Particular solution of the differential equation \( \displaystyle\frac{dy}{dx}+2y^2=0 \), given \( y=1 \), when \( x=1 \), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( y=\displaystyle\frac{1}{2x-1} \)
The variables separate:
\( \displaystyle \frac{dy}{y^2}=-2\,dx \)
\( \displaystyle -\frac{1}{y}=-2x+C \)
At x = 1, y = 1: −1 = −2 + C, so C = 1.
\( \displaystyle \frac{1}{y}=2x-1 \)
\( \displaystyle y=\frac{1}{2x-1} \)
Shortcut: Only Options (1) and (3) give y = 1 at x = 1. For Option (1), \( y'+2y^2=2+2(2x-1)^2\ne 0 \).