Question

`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`

Answer 1

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

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

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

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

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

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

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

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