For the differential equation given below,det | vedclass

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

Question

(A) The given differential equation is:$\left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x$The highest order derivative present

Vedclass · Accepted Answer

Order: $1$,Degree: $3$

Vedclass · Answer

(A) The given differential equation is:$\left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x$The highest order derivative present in the equation is $\frac{dy}{dx}$,which is a first-order derivative.Therefore,the order of the differential equation is $1$.The degree of a differential equation is the highest power of the highest order derivative when the equation is expressed as a polynomial in derivatives.Here,the highest power of the first-order derivative $\frac{dy}{dx}$ is $3$.Therefore,the degree of the differential equation is $3$.

For the differential equation given below,determine its order and degree (if defined):
$\left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x$

Similar Questions

The order and degree of the differential equation $(\frac{d^3y}{dx^3})^{2/3} - 3 \frac{d^2y}{dx^2} + 5 \frac{dy}{dx} + 4 = 0$ are respectively:

The order and degree of the differential equation $x\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + y^2 = 0$ are respectively:

The differential equation whose solution is $Ax^2 + By^2 = 1$,where $A$ and $B$ are arbitrary constants,is of:

For the differential equation $\left[1-\left(\frac{dy}{dx}\right)^2\right]^{5/2} = 8 \frac{d^2y}{dx^2}$,the order and degree are:

The order and degree of the differential equation $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$ are ...

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the differential equation given below,determine its order and degree (if defined):$\left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x$