The differential equation satisfied by the sy | vedclass

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

Question

(C) Given the equation of the parabola: $y^{2} = 4ax + 4a^{2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = 4a$$\Rightarrow a =

Vedclass · Accepted Answer

$y\left(\frac{dy}{dx}\right)^{2} + 2x\left(\frac{dy}{dx}\right) - y = 0$

Vedclass · Answer

(C) Given the equation of the parabola: $y^{2} = 4ax + 4a^{2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = 4a$$\Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.Substitute the value of $a$ back into the original equation:$y^{2} = 4\left(\frac{y}{2} \frac{dy}{dx}ight)x + 4\left(\frac{y}{2} \frac{dy}{dx}ight)^{2}$.Simplify the equation:$y^{2} = 2xy \frac{dy}{dx} + 4 \cdot \frac{y^{2}}{4} \left(\frac{dy}{dx}ight)^{2}$.$y^{2} = 2xy \frac{dy}{dx} + y^{2} \left(\frac{dy}{dx}ight)^{2}$.Dividing by $y$ (assuming $y 
eq 0$):$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}ight)^{2}$.Rearranging the terms,we get:$y \left(\frac{dy}{dx}ight)^{2} + 2x \frac{dy}{dx} - y = 0$.

The differential equation satisfied by the system of parabolas $y^{2} = 4a(x + a)$ is

Similar Questions

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

The differential equation representing the family of curves $x \sin x + y^3 = 4ax$ is

The differential equation satisfied by the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$ is

Form the differential equation of the family of hyperbolas having foci on the $x$-axis and centre at the origin.

If $c$ is a parameter,then the differential equation of the family of curves $x^2=c(y+c)^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module