Differential Equations Question 299
Question: Equation of curve through point $ (1,0) $ which satisfies the differential equation $ (1+y^{2})dx-xydy=0 $ , is
[WB JEE 1986]
Options:
A) $ x^{2}+y^{2}=1 $
B) $ x^{2}-y^{2}=1 $
C) $ 2x^{2}+y^{2}=2 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \frac{dx}{x}=\frac{ydy}{1+y^{2}} $
Integrating, we get $ \log |x|=\frac{1}{2}\log (1+y^{2})+\log c $
or $ |x|=c\sqrt{(1+y^{2})} $
But it passes through (1, 0), so we get $ c=1 $
$ \therefore $ Solution is $ x^{2}=y^{2}+1 $ or $ x^{2}-y^{2}=1 $ .
BETA
