The differential equation frac{d^3y}{dx^3} + | vedclass

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(A) The given differential equation is $\frac{d^3y}{dx^3} + 2\left[ 1 + \frac{d^2y}{dx^2} \right] = 1$. Expanding the equation,we get $\frac{d^3y}{dx^

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$3, 1$

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(A) The given differential equation is $\frac{d^3y}{dx^3} + 2\left[ 1 + \frac{d^2y}{dx^2} \right] = 1$. Expanding the equation,we get $\frac{d^3y}{dx^3} + 2 + 2\frac{d^2y}{dx^2} = 1$,which simplifies to $\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} + 1 = 0$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^3y}{dx^3}$,so the order is $3$. The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. The power of $\frac{d^3y}{dx^3}$ is $1$. Thus,the order is $3$ and the degree is $1$.

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

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