Solve $(A- \operatorname{diag}(x) ) \,\nabla_x f(x) - c f(x)=0, \, f(0)=1$ Let $f : \mathbb{R}^n \to \mathbb{R}$. How to solve the following differential equation
$$
(A-  \operatorname{diag}(x) ) \nabla_x f(x) - c f(x)=0,  \qquad  f(0)=1.
$$
where  $\operatorname{diag}(x)$  is a diagonal matrix with vector $x$ on the main diagonal, $A$ is some $n \times n$ matrix  and $ c \in \mathbb{R}^n$?
In the scalar case, this is easy to solve since it is a first-order linear ordinary differential equation (ODE) whose solution is given by
$$f(x) = a^{c} (a -  x)^{-c}.$$
I don't have have much experience solving matrix differential equations and would appreciate some references on this topic. 

Edit: Using the approach of NN2 the above PDE can be reformulated as:
\begin{align}
(A-  \operatorname{diag}(x) )  \nabla_x g(x)=c,  \qquad g(0)=0.
\end{align}
 A: My first answer isn't good. I edit the answer by taking account the comment of @loupblanc. I haven't had yet the solution.
I think we can simplify the problem by using this transformation.
From the equation, we have
$$
(A-  \operatorname{diag}(x) ) \frac{\nabla_x f(x)}{f(x)} - c=0
$$
$$ \iff
(A-  \operatorname{diag}(x) ) \nabla_x \ln{f(x)} - c=0
$$
Put $g(x) = \ln{f(x)}$
$$ \iff
(A-  \operatorname{diag}(x) ) \nabla_x g(x) - c=0
$$
A: We consider the ODE in the form obtained by NN2:
$(*)$ $\nabla_xg(x)=(A-diag(x))^{-1}c=[v_1,\cdots v_n]^T$. 
EDIT 1.
In general $(*)$ has no solutions except if $curl(V)=0$, that is if, for every $i<j$, 
$\dfrac{\partial v_i}{\partial x_j}=\dfrac{\partial v_j}{\partial x_i}$. This NS condition reflects the fact that $\dfrac{\partial^2 g}{\partial x_jx_i}=\dfrac{\partial^2 g}{\partial x_ix_j}$.
For example, for $n=2$, the conditions on $A=[a_{i,j}],c=[c_i]$ are
$a_{1,2}c_2=a_{2,1}c_1=0.$
For $n=3$, the conditions are 
$a_{2,1}c_1=a_{3,1}c_1=a_{1,2}c_2=a_{3,2}c_2=a_{1,3}c_3=a_{2,3}c_3=0$. 
In particular, if, for every $i$, $c_i\not= 0$, then $A$ is diagonal (at least when $n=2,3$) and the integration of the ODE is easy.
EDIT 2. To the OP. Since I have pity on you, I'll give you a basic example that may convince you that there is no solution except for exceptional choices of $A$ and $c$.
We choose $A=\begin{pmatrix}0&1\\1&0\end{pmatrix},c=[1,1]^T$.
The considered equation is $\Delta_x g(x)=(A-diag(x))^{-1}c$, that is
$[\dfrac{\partial g}{\partial x_1},\dfrac{\partial g}{\partial x_2}]^T=[\dfrac{-x_2-1}{x_1x_2-1},\dfrac{-x_1-1}{x_1x_2-1}]^T$. Then necessarily
$\dfrac{\partial^2 g}{\partial x_1x_2}=\dfrac{1+x_1}{(x_1x_2-1)^2}$ and $\dfrac{\partial^2 g}{\partial x_2x_1}=\dfrac{1+x_2}{(x_1x_2-1)^2}$. 
The last two quantities are not equal; therefore, there are no solutions $g$ that are $C^2$. 
