Particular solution of the differential equation \( dy+2xy^{2}dx=0 \), given that \( y=1 \), when \( x=0 \), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( y=\displaystyle\frac{1}{x^{2}+1} \)
The variables separate:
\( dy=-2xy^{2}\,dx \)
\( \displaystyle \frac{dy}{y^{2}}=-2x\,dx \)
Integrating both sides:
\( \displaystyle \int y^{-2}\,dy=-\int 2x\,dx \)
\( \displaystyle -\frac{1}{y}=-x^{2}+C \)
The initial condition y = 1 when x = 0 gives −1 = C, so
\( \displaystyle -\frac{1}{y}=-x^{2}-1 \)
\( \displaystyle \frac{1}{y}=x^{2}+1 \)
\( \displaystyle y=\frac{1}{x^{2}+1} \)
Shortcut: At x = 0, Options (2) and (4) give y = −1, so they fail the condition. For Option (3):
\( \displaystyle \frac{dy}{dx}=-\frac{2x}{(x^{2}+1)^{2}} \)
\( \displaystyle =-2xy^{2} \)
which is the given equation; Option (1) gives \( \frac{dy}{dx}=2x \), not \( -2xy^{2} \).