The degree and order of the differential equa | vedclass

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(B) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}=7\left(\frac{d^2y}{dx^2}\right)$.To find the degree,

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$3$ and $2$

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(B) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}=7\left(\frac{d^2y}{dx^2}\right)$.To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $3$:$\left[\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}\right]^3 = \left[7\left(\frac{d^2y}{dx^2}\right)\right]^3$$\Rightarrow \left[1+\left(\frac{dy}{dx}\right)^3\right]^7 = 343\left(\frac{d^2y}{dx^2}\right)^3$.The order of a differential equation is the highest derivative present,which is $\frac{d^2y}{dx^2}$,so the order is $2$.The degree is the power of the highest derivative after the equation is made a polynomial in derivatives,which is $3$.Thus,the degree is $3$ and the order is $2$.

The degree and order of the differential equation $\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}=7\left(\frac{d^2y}{dx^2}\right)$ respectively are

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