Differential Equations Question 118
Question: The differential equation of all circles which passes through the origin and whose centre lies on y-axis, is
[MNR 1986; DCE 2000]
Options:
A) $ (x^{2}-y^{2})\frac{dy}{dx}-2xy=0 $
B) $ (x^{2}-y^{2})\frac{dy}{dx}+2xy=0 $
C) $ (x^{2}-y^{2})\frac{dy}{dx}-xy=0 $
D) $ (x^{2}-y^{2})\frac{dy}{dx}+xy=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
The system of circles pass through origin and centre lies on y-axis is $ x^{2}+y^{2}-2ay=0 $
Therefore $ 2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0 $
Therefore $ 2a=2y+2x\frac{dx}{dy} $
Therefore, the required differential equation is $ x^{2}+y^{2}-2y^{2}-2xy\frac{dx}{dy}=0 $
Therefore $ (x^{2}-y^{2})\frac{dy}{dx}-2xy=0 $ .
BETA
