Finding the unknowns of a current equation leads to non-sense
$$ i_{(0+)}= i_{(0-)}= 0 $$
$$ v_{L_{(0-)}} = 0 $$
$$ v_{L_{(0+)}} = L \left[ \frac{di}{dt} \right]_{t=t_{0+}} $$
Note that  \$\left[ \frac{di}{dt} \right]_{t=t_{0+}} \neq 0 \$

From \$ i_{(0+)} \$ , \$ B = 0 \$

From the derivative of response \$i(t)\$:
$$\left[ \frac{di}{dt} \right]_{t=t_{o+}}= \frac{v_{L_{(0+)}}}{L} = A$$

From the KVL (\$V\$ is the source voltage and \$V_{C_{(0+)}}\$ is the voltage on capacitor in \$t=t_{o+}\$):

$$ v_{L_{(0+)}} =  -V_{C_{(0+)}} + V -Ri_{(0+)}$$

or 

$$ v_{L_{(0+)}} = V$$

Therefore:

$$ A=\frac{V}{L}$$