Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Differential Equations | Solutions[ c is an arbitrary constant] | ||
| A. | \( x\,dy=y\,dx \) | I. | \( y=e^x+\mathrm{c} \) |
| B. | \( \log\left(\displaystyle\frac{dy}{dx}\right)=x \) | II. | \( y=\mathrm{c}\ e^x \) |
| C. | \( \displaystyle\frac{y\,dx-dy}{y}=0 \) | III. | \( xy=\mathrm{c} \) |
| D. | \( x\,dy+y\,dx=0 \) | IV. | \( y=\mathrm{c}x \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
(A) \( x\,dy=y\,dx \)
This matches (IV) y = cx.
\( \displaystyle \frac{dy}{y}=\frac{dx}{x} \)
\( \displaystyle \log|y|=\log|x|+\log|c| \)
\( \displaystyle y=cx \)
(B) \( \log\left(\frac{dy}{dx}\right)=x \)
This matches (I) y = ex + c.
\( \displaystyle \frac{dy}{dx}=e^x \)
\( \displaystyle y=e^x+c \)
(C) \( \frac{y\,dx-dy}{y}=0 \)
This matches (II) y = c ex.
\( \displaystyle dy=y\,dx \)
\( \displaystyle \frac{dy}{y}=dx \)
\( \displaystyle \log|y|=x+c_1 \)
\( \displaystyle y=ce^x \)
(D) \( x\,dy+y\,dx=0 \)
This matches (III) xy = c. The left side is the exact differential of xy:
\( \displaystyle d(xy)=0 \)
\( \displaystyle xy=c \)
So the matching is A-IV, B-I, C-II, D-III.