Actually, for electric circuits, the voltage across an inductor is
$v_L = L \dfrac{di_L}{dt}$
By the passive sign convention, the inductor current enters the terminal that is the more positive when $v_L>0$.
Here's what you do. Pick a direction for the inductor current by drawing an arrow pointing into one of the terminals. Label that arrow $i_L$. Now, place a plus sign at the terminal the current enters and a minus sign at the other terminal. This is the polarity of the inductor voltage $v_L$.
By the formula above, if the inductor current is increasing with time, $v_L>0$. If the inductor current is decreasing with time, $v_L < 0$.
You can't pick the "wrong" direction or polarity. The signs will take care of themselves.
Is you problem similar to this one?
When the switch has been in position 2 for a long time, the series electric current is constant and clockwise through the resistor, inductor, and source. Since there is only one current, the series current, let's just label that $i$ and have the arrow pointing into the left end of the resistor.
According to the passive sign convention, the polarity of our voltage variables, $v_R, v_L$ have the positive sign at the left end of the resistor and the top end of the inductor.
Since the current is constant, $v_L = 0$, thus $i=E/R$ and $v_R = E$ for $t<0$.
If at $t=0$ the switch is (instantaneously) changed to position 1, The current through the inductor must be same at $t=0+$ as at $t=0-$, i.e., the current through the inductor is continuous.
With the same current as before, the voltage across the resistor $v_R(0+) = E$. But now, this means that $v_L(0+) = -E$ by KVL (sum of voltage drops around a closed loop is zero). This implies that the current through the inductor is decreasing (negative time rate of change) which is what you should expect!
We have the initial condition, $i(0+) = E/R$, and the form of $i(t)$ is the well known decaying exponential:
$i(t) = \dfrac{E}{R}e^{-\frac{t}{\tau}}, \tau = \dfrac{L}{R}, t>0$
