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The degree of the differential equation (1 - ( dy dx )^2 )^3/2 = k d^2 y dx^2 is :

CUET UG 2022 — Mathematics Applied-Mathematics

2022mcqmedium

The degree of the differential equation (1−(dydx)2)3/2=kd2ydx2\left(1 - \left(\frac{dy}{dx}\right)^2\right)^{3/2} = k \frac{d^2 y}{dx^2}(1−(dxdy​)2)3/2=kdx2d2y​ is :

Official previous-year question

Held on 11 Aug 2022 · Verified 13 Jul 2026.

Options

  1. A

    1

  2. B

    2

  3. C

    3

  4. D

    32\frac{3}{2}23​

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