Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(III), (B)-(I), (C)-(IV), (D)-(II) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) |
(A)-(III), (B)-(I), (C)-(IV), (D)-(II) |
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) $\left(\frac{d^2y}{dx^2}\right)^2 - e^x\left(\frac{dy}{dx}\right)^4 + 1 = 0$ Highest derivative: $\frac{d^2y}{dx^2}$ → order = 2; highest power = 2 → degree = 2 → (III) (B) $\left(\frac{dy}{dx}\right)^2 + xy = 0$ Highest derivative: $\frac{dy}{dx}$ → order = 1; highest power = 2 → degree = 2 → (I) (C) $\left(1+\frac{dy}{dx}\right)^{3/2} = 4\left(\frac{d^2y}{dx^2}\right)^2$ Fractional power present; remove root → square both sides: $\left(1+\frac{dy}{dx}\right)^3 = 16\left(\frac{d^2y}{dx^2}\right)^4$ → order = 2, degree = 4 → (IV) (D) $\sqrt{\frac{d^2y}{dx^2}+1} = \frac{dy}{dx}$ Square both sides → $\frac{d^2y}{dx^2} + 1 = \left(\frac{dy}{dx}\right)^2$ → order = 2, degree = 1 → (II) Final Matching: (A)–(III), (B)–(I), (C)–(IV), (D)–(II) |