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Question 17 of 43

Q.The order and degree of the differential equation [d2ydx2]32−(dydx)−4=0\left[\frac{d^2y}{dx^2}\right]^{\frac{3}{2}} - \sqrt{\left(\frac{dy}{dx}\right)} - 4 = 0 are respectively.

(a) 11 and 44
(b) 22 and 66
(c) 22 and 44
(d) 33 and 66
Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2022MCQ· 1mImportance★★★★★
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Order =2= 2, degree =6= 6.

The equation is

[d2ydx2]3/2−dydx−4=0.\left[\frac{d^2y}{dx^2}\right]^{3/2} - \sqrt{\frac{dy}{dx}} - 4 = 0.

Order = order of the highest derivative present =d2ydx2= \dfrac{d^2y}{dx^2}, so order =2= 2.

Degree requires the equation to be a polynomial in the derivatives, free of fractional powers. Move terms and square to clear 32\tfrac{3}{2}:

[d2ydx2]3/2=dydx+4  ⇒  [d2ydx2]3=dydx+8dydx+16.\left[\frac{d^2y}{dx^2}\right]^{3/2} = \sqrt{\frac{dy}{dx}} + 4 \;\Rightarrow\; \left[\frac{d^2y}{dx^2}\right]^{3} = \frac{dy}{dx} + 8\sqrt{\frac{dy}{dx}} + 16.

A square-root term remains, so isolate and square once more:

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