# Find the general solution $\sec^2x\;\tan y\;dx+\sec^2y\;\tan x\;dy=0$

$\begin{array}{1 1} \tan y \tan x = C \\ \Large\frac{\tan y}{\tan x} = \normalsize C \\ \Large\frac{\tan x}{\tan y} = \normalsize C \\1 - \Large\frac{\tan y}{\tan x} = \normalsize C\end{array}$

Toolbox:
• $\int\tan x = \sec^2x$
Step 1:
Given : $\sec^2x\tan ydx + \sec^2y\tan x dy = 0$
$\sec^2y\tan xdy = - \sec^2x\tan ydx$
Now seperating the variables we get
$\large\frac{\sec^2y dy}{\tan y} = -\frac{ \sec^2x dx}{\tan x}$
Step 2:
Let $\tan y$ be t hence $dt = \sec^2ydy$ and let $u = \tan x$ and $dt = \sec^2x dx$
Substituting this and integrating we get
$\int\large\frac{ dt}{t} =- \int\large\frac{ du}{u}$
$\log t = - \log u + \log C$
$\log t + \log u =\ log C$
$\log tu = \log C$
$tu = C$
Step 3:
substituting for t and u we get
$\tan y \tan x = C$
This is the required general solution.