The order and degree of the differential equa | vedclass

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

Question

(A) The given differential equation is $e^{\frac{d^2 y}{d x^2}} = x$. To find the order,we identify the highest order derivative present in the equati

Vedclass · Accepted Answer

$2$ and not defined

Vedclass · Answer

(A) The given differential equation is $e^{\frac{d^2 y}{d x^2}} = x$. To find the order,we identify the highest order derivative present in the equation,which is $\frac{d^2 y}{d x^2}$. Thus,the order is $2$. To find the degree,the differential equation must be a polynomial in terms of its derivatives. Since the term $\frac{d^2 y}{d x^2}$ appears in the exponent of $e$,the equation cannot be expressed as a polynomial in its derivatives. Therefore,the degree of this differential equation is not defined.

The order and degree of the differential equation $e^{\frac{d^2 y}{d x^2}} = x$ are . . . . . . respectively.

Similar Questions

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

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

Determine the order and degree (if defined) of the differential equation $\frac{d^4y}{dx^4} + \sin(y''') = 0$.

The degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^5+\left(\frac{d y}{d x}\right)^2+\cos \left(\frac{d y}{d x}\right)+1=0$ is . . . . . . .

The degree of the differential equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module