If the order and degree of the differential e | vedclass

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

Question

(C) Given the family of curves: $y^2=4a(x+a)$ Step $1$: Differentiate with respect to $x$: $2y \frac{dy}{dx} = 4a$ $\implies a = \frac{y}{2} \frac{dy}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the family of curves: $y^2=4a(x+a)$ Step $1$: Differentiate with respect to $x$: $2y \frac{dy}{dx} = 4a$ $\implies a = \frac{y}{2} \frac{dy}{dx}$ Step $2$: Substitute the value of $a$ back into the original equation: $y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$ $y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$ $y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$ Step $3$: Identify the order and degree: The highest order derivative is $\frac{dy}{dx}$,so the order $m = 1$. The highest power of the highest order derivative is $2$,so the degree $n = 2$. Step $4$: Calculate $m+n^2$: $m+n^2 = 1 + 2^2 = 1 + 4 = 5$.

If the order and degree of the differential equation corresponding to the family of curves $y^2=4a(x+a)$ (where $a$ is a parameter) are $m$ and $n$ respectively,then $m+n^2=$

Similar Questions

The differential equation of the family of circles passing through the origin and having centre on the $X$-axis is:

Form the differential equation of the family of lines $y = mx + \frac{4}{m}$ by eliminating the arbitrary constant $m$.

The order of the differential equation corresponding to the family of parabolas whose axes are along the $X$-axis and whose foci are at the origin,is

The differential equation obtained from the function $y=a(x-a)^{2}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module