Differential Equations Question 305
Question: The differential equation $ (1+y^{2})xdx-(1+x^{2})ydy=0 $ Represents a family of:
Options:
A) Ellipses of constant eccentricity
B) Ellipses of variable eccentricity-
C) Hyperbolas of constant eccentricity
D) Hyperbolas of variable eccentricity
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given $ \frac{xdx}{1+x^{2}}=\frac{ydy}{1+y^{2}} $ Integrating we get, $ \frac{1}{2}\log (1+x^{2})=\frac{1}{2}\log (1+y^{2})+a $
$ \Rightarrow 1+x^{2}=c(1+y^{2}),c=e^{2a} $
$ x^{2}-cy^{2}=c-1\Rightarrow \frac{x^{2}}{c-1}-\frac{y^{2}}{( \frac{c-1}{c} )}=1 $
- (1) Clearly $ c>0 $ as $ c=e^{2a} $ Hence, the equation (i) gives a family of hyperbolas with eccentricity $ =\sqrt{\frac{c-1+\frac{c-1}{c}}{c-1}}=\sqrt{\frac{c^{2}-1}{c-1}}=\sqrt{c+1} $ if $ c\ne 1 $ Thus eccentricity varies form number to member of the family as it depends on c. If c=1, it is a pair of lines $ x^{2}-y^{2}=0 $
BETA
