The solution of the differential equation \( \displaystyle\frac{dy}{dx}=x^2\,y^2 \) with \( y=1 \) at \( x=0 \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( y(x^3-3)+3=0 \)
The variables separate:
\( \displaystyle \frac{dy}{y^2}=x^2\,dx \)
Integrating both sides,
\( \displaystyle -\frac{1}{y}=\frac{x^3}{3}+C \)
Putting x = 0, y = 1: −1 = 0 + C, so C = −1.
\( \displaystyle -\frac{1}{y}=\frac{x^3}{3}-1 \)
\( \displaystyle -\frac{1}{y}=\frac{x^3-3}{3} \)
\( \displaystyle y(x^3-3)=-3 \)
\( \displaystyle y(x^3-3)+3=0 \)
Shortcut: Put x = 0, y = 1 in the options: Option (1) gives −6, Option (2) gives 0, Option (3) gives 6 and Option (4) gives 0. To choose between (2) and (4), note that \( \frac{dy}{dx}=x^2y^2\ge 0 \), so y never decreases; Option (2) gives \( y=\frac{3}{x^3+3} \), which decreases for x > 0, so Option (4) is the answer.