The particular solution of the differential equation \( \displaystyle\frac{dy}{dx}=-2xy^2 \) given that \( y=1 \), when \( x=0 \), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( y=\displaystyle\frac{1}{x^2+1} \)
The variables separate:
\( \displaystyle \frac{dy}{y^2}=-2x\,dx \)
\( \displaystyle \int y^{-2}\,dy=-\int 2x\,dx \)
\( \displaystyle -\frac{1}{y}=-x^2+C \)
Put x = 0, y = 1:
\( \displaystyle -1=0+C\ \Rightarrow\ C=-1 \)
\( \displaystyle -\frac{1}{y}=-x^2-1 \)
\( \displaystyle \frac{1}{y}=x^2+1 \)
\( \displaystyle y=\frac{1}{x^2+1} \)
Shortcut: Test the options. y(0) = 1 is satisfied only by Options (2) and (4) (Options (1) and (3) give −1). For Option (4):
\( \displaystyle \frac{dy}{dx}=-\frac{2x}{(x^2+1)^2} \)
\( \displaystyle =-2x\left(\frac{1}{x^2+1}\right)^2=-2xy^2 \)
For Option (2), dy/dx = 2x, which is not \( -2x(x^2+1)^2 \).
NCERT (2026-27 edition):
Differential Equations with Variables Separable
