The current in an $LR$ circuit I am trying to understand $LR$ circuits better. I am under the impression that inductors are resistant to change in current.

So, the premise here is that the circuit was switched to side 1 for a long time and switched on to side 2.
What I thought was the 4H inductor would be "okay" with the 5A current when the switch is in side 1. When the switch is switched on to side 2, i.e. when the current should change to 2.5A, the 4H inductor would behave as if 5A was passing through it, but the current is changing gradually to 2.5A. So, without that 1H inductor, the current should be
$$I=2.5(1+e^{-(4t/5)})$$
Because the 1H inductor is then applied to the circuit, then the current should be
$$I=2.5(1+e^{-(4t/4)})(1-e^{4t})$$
I know I took a bunch of wrong assumptions, but how should we find the answer?
 A: The problem is not with you. This is an example where the assumption of ideal circuit behavior leads to inconsistencies.
Assuming ideal inductors, immediately after switching the current in the 1 H inductor has to be zero, as it was before switching, while the current in the 4 H inductor in series with it has to be 5 A, as it was before switching. This gives the inconsistent result of initially having two different values of current for a series circuit.
Hope this helps.
A: I need a labeled diagram.

There are two kinds of switches, make-before-break, which means that the two contacts and the pole will all be shorted together before completing the switching. And break-before-make where the two contacts and the pole are all disconnected before the action is completed.
As you say, after sitting for a long time in POS1, $V_{L1} = 5$A, $V_{L2} = 0$, $I_{L1} = 0$ and, $I_{L2} = 0$.
Make-before-break switching to POS2:
During the transition the currents and voltages remain the same. When the POS contact opens (goes to high resistance) then sparks happen.(Actually!) It happens as a result of the defining relationship for the inductor.$$v_{L} = L\frac {di}{dt}\frac {di}{dt}\rightarrow \infty $$, so the voltage will go to infinity across $L_{1}$ because the inductor is trying to maintain 4A if current. In doing so, the switch will arc between position 1 and the pole. The arc will allow 4A of current to flow. as the current stabilizes in the inductor, the voltage across the inductor and switch drop until the arc cannot be maintained an current will cease to flow.
Because S1 is in position 2, the high-voltage across $L_{1}$ will appear across $L_{2}$. $L_{2}$ will happily respond but not allow current to change instantly. However the current in $L_{2}$ will change quickly while the high voltage is present (micro seconds) then after the current in all the components will increase according to $2.5(1-e^{t/\tau})$.
During the transition, the currents in L1 and L2 will equalize. What value that is depends in the resistance in the arc.
The arc makes it difficult to predict. The voltage across the switch can rise to as high as 1000V/mm of gap. The resistance of the arc is almost 0 so the voltage goes immediately back to zero. This should extinguish the arc.
Break-before-make switching to POS2:
During the transition, all the switch contacts are open, so by time the switch reaches POS2, the arc may be extinguished. If not my previous comments apply.
There is always a small amount of inductance in electrical wiring. Even circuits that switch microamps will cause switch arcing and is the reason that equipment in explosive environments must has there switches in a sealed case making them intrinsically safe
