Question

y sec² x + (y + 7) tan x $\frac{dy}{dx}$ = 0

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y{\sec }^{2}x+\left(y+7\right)\tan x\frac{dy}{dx}=0 \\ \Rightarrow y{\sec }^{2}x=-\left(y+7\right)\tan x\frac{dy}{dx} \\ \Rightarrow \left(\frac{-y-7}{y}\right)dy=\frac{{\sec }^{2}x}{\tan x}dx \\ \Rightarrow \left(-1-\frac{7}{y}\right)dy=\frac{{\sec }^{2}x}{\tan x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \left(-1-\frac{7}{y}\right)dy=\int \frac{{\sec }^{2}x}{\tan x}dx \\ \Rightarrow -y-7\log \left|y\right|=\log \left|\tan x\right|+\log C \\ \Rightarrow -y=\log \left|\tan x\right|+\log \left|y^7\right|+\log C \\ \Rightarrow -y=\log \left|C{y}^7\tan x\right| \\ \Rightarrow e^{-y}=C{y}^7\tan x \\ \Rightarrow y^7\tan x=\frac{e^{-y}}{C} \\ \Rightarrow y^7\tan x=k{e}^{-y},\mathrm{where}k=\frac{1}{C} \end{gathered}$$