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

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

Question

(C) The given differential equation is $\left(1+\left(\frac{dy}{dx}\right)^{2}\right)^{2/3} = \frac{d^{2}y}{dx^{2}}$.To find the degree,we must elimin

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The given differential equation is $\left(1+\left(\frac{dy}{dx}\right)^{2}\right)^{2/3} = \frac{d^{2}y}{dx^{2}}$.To find the degree,we must eliminate the fractional exponent by cubing both sides:$\left(1+\left(\frac{dy}{dx}\right)^{2}\right)^{2} = \left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.The order $n$ is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so $n = 2$.The degree $m$ is the power of the highest derivative after rationalizing,which is $3$,so $m = 3$.Now,calculate the required value: $\frac{m+n}{m-n} = \frac{3+2}{3-2} = \frac{5}{1} = 5$.

If $m$ and $n$ are the degree and order of the differential equation $\left(1+y_{1}^{2}\right)^{2 / 3}=y_{2}$,then the value of $\frac{m+n}{m-n}$ is

Similar Questions

If $a$ and $b$ are the order and degree respectively of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$,then $a-b=$

The order and degree of the differential equation $\left(y^{\prime \prime \prime}\right)^3+\left(y^{\prime \prime}\right)^4+\left(y^{\prime}\right)^4+y=7$ are . . . . . . respectively.

The order and degree of the differential equation $\frac{d^2y}{dx^2} = \cos \left( \frac{dy}{dx} \right) + xy$ are respectively-

Degree of the differential equation $e^{\frac{dy}{dx}} + (\frac{dy}{dx})^3 = x$ is

Find the degree of the differential equation $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module