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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

The differential equation for the family of curves $x^2+y^2-2ay=0$ where a is an arbitary constant is given $y'= \large\frac{dy}{dx}$

$(a)\;(x^2-y^2)y'=2 xy \\ (b)\;2 (x^2+y^2)y'=xy \\ (c)\;2(x^2-y^2)y'=xy \\ (d)\;(x^2+y^2)y'=2xy $

1 Answer

$x^2+y^2-2ay=0$
$2x+2y \large\frac{dy}{dx}$$-2 ay^{1}$$\large\frac{dy}{dx}$$=y^1$
$x+yy'=ay'$
$a= \large\frac{x}{y'}$$+y$
$x^2+y^2-2 \bigg(\large\frac{x}{y'}+y \bigg)y=0$
$x^2-y^2-\large\frac{2xy}{y'}$$=0$
$(x^2-y^2)y'= 2xy$
Hence a is the correct answer.
answered Feb 4, 2014 by meena.p
 

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