> 
> $$y^\prime=(y-1)^2, \qquad y(0)=1.01$$
> 
> 
> 

I have tried: 
$$\begin{align}
\int \frac{dy}{(y-1)^2} &= \int dx \\[4pt]
-(y-1)^{-1} &= x \\
y &= 1-\frac{1}{x}+C
\end{align}$$

Then I cannot find the $C$, because if $x=0$, then $y= -\infty$.

What's wrong?