General solution of the differential equation \( \displaystyle\frac{3y\,dx-2x\,dy}{y}=0 \) is :
[where \( c \) is an arbitrary constant]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( x^3=cy^2 \)
The equation gives \( 3y\,dx=2x\,dy \). Dividing by xy separates the variables:
\( \displaystyle \frac{3\,dx}{x}=\frac{2\,dy}{y} \)
Integrating,
\( \displaystyle 3\log|x|=2\log|y|+\log|c| \)
\( \displaystyle \log|x^3|=\log|cy^2| \)
\( \displaystyle x^3=cy^2 \)
Why the other options are incorrect
Option (1) \( y=cx^2 \) — Incorrect. For \( y=cx^2 \), dy/dx = 2y/x, which gives \( x\,dy=2y\,dx \), not \( 3y\,dx=2x\,dy \).
Option (3) \( x=cy^2 \) — Incorrect. For \( x=cy^2 \), dx/dy = 2x/y, which gives \( y\,dx=2x\,dy \); the coefficient 3 is lost.
Option (4) \( y^3=cx^2 \) — Incorrect. This swaps the roles of x and y: \( y^3=cx^2 \) solves \( 3x\,dy=2y\,dx \).