Match List-I with List-II List-

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Match List-I with List-II

List-I (Differential Equations)	List-II (Order and degree)
(A) $ydx + x \log(y/x)dy - 2xdy = 0$	(I) $\text{Order: 2, Degree: 1}$
(B) $\left(\frac{d^3 y}{dx^3}\right)^2 + 3\frac{d^2 y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^2$	(II) $\text{Order: 1, Degree: 1}$
(C) $\frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y$	(III) $\text{Order: 3, Degree: 2}$
(D) $\left(\frac{ds}{dt}\right)^4 + 2s \frac{d^2 s}{dt^2} = 0$	(IV) $\text{Order: 1, Degree: Not defined}$

Choose the correct answer from the options given below:

Options:

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

(A)-(IV), (B)-(III), (C)-(I), (D)-(III)

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Correct Answer:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Explanation:

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

List-I (Differential Equations)	List-II (Order and degree)
(A) $ydx + x \log(y/x)dy - 2xdy = 0$	(II) $\text{Order: 1, Degree: 1}$
(B) $\left(\frac{d^3 y}{dx^3}\right)^2 + 3\frac{d^2 y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^2$	(III) $\text{Order: 3, Degree: 2}$
(C) $\frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y$	(IV) $\text{Order: 1, Degree: Not defined}$
(D) $\left(\frac{ds}{dt}\right)^4 + 2s \frac{d^2 s}{dt^2} = 0$	(I) $\text{Order: 2, Degree: 1}$

List-I and List-II matching:

(A) $y dx + x \log(y/x) dy - 2x dy = 0$

Highest derivative: $dy/dx$ → Order = 1, appears linearly → Degree = 1 → (II)

(B) $(d^3y/dx^3)^2 + 3 d^2y/dx^2 + 2 (dy/dx)^4 = y^2$

Highest derivative: $d^3y/dx^3$ → Order = 3, appears as square → Degree = 2 → (III)

Highest derivative: $dy/dx$ appears inside log → Degree = Not defined → Order =1 → (IV)

(D) $(d s/dt)^4 + 2s d^2 s/dt^2 = 0$

Highest derivative: $d^2s/dt^2$ → Order = 2, appears linearly → Degree = 1 → (I)