The degree of the differential equation 1+lef | vedclass

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(D) Given the differential equation: $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\left(\frac{d^2y}{dx^2}+1\right)^{1/3}$.To find

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$6$

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(D) Given the differential equation: $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\left(\frac{d^2y}{dx^2}+1\right)^{1/3}$.To find the degree,we must eliminate the fractional exponent by cubing both sides:$\left[1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2\right]^3 = \frac{d^2y}{dx^2}+1$.Expanding the left side using the binomial expansion $(a+b+c)^3$,the highest power of the highest order derivative $\frac{d^2y}{dx^2}$ will come from the term $\left(\left(\frac{d^2y}{dx^2}\right)^2\right)^3 = \left(\frac{d^2y}{dx^2}\right)^6$.Since the highest order derivative is $\frac{d^2y}{dx^2}$ and its highest power after rationalizing the equation is $6$,the degree of the differential equation is $6$.

The degree of the differential equation $1+\left(\frac{dy}{dx}\right)^2+\left(\frac{d^2y}{dx^2}\right)^2=\sqrt[3]{\frac{d^2y}{dx^2}+1}$ is

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