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Question 18 of 40

Q.The order and degree of the differential equation [1+(dydx)3]2/3=8(d3ydx3)\left[1 + \left(\dfrac{dy}{dx}\right)^3\right]^{2/3} = 8\left(\dfrac{d^3y}{dx^3}\right) are respectively ______.

(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022MCQ· 1mImportance★★★★★
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The highest derivative present is d3ydx3\dfrac{d^3y}{dx^3}, so the order is 3. Cubing both sides removes the 2/32/3 power and leaves (d3ydx3)3\left(\dfrac{d^3y}{dx^3}\right)^3, so the degree is 3.

The equation is

[1+(dydx)3]2/3=8(d3ydx3).\left[1 + \left(\frac{dy}{dx}\right)^3\right]^{2/3} = 8\left(\frac{d^3y}{dx^3}\right).

Order: the highest-order derivative appearing is d3ydx3\dfrac{d^3y}{dx^3}, so the order is 33.

Degree: the degree is defined only when the equation is a polynomial in all the derivatives, so we must clear the fractional exponent 2/32/3. Cubing both sides: …

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