A) \[{{x}^{2}}-{{y}^{2}}-2ax=0\]
B) \[{{y}^{2}}-{{x}^{2}}=2xyy\]
C) \[{{x}^{2}}+{{y}^{2}}+2y=0\]
D) none of the above
Correct Answer: B
Solution :
The given equation is \[{{x}^{2}}+{{y}^{2}}-2ax=0\] ?(i) On differentiating Eq. (i) with respect to\[x,\] we get \[2x+2y\frac{dy}{dx}-2a=0\] \[\Rightarrow \] \[a=x+y\frac{dy}{dx}\] On putting the value of a in Eq. (i), we get \[{{x}^{2}}+{{y}^{2}}-2x\left( x+y\frac{dy}{dx} \right)=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2{{x}^{2}}-2xy\frac{dy}{dx}=0\] \[\Rightarrow \]\[{{y}^{2}}-{{x}^{2}}-2xyy=0\] \[\left( \because \,y=\frac{dy}{dx} \right)\] \[\Rightarrow \]\[{{y}^{2}}-{{x}^{2}}=2xyy\] Which is required differential equation. Note: To find the differential equation of any given equation, we have to remove parameters.
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