The degree of the differential equation 1+lef | vedclass

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

Question

(D) Given the differential equation: $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\left(\frac{d^2y}{dx^2}+1\right)^{1/3}$.To find

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Given the differential equation: $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\left(\frac{d^2y}{dx^2}+1\right)^{1/3}$.To find the degree,we must eliminate the fractional exponent by cubing both sides:$\left[1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2\right]^3 = \frac{d^2y}{dx^2}+1$.Expanding the left side using the binomial expansion $(a+b+c)^3$,the highest power of the highest order derivative $\frac{d^2y}{dx^2}$ will come from the term $\left(\left(\frac{d^2y}{dx^2}\right)^2\right)^3 = \left(\frac{d^2y}{dx^2}\right)^6$.Since the highest order derivative is $\frac{d^2y}{dx^2}$ and its highest power after rationalizing the equation is $6$,the degree of the differential equation is $6$.

The degree of the differential equation $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\sqrt[3]{\frac{d^2y}{dx^2}+1}$ is

Similar Questions

Determine the order and degree (if defined) of the differential equation $y^{\prime} + 5y = 0$.

The order of the differential equation,whose solution is $y=(C_1+C_2) e^x+C_3 e^{x+C_4}$,is

The differential equation $\frac{d^3y}{dx^3} + 2\left[ 1 + \frac{d^2y}{dx^2} \right] = 1$ has order and degree as:

The degree and the order of the differential equation $\frac{d^{2} y}{d x^{2}}=\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}$ respectively are

The order and degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0$ are . . . . . . and . . . . . . respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module