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(A) The given differential equation is $y^{\prime \prime} + 2y^{\prime} + \sin(y) = 0$. The highest order derivative present in the equation is $y^{\p

Vedclass · Accepted Answer

Order $2$,Degree $1$

Vedclass · Answer

(A) The given differential equation is $y^{\prime \prime} + 2y^{\prime} + \sin(y) = 0$. The highest order derivative present in the equation is $y^{\prime \prime}$,which represents the second derivative. Therefore,the order of the differential equation is $2$. $A$ differential equation is a polynomial in derivatives if the derivatives are not arguments of transcendental functions like $\sin$,$\cos$,$e^x$,etc. In this equation,the term $\sin(y)$ involves $y$,but the derivatives $y^{\prime \prime}$ and $y^{\prime}$ are not inside a transcendental function. Thus,the equation is a polynomial in its derivatives. The highest power of the highest order derivative $y^{\prime \prime}$ is $1$. Therefore,the degree of the differential equation is $1$.

Determine the order and degree (if defined) of the differential equation $y^{\prime \prime} + 2y^{\prime} + \sin(y) = 0$.

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