Question

The solution of the differential equation ${\sec }^{2}x\cdot \tan ydx+{\sec }^{2}y\cdot \tan xdy=0$ is

• A. $\tan x\cdot \cot y=C$
• B. $\cot x\cdot \tan y=C$
• C. $\tan x\cdot \tan y=C$
• D. $\sin x\cdot \cos y=C$

Answer 1

Answer: (C)

$\tan x\cdot \tan y=C$

Given, ${\sec }^{2}x\cdot \tan ydx+{\sec }^{2}y\cdot \tan xdy=0$

On separating the varaibales, we get

$\Rightarrow {\sec }^{2}x\cdot \tan ydx=-{\sec }^{2}y\cdot \tan xdy$

$\Rightarrow \frac{{\sec }^{2}x}{\tan x}dx=-\frac{{\sec }^{2}y}{\tan y}dy$

On integration both the sides, we get

$\int \frac{{\sec }^{2}x}{\tan x}dx=-\int \frac{{\sec }^{2}y}{\tan y}dy$

Let $\tan x=u\Rightarrow {\sec }^{2}x=\frac{du}{dx}$

$\Rightarrow dx=\frac{du}{{\sec }^{2}x}$ and $\tan y=v$

$\Rightarrow {\sec }^{2}y=\frac{dv}{dy}$

$\Rightarrow dy=\frac{dv}{{\sec }^{2}y}$

$\therefore \int \frac{{\sec }^{2}x}{u}\cdot \frac{du}{{\sec }^{2}x}=-\int \frac{{\sec }^{2}y}{v}\cdot \frac{dv}{{\sec }^{2}y}$

$\Rightarrow \int \frac{du}{u}=\int \frac{dv}{v}$

$\Rightarrow \log |u|=-\log |v|+\log |C|$

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

$\Rightarrow \log |\tan x\cdot \tan y|=\log |C|$

$(\because \log m+\log n=\log mn)$

$(\because \log m=\log n\Rightarrow m=n)$

$\Rightarrow \tan x\cdot \tan y=C$