c$
$\log \tan x+7\log y+y=c$
$\log y^7\tan x{e}^y=\log k$
$y^7\tan x=k{e}^{-y}$