assume we have a mixing problem with a salt solution coming into a water tank. the flow rate in and flow out rate are equal (5 L/hour), the concentration of salt flowing in is 1 g/L and the tank contains 100 L of water. this is described by the ODE,
$\displaystyle\frac{dS}{dt} = 5\times 1 - 5\displaystyle\frac{S}{100}$
if the tank contains no salt initially it takes ~1 hour to add 5 g of salt to it. however, if tank starts with 5 g of salt and pure water is flowed into the tank then after 1 hour there will be ~4.7 grams of salt in tank, not roughly 0 grams. this asymmetry is counterintuitive - is there a physical explanation for it?
if we instead had 95 grams of salt in the tank to start with and flowed in pure water at the rates described above, then after 1 hour there would be 90 grams of salt in the tank, as expected (since going from 0 grams to 5 grams took 1 hour when we flowed salt in.) put another way, the solution equation $S(t)$ to the ODE is not linear. what is the intuition behind the initial amount of salt making such a big difference to the rate, physically?
There is no asymmetry other than you can add salt much more quickly than you can remove it.
Keep in mind, the time constant here is 20 hours, so not much happens in an hour, so for periods of an hour, the rate of change of salt is well approximated by $-{1 \over 20} S + u$, where $u=5$ when you add salt but $u=0$ when adding pure water, and $S$ is the initial amount of salt.
If you add 5 l in an hour at a concentration of 1 g/l, then you add $\approx -{1 \over 20} 0 + 5$ g in an hour.
The tank loses 5 l in an hour, so if you start with 5 g in 100 l, you would expect to add (well, lose) $\approx -{1 \over 20} 5 + 0$ g in an hour.
In the second case we have an initial 95 g and add pure water, so you would expect to add (well, lose) $\approx -{1 \over 20} 95 + 0$ g in an hour.
The system is linear as a map from the initial state $S(0)$ and the rate of salt addition $t \mapsto u(t)$ to the function $t \mapsto S(t)$.