Differential Equations Question 27
Question: The solution of the differential equation $ \cos y\log (\sec x+\tan x)dx=\cos x\log (\sec y+\tan y)dy $ is
[AI CBSE 1990]
Options:
A) $ {{\sec }^{2}}x+{{\sec }^{2}}y=c $
B) $ \sec x+\sec y=c $
C) $ \sec x-\sec y=c $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ \cos y\log (\sec x+\tan x)dx=\cos x\log (\sec y+\tan y)dy $
Therefore $ \int _{{}}^{{}}{\sec y\log (\sec y+\tan y)dy} $
$ =\int _{{}}^{{}}{\sec x\log (\sec x+\tan x)dx} $
Put $ \log (\sec x+\tan x)=t $ and $ \log (\sec y+\tan y)=z $
$ \frac{{{[\log (\sec x+\tan x)]}^{2}}}{2}=\frac{{{[\log (\sec y+\tan y)]}^{2}}}{2}+c $ .
BETA
