The solution of the differential equation \( x\displaystyle\frac{dy}{dx}=2x^2+1\ ;x\ne 0 \) with \( y=1 \) when \( x=1 \), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( y=x^2+\log_e|x| \)
Dividing by x (x ≠ 0) separates the variables:
\( \displaystyle \frac{dy}{dx}=2x+\frac{1}{x} \)
\( \displaystyle y=x^2+\log_e|x|+C \)
Put x = 1, y = 1: \( 1=1+0+C \), so C = 0.
\( \displaystyle y=x^2+\log_e|x| \)
Shortcut: Only Options (3) and (4) give y = 1 at x = 1. Differentiating Option (4) gives
\( \displaystyle x\frac{dy}{dx}=x\left(2x+\frac{1}{x}\right)=2x^2+1 \), as required.
NCERT (2026-27 edition):
Differential Equations with Variables Separable
