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: Plan.
We solve each differential equation in List-I and match its general solution with an item in List-II.

Step 2: Solve A.
\(x\,dy = y\,dx\) gives \(\frac{dy}{y} = \frac{dx}{x}\). Integrating, \(\ln y = \ln x + \ln c\), so \(y = cx\). This is IV.

Step 3: Solve B.
\(\log\left(\frac{dy}{dx}\right) = x\) means \(\frac{dy}{dx} = e^{x}\). Integrating, \(y = e^{x} + c\). This is I.

Step 4: Solve C.
\(\frac{y\,dx - dy}{y} = 0\) means \(y\,dx = dy\), so \(\frac{dy}{y} = dx\). Integrating, \(\ln y = x + \ln c\), so \(y = c\,e^{x}\). This is II.

Step 5: Solve D.
\(x\,dy + y\,dx = 0\) is the same as \(d(xy) = 0\). So \(xy = c\). This is III.

Step 6: Compare the options.
We found A-IV, B-I, C-II, D-III. Option 1 has B and C wrongly swapped, so it is wrong. Option 3 has A-III and D-IV, which is wrong. Option 4 also has A-III and D-IV, so it is wrong. Only option 2 matches.

Final Answer:
The correct matching is A-IV, B-I, C-II, D-III, which is option 2.