Help with ODE - Cannot get to computed solution - Possible problem with absolute values? I'm struggling with the following ODE, which computed gives a different answer than the one I got doing it analytically. Maybe I have made a silly mistake or I do not get how to work with the absolute values in the correct manner.
The ODE is the following:
$\dot{x} = x^2 -1$
Analytically, I get after separating values and performing a partial fraction decomposition the following answer:
$x(t) = \dfrac{1+e^{2t+c}}{1-e^{2t+c}}$
When I compute it in Wolfram Alpha I get:
$x(t) = \dfrac{1-e^{2t+c}}{e^{2t+c}+1}$
I have checked the integrals separately and compared the solutions to the Phase Plane.
When plotting my solution I think I am getting the solutions for $x_0 < -1 $ and $x_0 > 1$ because they look like it.
I think the Wolfram Alpha solution delivers the solution for $-1<x_0<1$.
Can someone please enlighten me and tell me what I am not getting?
Can I get from one solution to the other?
Is it because of the absolute values I get when integrating?
Thank you all in advance,
Cheers
Santiago
 A: You're right that the problem lies in handling the absolute values. WA may not worry about that, since it's working more formally, possibly allowing complex-valued solutions. But anyway, the correct way of dealing with real-valued solutions is the following:
First, $x(t)=-1$ and $x(t)=1$ are constant solutions. All other solutions necessary satisfy $x \neq \pm 1$ (since different solution curves cannot cross), which means that $x^2 - 1 \neq 0$ and it's safe to separate the variables:
$$
\int\frac{dx}{x^2-1} = \int dt
.
$$
This gives
$$
\frac12 \ln \left| \frac{x-1}{x+1} \right| = t + C
,
$$
where $C \in \mathbf{R}$ is arbitrary,
which in turn implies that
$$
\frac{x-1}{x+1} = \pm e^{2(t+C)} = \pm e^{2C} e^{2t} = D e^{2t}
,
$$
where $D = \pm e^{2C}$ is an arbitrary nonzero real constant.
So we get
$$
x=-1
\quad\text{or}\quad
x=1
\quad\text{or}\quad
x=\frac{1+D e^{2t}}{1-D e^{2t}}
\quad
(D \neq 0)
,
$$
which can be further simplified to
$$
x=-1
\quad\text{or}\quad
x=\frac{1+D e^{2t}}{1-D e^{2t}}
\quad
(D \in \mathbf{R})
.
$$
This will allow you to find the solution for any initial value, say $x(0) = x_0 \in \mathbf{R}$. If $x_0=-1$, then $x(t)=-1$, otherwise there is a unique $D \in \mathbf{R} \setminus \{ 1 \}$, namely $D=(x_0-1)/(x_0+1)$, such that the solution is given by $x(t)=\frac{1+D e^{2t}}{1-D e^{2t}}$ (for all $t$ in the maximal interval of existence).
(Notice that if $x_0>1$ so that $0<D<1$, then the solution blows up at a finite positive time when the denominator becomes zero, and if $x_0<-1$ so that $D>1$, then the solution blows up at a finite negative time. The solution with $D=1$ is singular at $t=0$, which is why that value of $D$ cannot occur if the initial value is given at $t=0$.)
