Basic ODE Question ODEs is by far my weakest area so I thought to ask here.
Fix $a_1,\dots,a_d,b,x_0 \in \mathbb{R}^d$.
I'm trying to solve the system:
$$
\dot{x}_t = \left(\tanh(a \cdot x_t + b_1),\dots,\tanh(a \cdot x_t + b_d)
\right)
$$
So Far: (What I've got)
In that case I can solve
$$
\dot{x}_t = \tanh(ax_t +b) ; x_0=x_0.
$$
to be
$$
x_t = \frac{\log(\cosh(at +b))}{a} + c(x_0),
$$
where $c(x_0)$ is a constant depending on the initial condition $x_0$.  
Question:
How can we solve the multi-dimensional version?  Unlike the 1-d version I don't know what do next.  
 A: For $d=1$.
You can solve it by using the fact that this ODE is "separable", namely 
$$
\frac{dx}{\tanh(a x +b) } = dt
$$
This gives you (integrate between $x_0$ and $x$ and between $0$ and $t$, then use $y = a x + b$)
$$
\int_{a x_0 + b}^{a x +b}\frac{dy}{\tanh(y) } = t
$$
The final step is to know that $1/\tanh(z) = \coth(z)$ and that $\int \coth(z) dz = \ln(\sinh(z))$. Now apply this fact to the above definite integral to obtain the solution.
EDIT:
For $d>1$, a strategy could be to use the scalar $z(t) = x(t) \cdot a$. Multiply by $a$ the RHS and LHS of the ODE system to obtain
$$
\dot{z}(t) = a_1 \tanh(z(t) +b_1)+...+ a_d \tanh(z(t) +b_d)
$$
with the initial condition $z(0) = a\cdot x_0$.
You still have a single ODE for $z$, but in general it seems difficult to find a general solution, unless $b_1 = ...= b_d$. If all the $b_i$ are the same, then you have the ODE:
$$
\dot{z}(t) = A \tanh(z(t) +B)
$$
with $A=a_1+...+a_d$ and $B=b_1,...,b_d$. This can be solved with a method analogous to the one shown for the $d=1$ case.
EDIT #2 (since I was asked):
For $d=1$ you have $x(t)=[\sinh^{-1}( e^{a t} \sinh(b+a x_0)) -b]/a$
For $d>1$ you have $z(t) = \sinh^{-1}(e^{A t} \sinh(B+z_0))-B$ 
