The order and the degree of the differential | vedclass

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(D) To find the order and degree,we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and

Vedclass · Accepted Answer

$2, 6$

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(D) To find the order and degree,we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$).Given equation: $(1 + (y'')^2)^{1/2} = (x + (y')^3)^{1/3}$.Raising both sides to the power of $6$:$((1 + (y'')^2)^{1/2})^6 = ((x + (y')^3)^{1/3})^6$$(1 + (y'')^2)^3 = (x + (y')^3)^2$.Expanding the left side: $1 + 3(y'')^2 + 3(y'')^4 + (y'')^6 = (x + (y')^3)^2$.The highest order derivative present is $y'' = \frac{d^2y}{dx^2}$,so the order is $2$.The exponent of the highest order derivative after rationalizing the equation is $6$. Therefore,the degree is $6$.

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

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