Match List-I with List-II
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List-I (Differential Equations) |
List-II (Order and degree) |
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(A) $ydx + x \log(y/x)dy - 2xdy = 0$ |
(I) $\text{Order: 2, Degree: 1}$ |
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(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) $\text{Order: 1, Degree: 1}$ |
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(C) $\frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y$ |
(III) $\text{Order: 3, Degree: 2}$ |
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(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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
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List-I (Differential Equations) |
List-II (Order and degree) |
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(A) $ydx + x \log(y/x)dy - 2xdy = 0$ |
(II) $\text{Order: 1, Degree: 1}$ |
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(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) $\text{Order: 3, Degree: 2}$ |
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(C) $\frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y$ |
(IV) $\text{Order: 1, Degree: Not defined}$ |
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(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)
(C) $dy/dx + \log(dy/dx) + x = y$
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)