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) \(x^2\) |
- A (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
- B (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
- C (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
- D (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
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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