Match the LIST-I with LIST-II
[ \( a \) is an arbitrary constant]
| LIST-I | LIST-II | ||
| Differential Equation | Family of curves | ||
| (A) | \( xdy-ydx=0 \) | (I) | \( x^2+y^2=a^2 \) |
| (B) | \( 2ydx-xdy=0 \) | (II) | \( y=ax \) |
| (C) | \( xdx+ydy=0 \) | (III) | \( y^2=16x+a \) |
| (D) | \( ydy-8dx=0 \) | (IV) | \( x^2=ay \) |
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 equation is solved by separating the variables.
(A) \( xdy-ydx=0 \)
This matches (II) y = ax.
\( \displaystyle \frac{dy}{y}=\frac{dx}{x} \)
\( \displaystyle \log|y|=\log|x|+\log|a| \)
\( \displaystyle y=ax \)
(B) \( 2ydx-xdy=0 \)
This matches (IV) \( x^2=ay \).
\( \displaystyle \frac{dy}{y}=\frac{2\,dx}{x} \)
\( \displaystyle \log|y|=2\log|x|+\log|c| \)
\( \displaystyle y=cx^2 \)
Writing \( a=\frac{1}{c} \), this is \( x^2=ay \).
(C) \( xdx+ydy=0 \)
This matches (I) \( x^2+y^2=a^2 \).
\( \displaystyle \frac{x^2}{2}+\frac{y^2}{2}=c \)
\( \displaystyle x^2+y^2=2c=a^2 \)
(D) \( ydy-8dx=0 \)
This matches (III) \( y^2=16x+a \).
\( \displaystyle \frac{y^2}{2}=8x+c \)
\( \displaystyle y^2=16x+2c=16x+a \)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).
Shortcut: Differentiate the curves instead: \( y=ax \) gives \( \frac{dy}{dx}=a=\frac{y}{x} \), which is (A); \( x^2+y^2=a^2 \) gives \( x+y\frac{dy}{dx}=0 \), which is (C). Only Option (2) has (A)-(II) and (C)-(I).