The particular solution of the differential equation $xdy = (2x^2 + 1)dx, x ≠0$, given that $y = 1$ when $x = 1$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $y = x^2 + \log|x|$
$x\,dy=(2x^2+1)\,dx,\ x\ne 0$
$\Rightarrow \frac{dy}{dx}=2x+\frac{1}{x}$
$\Rightarrow y=\int\!\left(2x+\frac{1}{x}\right)dx=x^2+\ln|x|+C$
Given $y=1$ at $x=1$:
$1=1+\ln 1+C\Rightarrow C=0$
$y=x^2+\ln|x|$