The differential equation corresponding to th | vedclass

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

Question

(A) The equation of a parabola with axis parallel to the $y$-axis and axis of symmetry $x=1$ is given by $(x-1)^2 = 4a(y-k)$,where $a$ and $k$ are arb

Vedclass · Accepted Answer

$(x-1) \frac{d^2 y}{d x^2} - \frac{d y}{d x} = 0$

Vedclass · Answer

(A) The equation of a parabola with axis parallel to the $y$-axis and axis of symmetry $x=1$ is given by $(x-1)^2 = 4a(y-k)$,where $a$ and $k$ are arbitrary constants.Alternatively,we can write this as $y = A(x-1)^2 + B$,where $A$ and $B$ are arbitrary constants.Differentiating with respect to $x$ once: $\frac{dy}{dx} = 2A(x-1)$.Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2A$.From the first derivative,$A = \frac{1}{2(x-1)} \frac{dy}{dx}$.Substituting this into the second derivative: $\frac{d^2y}{dx^2} = 2 \left( \frac{1}{2(x-1)} \frac{dy}{dx} \right) = \frac{1}{x-1} \frac{dy}{dx}$.Rearranging gives: $(x-1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

The differential equation corresponding to the family of parabolas whose axis is along $x=1$ is

Similar Questions

The differential equation for which $y = ax^2 + bx + c$ is the general solution is:

The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$ where $A, B$ are arbitrary constants is

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

The order of the differential equation whose solution is $a e^{x} + b e^{2x} + c e^{3x} + d = 0$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module