Match List-I with List-II

List-I	List-II
Differential Equation	General solution
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(A) $\dfrac{dy}{dx} = \dfrac{y}{x}; x \neq 0$	(I) $y = cx; c \text{ is an arbitrary constant}$
(B) $x dx - y dy = 0; y \neq 0, x \neq 0$	(II) $x^2 - y^2 = c; c \text{ is an arbitrary constant}$
(C) $\dfrac{(x^2 - 1)}{y^2 + 1}\dfrac{dx}{dy} = 1$	(III) $2x + 3y = c; c \text{ is an arbitrary constant}$
(D) $2 dx + 3 dy = 0$	(IV) $(x^3-y^3) = c + 3(x+y); c \text{ is an arbitrary constant}$

Choose the correct answer from the options given below:

(A) $\dfrac{dy}{dx} = \dfrac{y}{x}$

Separating variables:

$\dfrac{dy}{y} = \dfrac{dx}{x}$

$\ln|y| = \ln|x| + \ln|c|$

$\ln|y| = \ln|cx|$

$y = cx$

(A) → (I)

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(B) $x\,dx - y\,dy = 0$

$x\,dx = y\,dy$

$\dfrac{x^2}{2} = \dfrac{y^2}{2} + \dfrac{c}{2}$

$x^2 - y^2 = c$

(B) → (II)

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(C) $\dfrac{(x^2 - 1)}{(y^2 + 1)} \cdot \dfrac{dx}{dy} = 1$

$(x^2 - 1)\,dx = (y^2 + 1)\,dy$

$\dfrac{x^3}{3} - x = \dfrac{y^3}{3} + y + C_1$

$x^3 - 3x = y^3 + 3y + c$

$x^3 - y^3 = 3x + 3y + c$

$(x^3 - y^3) = c + 3(x + y)$

(C) → (IV)

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(D) $2\,dx + 3\,dy = 0$

$2x + 3y = c$

(D) → (III)

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

Therefore, the correct answer is Option 4.