The differential equation of all parabolas wh | vedclass

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

Question

(C) The equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^2 = 4a(y-k)$. Since the axis is the $y$-axis,the vertex lies

Vedclass · Accepted Answer

$x \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$

Vedclass · Answer

(C) The equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^2 = 4a(y-k)$. Since the axis is the $y$-axis,the vertex lies on the $y$-axis,so $h=0$. Thus,the equation is $x^2 = 4a(y-k)$.Differentiating with respect to $x$:$2x = 4a \frac{dy}{dx} \implies 4a = \frac{2x}{dy/dx}$.Differentiating again with respect to $x$:$2 = 4a \frac{d^2y}{dx^2}$.Substituting the value of $4a$ from the first derivative into the second derivative equation:$2 = \left( \frac{2x}{dy/dx} \right) \frac{d^2y}{dx^2}$.Simplifying:$1 = \frac{x}{dy/dx} \cdot \frac{d^2y}{dx^2}$$\implies \frac{dy}{dx} = x \frac{d^2y}{dx^2}$$\implies x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

The differential equation of all parabolas whose axis is the $y$-axis is:

Similar Questions

The differential equation of $y = \sec(\tan^{-1}x)$ is

Verify that the given function $y = Ax$ is a solution of the differential equation $xy' = y$ $(x \neq 0)$.

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

If $\frac{d^2y}{dx^2} = 0$,then:

The order of the differential equation whose solution is $y=a \cos x+b \sin x+c e^{-x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module