CUET · MATHS · PYQ PAPER 2023
Match \(List - I\) with \(List - II\)
| \(List - I\) | \(List - II\) |
| General solution | Differential 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:
- A \((A)-(I V),(B)-(I),(C)-(I I),(D)-(I I I)\)
- B \((A) - (II), (B) - (I), (C) - (IV), (D) - (III)\)
- C \((A) - (IV), (B) - (II), (C) - (III), (D) - (I)\)
- D \((A)-(I V),(B)-(I I),(C)-(I),(D)-(I I I)\)
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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