Solve a system of ordinary differential equations I have the following set of ordinary differential equations:
$$
\begin{cases}
x_1'(s) &= -e^{-s} (1-x_1(s)) - x_2(s) + x_1(s) x_3(s)\\
x_2'(s) &= -x_2(s) + x_1(s)^2\\
x_3'(s) &= -x_3(s) + x_1(s),
\end{cases}
$$
with boundary condition $x_2(0)=x_3(0)=0$.
I am looking for a method to find an exact solution for this type of ODE. In particular I am interested in finding a solution for $x_1$ which satisfies $x_1(\infty) = 0$, but it already excelent if the solution of $x_1$ is simply given in function of a general boundary condition $x_1(0) = a$.
EDIT I have found this link of methods for solving this type of problems, but my problem does not seem to fit in any of the suggested methods.
 A: $$
\begin{cases}
x_1'(s) &= x_2(s) - x_1(s) x_3(s)\\
x_2'(s) &= -x_2(s) + x_1(s)^2\\
x_3'(s) &= -x_3(s) + x_1(s),
\end{cases}
$$
We express $x_1$ and $x_2$ as functions of $x_3$. For simplification let :
$$x_3=X$$
From the third equation :
$$x_1=X'-X$$
From the first equation $$x_2=X''-X'+(X'-X)X$$
From the second equation :
$$X'''-X''+(X''-X')X+(X'-X)X'=-(X''-X'+(X'-X)X)+(X'-X)^2$$
Now we have a third order ODE with only one function $X(s)$.
$$X'''+X''X+(-X'+X-1)X'-X^2=0$$
This is an autonomous ODE. The change of function is :
$$X'(s)=F(X) \quad;\quad X''(s)=F'(X)F(X) \quad;\quad X'''(s)=F''F^2+F'^2F$$
$$F''F^2+F'^2F+XF'F+(-F+X-1)F-X^2=0$$
This is a second order non-linear ODE difficult to solve. Probably there is no closed form general solution with the standard functions. 
Some particular solutions might be found. For example $F(X)=X$ is a particular solution. But this solution which is not compatible with the boundary condition $x_3(0)=0$ must be rejected.
If the question comes from an academic exercise, a typo might be suspected in the given equations. Especially the third equation is doubtful. Are you sure that nothing is missing or wrong in the third equation ? 
A: Some terms where missing in the first edition of the question. The wording becomes clear with the original integro-differential equation from which the system of ODEs is derived.
$$x'(s)=e^{-s}(1-x(s))+\int_0^se^{-(s-u)}x^2(u)du-\int_0^se^{-(s-u)}x(u)x(s)du
$$
$$\begin{cases}
x_1=x(s)\\
x_2=\int_0^se^{-(s-u)}x^2(u)du \\
x_3=\int_0^se^{-(s-u)}x(u)du\
\end{cases}$$
$$
\begin{cases}
x_1'(s) &= e^{-s} (1-x_1(s)) + x_2(s) - x_1(s) x_3(s)\\
x_2'(s) &= -x_2(s) + x_1(s)^2\\
x_3'(s) &= -x_3(s) + x_1(s)
\end{cases}
$$
We express $x_1$ and $x_2$ as functions of $x_3$. For simplification let :
$$x_3=X$$
From the third equation :
$$x_1=X'+X$$
From the first equation $$x_2 =(X''+X') -e^{-s} (1-X'-X)  + (X'+X)X$$
From the second equation :
$$
(X'''+X'')+e^{-s}(1-X'-X)-e^{-s}(-X''-X')+(X''°X')X+(X'+X) = -(X''+X')+e^{-s}(1-X'-X)-(X'+X)X+(X'+X)^2
$$
Now we have a third order ODE with only one function $X(s)$, after simplification :
$$ X'''+(2+X+e^{-s})X'' +(1+X+e^{-s})X' =0 $$
This is a non-linear third order ODE. It can be simplified with the change :
$$t=e^{-s}\quad;\quad X(s)=Y(t)$$
$X'(s)=Y'(t)\frac{dt}{ds}=Y'(t)(-t)=-tY'(t)$
$X''(s)=(-Y'(t)-tY''(t))(-t)=tY'(t)+t^2Y''(t)$
$X'''(s)=(Y'(t)+3tY''(t)+t^2Y'''(t))(-t)= -tY'(t)-3t^2Y''(t)-t^3Y'''(t)$
$ (-tY'-3t^2Y''-t^3Y''')+(2+Y+t)(tY'+t^2Y'') +(1+Y+t)(-tY') =0 $ 
After simplification :
$$tY'''+(-Y-t+1)Y''=0$$
The general solution seems arduous to find. But we don't need it to find some particular solutions. Obviously $Y''=0$ leads to solutions of the form :
$$Y(t)=c_1t+c_2$$
where $c_1$ and $c_2$ are arbitrary constants.
$$X(s)=c_1\;e^{-s}+c_2$$
Putting it into the above equations for $x_1$ and $x_2$ leads to 
$$\begin{cases}
x_1=c_2 \\
x_2=c_1\;e^{-s}+c_2\\
x_3=c_1\;e^{-s}+c_2
\end{cases}$$
The conditions $x_2(0)=x_3(0)=0$ leads to $c_1+c_2=0$. Then putting it into the integro-differential equation shows that $c_2=1$. Thus the answer is 
$$\begin{cases}
x_1(s)=1 \\
x_2(s)=1-e^{-s}\\
x_3(s)=1-e^{-s}
\end{cases}$$
This is a very simple solution. 
This rightly answers to the question insofar the third condition is $x_1(0)=1$. No condition on $x_1$ was specified in the wording of the problem. 
Note : If there was a  third specified condition $x_1(0)\neq 1$ the above very simple solution would be no longer convenient. In this case we have to fully solve the ODE $\:tY'''+(-Y-t+1)Y''=0$ in order to have the general solution which must involve three arbitrary  constants $c_1$ , $c_2$ and $c_3$ to be determined according to the boundary conditions. This would be more arduous.
A: Just a beginning, you can eliminate $x_3$
$$
\begin{cases}
x_1' &= x_2 - x_1 x_3\\
x_2' &= -x_2 + x_1^2\\
x_3' &= -x_3 + x_1,
\end{cases}
$$
Sum the first and the second: $$x_1'+x_2'=x_1^2-x_1 x_3=x_1(x_1-x_3)$$
So we have 
$$
\begin{cases}
x_3' &= {x_1' +x_2' \over x_1} \\
x_3 &= x_1 - {x_1' +x_2' \over x_1}  ,
\end{cases}
$$
