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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The differential equation of the family of curves $x^2+y^2-2ay=0$,where a is arbitrary constant,is\[(A)\;(x^2-y^2)\frac{dy}{dx}=2xy \quad (B)\;2(x^2+y^2)\frac{dy}{dx}=xy \quad (C)\;2(x^2-y^2)\frac{dy}{dx}=xy \quad (D)\;2(x^2+y^2)\frac{dy}{dx}=2xy\]

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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$
or $x+y \large\frac{dy}{dx}$$-a \large \frac{dy}{dx}=0$
$=>a=\large\frac{x+y(\Large\frac{dy}{dx})}{\Large\frac{dy}{dx}}$
Substitute the value a in equ (1)
$x^2+y^2=2y \bigg(\large\frac{x+y(\Large\frac{dy}{dx})}{\Large\frac{dy}{dx}}\bigg)$
$=>\large\frac{dy}{dx}$$(x^2+y^2)$$=-2xy$
Hence option $D$ is the correct answer
answered May 17, 2013 by meena.p
 

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