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# The differential equation of the family of curves $x^2+y^2-2ay=0$,where a is arbitrary constant,is$(A)\;(x^2-y^2)\frac{dy}{dx}=2xy \quad (B)\;2(x^2+y^2)\frac{dy}{dx}=xy \quad (C)\;2(x^2-y^2)\frac{dy}{dx}=xy \quad (D)\;2(x^2+y^2)\frac{dy}{dx}=2xy$

Toolbox:
• The general solution of a differential equation is a relation between dependent and independent variables having $'n'$ arbitary constant.
• A general solution may have more than one form but arbitary constants must be the same.
Given $x^2+y^2-2ay=0$-----(1)
Differentiating w.r.t x to x we get
$2x+2y.\large\frac{dy}{dx}$$-2a.\large\frac{dy}{dx}=0 Dividing through out by 2 we get x+\large\frac{dy}{dx}$$(y-a)=0$