Does it mean that there no analytical solution if Mathematica cannot find analytical solution?

Question

I'm solving simple but coupled ODEs recently. I use both MATLAB symbolic computation and Mathematica.

For example, my coupled ODE is the following \begin{align*} &\dot{x}(t)=y(t)-\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ &\dot{y}(t)=-y(t)+\rho b\frac{x(t)}{1-\rho(1-e^{-t})}\\ \end{align*} where $\rho\in(0,1)$ and $b\in(0,1)$ are given constants and the initial value of this set of ODEs are $x(0)=a$, there $a\in(0,1)$ and $y(0)=0$.

This expression looks simple but these 2 equations are coupled.

First, I tried MATLAB, it generate "Warning: Explicit solution could not be found." explicitly.

Then I tried Mathematica,

system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])), 
   y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};

Then I try to solve it via sol = DSolve[system, {x, y}, t].

The thing that I don't understand is that after I press Shift+Enter, Mathematica only makes my input look nicer, but didn't produce any result or generating any warning message like MATLAB. So I couldn't tell whether it is because Mathematica also couldn't find the analytical solution like MATLAB, or it just does not even try to solve the problem since I input something wrong?

This simple coupled ODE drives me crazy these days. Any suggestion, input is deeply appreciated.

If math software couldn't find analytical solution, then is it still possible to analyze the monotonicity of the solution? For example, in this case, it's easy to analyze the case when $t=\infty$ by setting $\dot{x}=0=\dot{y}$ and solve the equations. I can also numerically plot the solution to see the trend, e.g., $x(t)$ is decreasing. But without the explicit functional form of $x(t)$, how can we prove the monotonicity, stuff like that? In general, if the math software fails to find analytical solution, what should we do next?

Answer 1

Here, we take advantage of the observation mentioned in a comment that the sum of x and y is conserved. If we define

s[t] == x[t] + y[t]
d[t] == x[t] - y[t]

your differential equations become

eqns = {s'[t] == 0,
        d'[t] == -(1 + (r b)/(1 - (1 - E^-t) r)) d[t] + (1 - (r b)/(1 - (1 - E^-t) r)) s[t]}

Mathematica knows how to solve this:

First@DSolve[eqns, {s[t], d[t]}, t]

Answer 2

As noted in the comments above, you have $\dot{x} + \dot{y} = 0$, which implies that $x(t) + y(t)$ is a constant; call it $c_3$. (Note that in particular, $c_3 = x(0) + y(0)$.) You can therefore replace y[t] with c3 - x[t] in your first equation above, and Mathematica can solve that explicitly:

reducedsystem = {x'[t] == c3 - x[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))}
DSolve[reducedsystem, x[t], t]

(* {{x[t] -> -((c3 E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) (-E^t + c1 (-1 + E^t))^(1 - (c1 c2)/(-1 + c1)))/(1 + c1 (-1 + c2))) 
               + E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}} *)

I'm a little surprised that Mathematica can't do this on its own, to be honest; but here we are.

Answer 3

The non-trivial solution for this system can be obtained by doing this:

system = {x'[t] == y[t] - c1*c2*x[t]/(1 - c1*(1 - Exp[-t])), 
   y'[t] == -y[t] + c1*c2*x[t]/(1 - c1*(1 - Exp[-t]))};

DSolve[First[system /. y -> (-x[#] &)], x[t], t]

(*
==> {{x[t] -> 
   E^(-t + (c1 c2 Log[-c1 - E^t + c1 E^t])/(-1 + c1)) C[1]}}
*)

All I did here is use the fact that the equations become identical if $y=-x$. That allows us to drop one of the equations.