The differential equation representing the fa | vedclass

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

Question

(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a$ is an arbitrary consta

Vedclass · Accepted Answer

$x \frac{dy}{dx} - 2y = 0$

Vedclass · Answer

(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a$ is an arbitrary constant.To find the differential equation,we differentiate both sides with respect to $x$:$\frac{d}{dx}(x^2) = \frac{d}{dx}(4ay)$$2x = 4a \frac{dy}{dx}$$x = 2a \frac{dy}{dx}$From this,we get $2a = \frac{x}{dy/dx}$.Substitute $2a$ back into the original equation $x^2 = 2a(2y)$:$x^2 = \left( \frac{x}{dy/dx} \right) (2y)$$x^2 \frac{dy}{dx} = 2xy$$x \frac{dy}{dx} = 2y$$x \frac{dy}{dx} - 2y = 0$.Thus,the correct option is $A$.

The differential equation representing the family of parabolas having vertex at the origin and axis along the positive $Y$-axis is

Similar Questions

The order of the differential equation obtained by eliminating arbitrary constants in the family of curves $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$ is

The difference between the degree and the order of the differential equation that represents the family of curves given by $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right)$,where $a>0$,is:

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

The differential equation satisfied by the family of curves $y = ax \cos \left( \frac{1}{x} + b \right)$,where $a$ and $b$ are parameters,is

The differential equation corresponding to all the circles lying in the first quadrant and touching the coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module