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

$A$ differential equation representing the family of parabolas with axis parallel to the $y$-axis and whose length of latus rectum is the distance of the point $(2, -3)$ from the line $3x + 4y = 5$,is given by:

The differential equation for which $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$ is the general solution,is

The differential equation for the line $y = mx + c$ is (where $m$ and $c$ are arbitrary constants).

The differential equation of the family of all circles of radius $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module