Question

Match the LIST-I with LIST-II

LIST-I Differential Equations	LIST-II Solutions[ c is an arbitrary constant]
A. \(x\,dy = y\,dx\)	I. \(y = e^{x} + c\)
B. \(\log\left(\frac{dy}{dx}\right) = x\)	II. \(y = c\,e^{x}\)
C. \(\frac{y\,dx - dy}{y} = 0\)	III. \(xy = c\)
D. \(x\,dy + y\,dx = 0\)	IV. \(y = cx\)

Choose the correct answer from the options given below:

• A. A-IV, B-II, C-I, D-III
• B. A-IV, B-I, C-II, D-III
• C. A-III, B-II, C-I, D-IV
• D. A-III, B-I, C-II, D-IV

Hint:

Solve each equation by separating variables and compare with List-II.

Answer 1

The Correct Option is B

Step 1: Work backwards from List-II.
Instead of solving, we differentiate each solution in List-II and see which equation it satisfies.

Step 2: Differentiate I and II.
For $y = e^{x} + c$: $\frac{dy}{dx} = e^{x}$, so $\log\frac{dy}{dx} = x$. This is equation B.
For $y = c\,e^{x}$: $\frac{dy}{dx} = c\,e^{x} = y$, so $dy = y\,dx$, which is equation C.

Step 3: Differentiate III and IV.
For $xy = c$: $x\,dy + y\,dx = 0$. This is equation D.
For $y = cx$: $dy = c\,dx$, and $c = \frac{y}{x}$, so $x\,dy = y\,dx$. This is equation A.

Step 4: Read off the pairs.
A-IV, B-I, C-II, D-III. Only option 2 lists exactly these pairs.

Final Answer:
This matches option 2. \[ \boxed{\text{Option 2}} \]