How to solve the system of differential equations? Let $x_i=x_i(t,s),i=0,1,2$ be  functions of $t,s$ and consider the system of differential equations
\begin{cases}
\frac{\partial x_0}{\partial t}=0,\\
\frac{\partial x_1}{\partial t}=x_0,\\
\frac{\partial x_2}{\partial t}=2x_1,\\
\frac{\partial x_0}{\partial s}=2x_1,\\
\frac{\partial x_1}{\partial s}=x_2,\\
\frac{\partial x_2}{\partial s}=0,
\end{cases}
How to solve it?
My attemt.
First of all  I have solved the first 3 equations and get 
\begin{align}
 &x_{{0}}=C_1 \left( s \right) ,\\
&x_{{1}}  =C_1 \left( s \right) t+C_2(s),\\
&x_{{2}}  =C_1 \left( s
 \right) {t}^{2}+2\,C_2 \left( s \right) t+C_3 \left( s
 \right),  
\end{align}
and substitute into the last 3 equation 
$$
\begin{cases}
\displaystyle {\frac { d}{{ d}s}}C_1 \left( s
 \right) =2\,C_1 \left( s \right) t+2\,C_2 \left( s
 \right),\\
\\
 \displaystyle t  {\frac { d}{{ d}s}}C_1 \left( s \right)+{\frac { d}{{ d}s}}C_2 \left( s \right) =C_1 \left( s \right) {t}^{2}+2\,C_2 \left( s \right) t+C_3 \left( s \right) ,\\ \\
 \displaystyle {t}^{2}\, {\frac { d}{{ d}s}}C_1
 \left( s \right)  +2\,t  {\frac { d}{{ d}s}
}C_2 \left( s \right) +{\frac { d}{{ d}s}}C_3 \left( s \right) =0,
\end{cases}
$$
or normalize it:
$$
\begin{cases}
\displaystyle{\frac { d}{{ d}s}}C_1 \left( s
 \right)=2\,C_1 \left( s \right) t+2\,C_2 \left( s \right) ,\\ \\
\displaystyle{\frac { d}{{ d}s}}C_2 \left( s
 \right)=-C_1 \left( s \right) {t}^{2}+C_2 \left( s \right) ,\\ \\
\displaystyle{\frac { d}{{ d}s}}C_3 \left( s
 \right)=-2\,C_2 \left( s \right) {t}^{
2}-2\,C_2 \left( s \right) t 
\end{cases}
$$
Then I solve the system for $C_1(s),C_2(s),C_3(s)$ and get 
\begin{align*}
&C_1 \left( s \right) =-{\frac {2\,C_1'\,{s}^{2}{t}^{2}+4\,C_2'\,s{t}^{2}+2\,C_1'\,st+4\,C_3'\,{
t}^{2}+2\,C_2'\,t+C_1'}{4{t}^{3}}},\\
&C_2 \left( s
 \right) =\frac{1}{2}\,C_1'\,{s}^{2}+C_2'\,s+C_3', \\
&C_3
 \left( s \right) =-\,{\frac {2\,C_1'\,{s}^{2}{t}^{2}+4\,C_2\,s{t}^{2}-2\,C_1'\,st+4\,C_3'\,{t}^{2}-2\,C_2'\,t+C_1'}{4t}}.
\end{align*}
Here $C_1',C_2',C_3'$ are constants.
Then I substitute it  into the above expression for $x_0,x_1,x_2$  and get 
\begin{align*}
&x_{{0}} \left( t,s \right) ={\frac { \left( -2\,C_1'\,{s}^{2}-4\,C_2'\,s-4\,C_3' \right) {t}^{2}+ \left( 
-2\,C_1'\,s-2\,C_2' \right) t-C_1'}{4{t}^{3}}}, \\&x_{{1}
} \left( t,s \right) ={\frac { \left( -2\,C_1'\,s-2\,{C_2'} \right) t-C_1'}{4{t}^{2}}},\\ &x_{{2}} \left( t,s \right) =-
\,{\frac {C_1'}{2t}}
\end{align*}
But obviously it is not solutions of the initial system!
Where is my mistake?
Note, the system have solution, for example $x_1^2-x_0 x_2$ is its first integral.
 A: $$
\frac{\partial x}{\partial t}=M_1 x\Rightarrow x = C_1e^{M_1 t}+\phi(s)\\
\frac{\partial x}{\partial s}=M_2 x\Rightarrow x = C_2e^{M_2 s}+\psi(t)\\
$$
then
$$
x = C_1e^{M_1 t}+C_2e^{M_2 s}
$$
but for this solution we have
$$
\frac{\partial x}{\partial t}-M_1 x = -M_1C_2e^{M_2 s}=0\Rightarrow C_2 = 0\\
\frac{\partial x}{\partial s}-M_2 x = -M_1C_1e^{M_1 t}=0\Rightarrow C_1 = 0\\
$$
hence the solution is $x = 0$
A: The first and last equations give $x_0(t,s) = x_0(s)$ and $x_2(t,s) = x_2(t)$. Therefore, the system becomes
\begin{aligned}
\partial_{t} x_1 &= x_0 \\
\partial_{t} x_2 &= 2x_1 \\
\tfrac{\text d}{\text d s} x_0 &= 2x_1 \\
\partial_{s} x_1 &= x_2
\end{aligned}
Applying $\partial_{t}$ to the third line yields $0 = 2\partial_{t} x_1$ and the first line gives $\partial_{t} x_1 = x_0$. Hence we know $x_0=0$ and $x_1(t,s) = x_1(s)$. The system
becomes
\begin{aligned}
\partial_{t} x_2 &= 2x_1 \\
0 &= 2x_1 \\
\tfrac{\text d}{\text d s} x_1 &= x_2
\end{aligned}
Since $x_1=0$ as well, we end up with $x_2=0$. The solution must be
$$
x_0 = x_1 = x_2 = 0.
$$
