Difficulties in KVL signs (RL circuit)
With the current flowing as shown,
$$v_L = L\frac{di}{dt}$$
and 
$$v_R = -iR$$
with \$v_R\$ being negative as expected.  

Using \$v_R\$ and \$v_L\$ with KVL yields
$$ v_R=v_L$$
or
$$ -v_R+v_L=0 $$
Note there is a negative sign as well.

Substituting the first 2 equations yields
$$ -(-iR) + L\frac{di}{dt} = 0$$
$$ iR + L\frac{di}{dt} = 0$$