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) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}+xy=0\]
Correct Answer: A
Solution :
The system of circles pass through origin and centre lies on y-axis is \[{{x}^{2}}+{{y}^{2}}-2ay=0\] Þ \[2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0\] Þ \[2a=2y+2x\frac{dx}{dy}\] Therefore, the required differential equation is \[{{x}^{2}}+{{y}^{2}}-2{{y}^{2}}-2xy\frac{dx}{dy}=0\]Þ\[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}-2xy=0\].You need to login to perform this action.
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