The differential equation of the family of ci | vedclass

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

Question

(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a p

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a parameter. From this equation,we have $2h = \frac{x^{2} + y^{2}}{x}$. Differentiating the equation $x^{2} + y^{2} - 2hx = 0$ with respect to $x$,we get: $2x + 2y \frac{dy}{dx} - 2h = 0$. Substituting the value of $2h$ from the first equation: $2x + 2y \frac{dy}{dx} - \left( \frac{x^{2} + y^{2}}{x} \right) = 0$. Multiplying the entire equation by $x$: $2x^{2} + 2xy \frac{dy}{dx} - x^{2} - y^{2} = 0$. Simplifying the expression: $x^{2} - y^{2} + 2xy \frac{dy}{dx} = 0$,which can be rewritten as $y^{2} = x^{2} + 2xy \frac{dy}{dx}$.

The differential equation of the family of circles passing through the origin and having their centres on the $x$-axis is

Similar Questions

The differential equation obtained from the function $y=a(x-a)^{2}$ is

By eliminating the arbitrary constants from $y=(a+b) \sin (x+c)-d e^{x+e+f}$,the differential equation obtained is of order

The differential equation representing the family of circles of constant radius $r$ is

The differential equation of all the ellipses centered at the origin and having axes as the coordinate axes is

The differential equation for the family of curves ${x^2} + {y^2} - 2ay = 0,$ where $a$ is an arbitrary constant,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module