Nonlinear ODE $\dot{X}(t)X(t)=B$ I am learning ODE now.
we know for ODE:
$$\dot{X}(t)=A(t)X(t),X(t_0)=x_0$$
where $X\in \mathbb{R}^{n\times 1},A\in\mathbb{R}^{n\times n}$.
The solution has the following form:
$$X(t)=e^{\int_{t_0}^{t}A(s)ds}x_0$$
My question is for the following ODE:
$$\dot{X}(t)X(t)=A(t),X(t_0)=x_0$$
where $X\in \mathbb{R}^{n\times n},A\in\mathbb{R}^{n\times n}$.
Do we have general experison? By the way, can we use this ODE to define $\log{A}$
?

We do not know that.  $X(t) = e^{\int_{t_0}^t A(s)\; ds} x_0$ is a solution to your first differential equation if the $A(s)$ all commute, but not otherwise.  What you need in general is a "time-ordered exponential".
A similar difficulty applies to your second equation.  If  $X(t)$ and $\dot{X}(t)$ commute, then $Y(t) = X(t)^2$ satisfies $\dot{Y}(t) =  2 \dot{X(t)} X(t) = 2 A(t)$, so 
$X(t)$ is a square root of $Y(t) = Y(0) + 2 \int_0^t A(s)\; ds$.  But if they don't commute, $\dot{Y}(t) = \dot{X(t)} X(t) + X(t) \dot{X(t)}$, and we don't know what to do with the second term.
