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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation of all circles which pass through origin and whose centers lie on y-axis as shown below:

$\begin{array}{1 1} (A)\; (x^2-y^2)\large\frac{dy}{dx}-2xy=0 \\ (B)\; (x^2+y^2)\large\frac{dy}{dx}-2xy=0 \\ (C)\; (x^2-y^2)\large\frac{dy}{dx}+2xy=0 \\ (D)\; (x^2+y^2)\large\frac{dy}{dx}+2xy=0\end{array} $

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1 Answer

+1 vote
Toolbox:
  • Equation of a family of circles in $(x-h)^2 +(y-k)^2=a^2$ where $(h,k)$ are the centers and $a$ is the radius.
  • If the given equation has 'n' arbitary constants, then the given equation will be of h order
We are asked to form the differential equations of all circles which pass through the orgin and whose centers lies on y-axis
Since it is given that the center lies on the y-axis, the sketch of the circle is as shown
The equation of the family of circles touching the x-axis at the origin is
$(x-0)^2+(y-a)^2=a^2$
=>$x^2+y^2-2ay=a^2$------(1)
Where a is a parameter
This equation has only one arbitary constant, hence let us differentiate it once w.r.t $x$
$2x+2y.\large\frac{dy}{dx}$$-2a \large\frac{dy}{dx}=$$0$
$=>a=\large\frac{x+y(dy/dx)}{dy/dx}$
Substituting this value of a in equ(1) we get,
$x^2+y^2=2y \bigg(\large\frac{x+y(dy/dx)}{dy/dx}\bigg)$
On simplification we get
=>$ (x^2-y^2)\large\frac{dy}{dx}$$=2xy$
or $ (x^2-y^2)\large\frac{dy}{dx}$$-2xy=0$ is the required equation
answered May 8, 2013 by meena.p
edited Mar 24, 2014 by thanvigandhi_1
 

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