We need to write the RHS of the DE in terms of unit step functions.

Now, taking Laplace transform of both sides gives us:

We need to find Inverse Laplace. First, we concentrate on the part and ignore the part for now.

Now

gives 1 = 2B gives

gives 1 = 4C gives

gives 1 = 3A + 3B + C gives

So

So since

then we have, using the Time-Displacement Theorem (see the Table of Laplace Transforms):

The graph of i(t) is as follows: