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 normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$. Find the differential equation of the family of all such lines, where $p$ and $\alpha$ are arbitrary constants.

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$, then $f(x)+g(x)+h(x)=$

Verify that the function $y=e^{-3x}$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}-6y=0$.

The differential equation corresponding to the family of parabolas $y^2=4a(x+a)$,where $a$ is the parameter,is

The differential equation having $y=(a+b) e^{cx+d}$ as its general solution,where $a, b, c, d$ are arbitrary constants,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module