# Y Sec 2 X + ( Y + 7 ) Tan X D Y D X = 0 - Mathematics

Sum

y sec^2 x + (y + 7) tan x(dy)/(dx)=0

#### Solution

We have,

$y \sec^2 x + \left( y + 7 \right)\tan x\frac{dy}{dx} = 0$

$\Rightarrow y \sec^2 x = - \left( y + 7 \right)\tan x\frac{dy}{dx}$

$\Rightarrow \left( \frac{- y - 7}{y} \right)dy = \frac{\sec^2 x}{\tan x}dx$

$\Rightarrow \left( - 1 - \frac{7}{y} \right)dy = \frac{\sec^2 x}{\tan x}dx$

Integrating both sides, we get

$\int\left( - 1 - \frac{7}{y} \right)dy = \int\frac{\sec^2 x}{\tan x}dx$

$\Rightarrow - y - 7\log \left| y \right| = \log \left| \tan x \right| + \log C$

$\Rightarrow - y = \log \left| \tan x \right| + \log\left| y^7 \right| + \log C$

$\Rightarrow - y = \log\left| C y^7 \tan x \right|$

$\Rightarrow e^{- y} = C y^7 \tan x$

$\Rightarrow y^7 \tan x = \frac{e^{- y}}{C}$

$\Rightarrow y^7 \tan x = k e^{- y},\text{ where }k = \frac{1}{C}$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 46 | Page 146