Second order ODE solution - help me spot a mistake Solve following ODE:
$$
(1-x)x''+2(x')^2=0; x(0)=2, x'(0)=-1
$$
$$
x''=\frac{-2(x')^2}{1-x}
$$
substitute $x'=u(x)$ and assume $u \neq 0$
$$
uu'=\frac{-2(u)^2}{1-x} \\
\frac{dx}{1-x}=-\frac{du}{2u} \\
-\ln{|1-x|}=-\frac{1}{2}\ln{|u|}+c \\
|1-x|=\sqrt(|u|)e^c
$$
undo substitution $x'=u(x)$ and use $x(0)=2, x'(0)=-1$:
$$
|1-2|=\sqrt(|-1|)e^c \\
1 = e^c \\
c = 0
$$
so
$$
|1-x|=\sqrt{|x'|}\\
x' = (1-x)^2 \\
\frac{dx}{dt}=(1-x)^2 \\
dt = \frac{dx}{(1-x)^2} \\
t + c = \frac{1}{1-x}
$$
use $x(0)=2$
$$
0 + c = \frac{1}{1-2} \\
c = -1
$$
finally:
$$
t - 1 = \frac{1}{1-x} \\
\frac{1}{t-1} = 1 - x \\
x = 1 - \frac{1}{t-1} \\
x(t) = \frac{t-2}{t-1}
$$
Wolfram Alpha however, says the ODE is $x(t)=\frac{t+2}{t+1}$. Where is my mistake?

The first stage of the solution should be kept with an undetermined constant up to the point 
$$
u(x)=C(x-1)^2.
$$
Then using $u(2)=-1$ gives $C=-1$ and thus
$$
x'(t)=-(x-1)^2,
$$
which has a different sign than the equation you got. The sign difference in the solution follows from here.
