CUET UG 2025 — Mathematics Calculus

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

Separating variables:

$$\dfrac{dy}{y} = \dfrac{dx}{x}$$ydy​=xdx​

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

$$\ln|y| = \ln|cx|$$ln∣y∣=ln∣cx∣

$$y = cx$$y=cx

(A) → (I)

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(B) $x\,dx - y\,dy = 0$xdx−ydy=0

$$x\,dx = y\,dy$$xdx=ydy

$$\dfrac{x^2}{2} = \dfrac{y^2}{2} + \dfrac{c}{2}$$2x2​=2y2​+2c​

$$x^2 - y^2 = c$$x2−y2=c

(B) → (II)

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(C) $\dfrac{(x^2 - 1)}{(y^2 + 1)} \cdot \dfrac{dx}{dy} = 1$(y2+1)(x2−1)​⋅dydx​=1

$$(x^2 - 1)\,dx = (y^2 + 1)\,dy$$(x2−1)dx=(y2+1)dy

$$\dfrac{x^3}{3} - x = \dfrac{y^3}{3} + y + C_1$$3x3​−x=3y3​+y+C1​

$$x^3 - 3x = y^3 + 3y + c$$x3−3x=y3+3y+c

$$x^3 - y^3 = 3x + 3y + c$$x3−y3=3x+3y+c

$$(x^3 - y^3) = c + 3(x + y)$$(x3−y3)=c+3(x+y)

(C) → (IV)

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

$$2x + 3y = c$$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.