Differential Equations Question 274
Question: The equation of the curve passing through the point $ ( 0,\frac{\pi }{4} ) $ whose differential equation is $ sinxcosydx+cosxsinydy=0 $ , is
Options:
A) $ secxsecy=\sqrt{2} $
B) $ cosxcosy=\sqrt{2} $
C) $ \sec x=\sqrt{2}\cos y $
D) $ cosy=\sqrt{2}\sec y $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The given differential equation is $ \sin x\cos ydx+\cos x\sin ydy=0 $
dividing by $ \cos x\cos y\Rightarrow \frac{\sin x}{\cos x}dx+\frac{\sin y}{\cos y}dy=0 $ Integrating, $ \int{\tan xdx+\int{\tan ydy=\log c}} $
Or $ \log \sec x\sec y=\log cor\sec x\sec y=c $ curve passes through the point $ ( 0,\frac{\pi }{4} ) $
$ \sec 0\sec \frac{\pi }{4}=c=\sqrt{2} $ Hence. The required equation of the curve is $ \sec x\sec y=\sqrt{2} $
BETA
