The order and degree of the differential equa | vedclass

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Question

(B) The given differential equation is $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$. The order of a differential equ

Vedclass · Accepted Answer

Order is $3$ and degree can't be defined

Vedclass · Answer

(B) The given differential equation is $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$. The order of a differential equation is the highest order derivative present in the equation. Here,the highest order derivative is $\frac{d^3 y}{d x^3}$,so the order is $3$. The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Since the term $\sin \left(\frac{d^3 y}{d x^3}\right)$ involves a transcendental function of a derivative,it cannot be expressed as a polynomial in derivatives. Therefore,the degree of this differential equation is not defined.

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

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