Match List-I with List-II
| List-I | List-II | ||
| Curve \( y=f(x) \) (where \( C \) is an arbitrary constant) | The curve \( y=f(x) \) satisfies the differential equation | ||
| (A) | \( y=\log x+C \) | (I) | \( x\displaystyle\frac{dy}{dx}-y=0 \) |
| (B) | \( y=e^{x}+C \) | (II) | \( x\displaystyle\frac{dy}{dx}-1=0 \) |
| (C) | \( y=Cx \) | (III) | \( x^{2}\displaystyle\frac{dy}{dx}+1=0 \) |
| (D) | \( y=\displaystyle\frac{1}{x}+C \) | (IV) | \( \displaystyle\frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}=0 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Each curve leads to its differential equation once the arbitrary constant C is eliminated by differentiating.
(A) \( y=\log x+C \)
This matches (II) \( x\frac{dy}{dx}-1=0 \).
\( \displaystyle \frac{dy}{dx}=\frac{1}{x} \)
\( \displaystyle x\frac{dy}{dx}-1=0 \)
(B) \( y=e^x+C \)
This matches (IV) \( \frac{d^2y}{dx^2}-\frac{dy}{dx}=0 \).
\( \displaystyle \frac{dy}{dx}=e^x \)
\( \displaystyle \frac{d^2y}{dx^2}=e^x \)
So \( \frac{d^2y}{dx^2}=\frac{dy}{dx} \).
(C) \( y=Cx \)
This matches (I) \( x\frac{dy}{dx}-y=0 \).
\( \displaystyle \frac{dy}{dx}=C=\frac{y}{x} \)
\( \displaystyle x\frac{dy}{dx}-y=0 \)
(D) \( y=\frac{1}{x}+C \)
This matches (III) \( x^2\frac{dy}{dx}+1=0 \).
\( \displaystyle \frac{dy}{dx}=-\frac{1}{x^2} \)
\( \displaystyle x^2\frac{dy}{dx}+1=0 \)
So the matching is (A) - (II), (B) - (IV), (C) - (I), (D) - (III).
Shortcut: Only (A) - (II) and (B) - (IV) are needed: they appear together only in Options (2) and (3), and (C) - (I) then decides Option (2).