The differential equation corresponding to th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of a family of circles touching the $Y$-axis at the origin $(0,0)$ is given by $(x-a)^2 + y^2 = a^2$,where $a$ is the radius of the c

Vedclass · Accepted Answer

$\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$

Vedclass · Answer

(A) The equation of a family of circles touching the $Y$-axis at the origin $(0,0)$ is given by $(x-a)^2 + y^2 = a^2$,where $a$ is the radius of the circle and $(a,0)$ is the center.Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$ $(i)$.Differentiating both sides with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2a$.Thus,$a = x + y \frac{dy}{dx}$ $(ii)$.Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$$2xy \frac{dy}{dx} = y^2 - x^2$$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

The differential equation corresponding to the family of circles in the plane touching the $Y$-axis at the origin is:

Similar Questions

If $Ax^3+Bxy=4$ (where $A$ and $B$ are arbitrary constants) is the general solution of the differential equation $F(x) \frac{d^2 y}{d x^2}+G(x) \frac{d y}{d x}-2 y=0$,then $F(1)+G(1)=$

The order of the differential equation corresponding to the family of parabolas whose axes are along the $X$-axis and whose foci are at the origin,is

The differential equation of the simple harmonic motion given by $x=A \cos (n t+\alpha)$ is

The differential equation for which $y^2 = 4a(x + a)$ (where $a$ is a parameter) is the general solution,is

The differential equation representing the family of circles having their centres on the $Y$-axis is (where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2y}{dx^2}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module