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CUET · MATHS · PYQ PAPER 2025

List-I (Differential equation)List-II (Integrating factor)
(A) \(x \frac{d y}{d x}-y=2 x^2\)(I) \(e^{-y}\)
(B) \(\frac{d y}{d x}+\frac{y}{x}=2 x\)(II) \(\frac{1}{x}\)
(C) \(x \frac{d y}{d x}+2 y=x^2 \log x\)(III) \(x\)
(D) \(\frac{d x}{d y}-x=y\)(IV) \(x^2\)
Choose the correct answer from the options given below:

  1. A (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
  2. B (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
  3. C (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
  4. D (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

(A) \(x \frac{d y}{d x}-y=2 x^2 \implies \frac{d y}{d x} - \frac{1}{x} y = 2x\) \(P(x) = -\frac{1}{x}\) I.F. \( = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = e^{\ln x^{-1}} = \frac{1}{x}\). So (A) - (II). (B) \(\frac{d y}{d x}+\frac{y}{x}=2 x\) \(P(x) = \frac{1}{x}\) I.F.…
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