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Q.The degree of the differential equation d3ydx3+y=1+dydx\dfrac{d^3y}{dx^3}+y=\sqrt{1+\dfrac{dy}{dx}} is

(a) 1
(b) 2
(c) 3
(d) 4
Tripura TbseHigher Secondary (+2 Stage) Examination 2025MCQ· 1mImportance★★★★★
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The degree of a differential equation is the power of the highest-order derivative once the equation is written as a polynomial in the derivatives (no radicals/fractional powers of derivatives).

Given: d3ydx3+y=1+dydx\dfrac{d^3y}{dx^3}+y=\sqrt{1+\dfrac{dy}{dx}}.

The right side has a square root, so the equation is not yet polynomial in the derivatives — we must first remove the radical by squaring both sides:

(d3ydx3+y)2=1+dydx\left(\dfrac{d^3y}{dx^3}+y\right)^2=1+\dfrac{dy}{dx}

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