If m and n are the order and degree of the di | vedclass

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

Question

(C) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = -4a(x-a)$, where $a$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = -4a(x-a)$, where $a$ is the parameter.$y^2 = -4ax + 4a^2$  $(i)$Differentiating with respect to $x$:$2y y' = -4a$$a = -\frac{y y'}{2}$Substituting the value of $a$ into equation $(i)$:$y^2 = -4\left(-\frac{y y'}{2}ight)x + 4\left(-\frac{y y'}{2}ight)^2$$y^2 = 2x y y' + 4\left(\frac{y^2 y'^2}{4}ight)$$y^2 = 2x y y' + y^2 y'^2$Dividing by $y$ (assuming $y 
eq 0$):$y = 2x y' + y y'^2$$y\left(\frac{dy}{dx}ight)^2 + 2x\left(\frac{dy}{dx}ight) - y = 0$The order $m$ of this differential equation is $1$ and the degree $n$ is $2$.Therefore, $mn - m + n = (1)(2) - 1 + 2 = 2 - 1 + 2 = 3$.

If $m$ and $n$ are the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis, then $m n-m+n=$

Similar Questions

Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2},$ where $a$ is an arbitrary constant.

The differential equation formed by eliminating $a$ and $b$ from the equation $y=e^x(a \cos x+b \sin x)$ is

The differential equation of the family of curves $y = ax + \frac{1}{a}$, where $a (\neq 0)$ is an arbitrary constant, has the degree

Verify that the given function $x^{2}=2 y^{2} \log y$ is a solution of the corresponding differential equation $(x^{2}+y^{2}) \frac{dy}{dx}-xy=0$.

If $m$ and $n$ are respectively the order and degree of the differential equation of the family of parabolas with focus at the origin and $X$-axis as its axis,then $m n-m+n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module