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  • The differential equation of the family of curves x^{2}+y^{2}-2 a y=0, where a is arbitrary constant, is (a) (x^{2}-y^{2} \frac{d y}{d x}=2 x y

The differential equation of the family of curves x^{2}+y^{2}-2 a y=0, where a is arbitrary constant, is
(a) \left(x^{2}-y^{2}\right) \frac{d y}{d x}=2 x y
(b) 2\left(x^{2}+y^{2}\right) \frac{d y}{d x}=x y
(c) 2\left(x^{2}-y^{2}\right) \frac{d y}{d x}=x y
(d) \left(x^{2}+y^{2}\right) \frac{d y}{d x}=2 x y

Answers (1)

The answer is the option (a)  \left(x^{2}-y^{2}\right) \frac{d y}{d x}=2 x y

Given:

x^{2}+y^{2}-2 a y=0.....(i)

\\2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0 \\\\ a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}

\\Put \ the \ value \ of 'a' \ in \ Eq. (i) \\ x^{2}+y^{2}-2 y \frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}=0 \\\\ \left(x^{2}+y^{2}\right) \frac{d y}{d x}-2 x y-2 y^{2} \frac{d y}{d x}=0 \\ \\\left(x^{2}-y^{2}\right) \frac{d y}{d x}-2 x y=0

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