question_answer

Differential equation of those circles which passes through origin and their centres lie on y-axis will be

A) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}+2xy=Q\]        

B) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=2xy\]

C) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=xy\]

D) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}+xy=0\]

Correct Answer: B

Solution :

Equation of circle passing through origin and centre lies on y-axis  \[{{(x-0)}^{2}}+\,\,{{(y-k)}^{2}}={{k}^{2}}\,\,\Rightarrow \,\,{{x}^{2}}-{{y}^{2}}+{{k}^{2}}-2ky={{k}^{2}}\] \[\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}-2ky=0\]                                  ... (i) Differentiate equation (i) w.r. to x \[2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0\] \[K\,\frac{dy}{dx}=x+y\frac{dy}{dx}\Rightarrow \,\,k=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\]              ? (ii) Put the value of k in (i) \[{{x}^{2}}+{{y}^{2}}-2\left( \frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}} \right)\,\,y=0\] Solving this equation \[({{x}^{2}}-{{y}^{2}})\,\frac{dy}{dx}=2xy\]