Question

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

Answer 1

$ \frac{ d }{ dx }\left( x ^2 y \right)= x -1$
Integrating both sides
$x^2 y=\int(x-1) d x=\frac{x^2}{2}-x+C$
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}{2 x^2} $
$ \frac{d y}{d x}=\frac{1}{x^2}-\frac{1}{x^3}=\frac{x-1}{x^3}$
$\therefore \frac{x-1}{x^3}>0$
$\therefore x \in(-\infty, 0) \cup(1, \infty)]$