The sum of the degree and order of the differ | vedclass

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(D) Given differential equation is $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$.To find the order and degree,we first eliminate the radical

Vedclass · Accepted Answer

$7$

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(D) Given differential equation is $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$.To find the order and degree,we first eliminate the radicals by raising both sides to the power of $10$ (the least common multiple of $2$ and $5$):$(\frac{d^2 y}{d x^2})^{1/2} = (\frac{dy}{d x}-5)^{1/5}$$(\frac{d^2 y}{d x^2})^{10/2} = (\frac{dy}{d x}-5)^{10/5}$$(\frac{d^2 y}{d x^2})^5 = (\frac{dy}{d x}-5)^2$.The order of the differential equation is the highest derivative present,which is $2$.The degree of the differential equation is the power of the highest derivative after making the equation a polynomial in derivatives,which is $5$.Therefore,the sum of the degree and order is $2 + 5 = 7$.

The sum of the degree and order of the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[5]{\frac{dy}{d x}-5}$ is

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