The particular solution of the differential equation \( \sec^{2}x\tan y\,dx-\sec^{2}y\tan x\,dy=0 \), \( y=\displaystyle\frac{\pi}{4} \) when \( x=\displaystyle\frac{\pi}{4} \) , is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \tan y=\tan x \)
Dividing by \( \tan x\tan y \) separates the variables:
\( \displaystyle \frac{\sec^2x}{\tan x}\,dx=\frac{\sec^2y}{\tan y}\,dy \)
Each side is of the form \( \frac{f'}{f} \), so integrating:
\( \displaystyle \log|\tan x|=\log|\tan y|+\log|C| \)
\( \displaystyle \tan x=C\tan y \)
Putting \( x=y=\frac{\pi}{4} \): \( 1=C\cdot 1 \), so C = 1.
\( \displaystyle \tan y=\tan x \)
Shortcut: Put \( x=y=\frac{\pi}{4} \) into the options: Option (2) gives 1 = −1 and Option (3) gives \( \tan\frac{\pi^2}{16}=-1 \), both false. Options (1) and (4) both pass, so check the equation with Option (4): y = x gives
\( \sec^2x\tan x\,dx-\sec^2x\tan x\,dx=0 \),
which holds.