Maths. Section ADifferential Equations

The particular solution of the differential equation xdy = (2x^2 + 1)…

The particular solution of the differential equation $xdy = (2x^2 + 1)dx, x \ne 0$, given that $y = 1$ when $x = 1$ is:

Answer & explanation

Correct answer: option 3

The correct answer is Option (3) → $y = x^2 + \log|x|$

The equation is in variables separable form. Since $x \ne 0$, dividing by x:

$ \displaystyle dy = \left(2x + \frac{1}{x}\right)dx $

Integrating both sides:

$y = x^2 + \log|x| + C$

Putting $y = 1$ when $x = 1$:

$1 = 1 + \log 1 + C$
$C = 0$

So the particular solution is $y = x^2 + \log|x|$.

Tip: Checking only the initial condition does not decide here: Option (1) also gives $y = 1$ at $x = 1$, but its derivative is $1 + \frac{1}{x}$, not $2x + \frac{1}{x}$.

NCERT (2026-27 edition):
Differential Equations with Variables Separable

$NCERT extract$