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CUET · MATHS · PYQ PAPER 2023

Match \(List - I\) with \(List - II\)
\(List - I\)\(List - II\)
General solutionDifferential Equation
(A) \(y=c x\)(I) \(y^{\prime}=\frac{1}{2 \sqrt{x}}-1\)
(B) \(y=e^{-3 x}+c\)(II) \(y^{\prime}+3 y=0\)
(C) \(x+y=\sqrt{x}+c\)(III) \(y^{\prime}=\frac{y^2}{1-x y}(x y \neq 1)\)
(D) \(x y=\log y+c\)(IV) \(x y^{\prime}=y\)

(c is an arbitrary constant and \(y^{\prime}=\frac{d y}{d x}\) )
Choose the correct answer from the options given below:

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

Answer & Solution

Correct Answer

(D) \((A)-(I V),(B)-(I I),(C)-(I),(D)-(I I I)\)

Step-by-step Solution

Detailed explanation

(A) \(\frac{d}{dx}(cx) = c\) \(y' = \frac{y}{x} \implies x y' = y\) (A) matches (IV) (C) \(\frac{d}{dx}(x+y) = \frac{d}{dx}(\sqrt{x}+c)\) \(1+y' = \frac{1}{2\sqrt{x}}\) \(y' = \frac{1}{2\sqrt{x}}-1\) (C) matches (I) (D) \(\frac{d}{dx}(xy) = \frac{d}{dx}(\log y+c)\)…
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