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Q.
The solution of the differential equation $\sec ^{2} x \cdot \tan y d x+\sec ^{2} y \cdot \tan x d y=0$ is
Bihar CECEBihar CECE 2013
Solution:
Given, $\sec ^{2} y \cdot \tan y d x+\sec ^{2} y \cdot \tan x d y=0$
On separating the variables, we get
$\Rightarrow \sec ^{2} x \cdot \tan y d x=-\sec ^{2} y \cdot \tan x d y$
$\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$
On integrating both the sides, we get
$\int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{2} x}{\tan y} d y$
Let $\tan x=u$
$\Rightarrow \sec ^{2} x=\frac{d u}{d x}$
$\Rightarrow d x=\frac{d u}{\sec ^{2} y}$ and $\tan y=v$
$\Rightarrow \sec ^{2} y=\frac{d v}{d y}$
$\Rightarrow d y=\frac{d u}{\sec ^{2} y}$
$\int \frac{\sec ^{2} x}{u} \cdot \frac{d u}{\sec ^{2} x}=-\int \frac{\sec ^{2} y}{v} \cdot \frac{d v}{\sec ^{2} y}$
$\Rightarrow \int \frac{d u}{u}=-\int \frac{d v}{v}$
$\Rightarrow \log |x|=-\log |v|+\log |c|$
$\Rightarrow \log |\tan x|=-\log |\tan y|+\log |c|$
$\Rightarrow \log |\tan x \cdot \tan y|=\log |c|$
$(\because \log m+\log n=\log m n)$
$\Rightarrow (\because \log m=\log n \Rightarrow m=n)$
which is the required general solution.