Question

Let $\frac{d}{dx}\left(x^{2}y\right)=x-1$ where $x\ne 0$ and $y=0$ when $x=1$. Find the smallest natural $x$ for which $\frac{dy}{dx}$ is positive.

Answer 1

Answer: 2

$\frac{d}{dx}\left(x^{2}y\right)=x-1$
Integrating both sides
$x^{2}y=\int (x-1)dx=\frac{x^2}{2}-x+C$
If $x=1,y=0$
$\therefore 0=\frac{1}{2}-1+C\Rightarrow C=\frac{1}{2}$
$x^{2}y=\frac{x^2}{2}-x-\frac{1}{2}$

$y=\frac{1}{2}-\frac{1}{x}+\frac{1}{2x^2}$
$\frac{dy}{dx}=\frac{1}{x^2}-\frac{1}{x^3}=\frac{x-1}{x^3}$
$\therefore \frac{x-1}{x^3}>0$
$\therefore x\in (-\infty ,0)\cup (1,\infty )]$