Circuit analysis - laplace transform
Assuming R1=5Ω , 

Initially, \$V_C\ + V_{R_1} = V_{R_2}\$. Since no current flows through C for \$t(0^-)\$ , \$i_{R_2}=1A\$. So, \$V_{R_2}=V_C=2V\$.

For \$t&gt;0\$,

$$E=L\frac{di}{dt} + i\cdot R_1 + \frac{1}{C}\int i dt + V_C(0)$$

Taking Laplace Transform,

$$\frac{E}{s} = L(sI(s)-I_L(0)) + I(s) \cdot R_1 + \frac{1}{Cs}I(s) + \frac{V_C(0)}{s}$$

Here \$I_L(0) = 1A, V_C(0) = 2V\$

Rearranging, 

$$\frac{E+LsI_L(0)-V_C(0)}{Ls^2 + s R_1 + \frac{1}{C} } = I(s)$$

Substitute the values and resolve into partial fractions and then take Inverse Transform to get the value of the current.

Edit: 

The Laplace Inverse of 1/7 is 1/7δ(t), which is a function that takes the value 0 for all \$t\neq 0\$.