\(\displaystyle{i}{\left({0}\right)}={A}_{{{1}}}{e}^{{{0}}}+{A}_{{{2}}}{e}^{{{0}}}={A}_{{{1}}}+{A}_{{{2}}}={0.12}{A}\)

We also know that \(\displaystyle{v}{\left({0}\right)}={3}{V}\) so we write

\(\displaystyle{\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}}=-{500}{A}_{{{1}}}{e}^{{-{500}{t}}}-{2000}{A}_{{{2}}}{e}^{{-{2000}{t}}}\)

\(\displaystyle{v}={L}{\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}}=-{25}{A}_{{{1}}}{e}^{{-{500}{t}}}-{100}{A}_{{{2}}}{e}^{{-{2000}{t}}}\)

\(\displaystyle{v}{\left({0}\right)}=-{25}{A}_{{{1}}}{e}^{{{0}}}-{100}{A}_{{{2}}}{e}^{{{0}}}=-{25}{A}_{{{1}}}-{100}{A}_{{{2}}}={3}{V}\)

Now we have two equations with two unknowns:

\(\displaystyle{A}_{{{1}}}+{A}_{{{2}}}={0.12}\)

\(\displaystyle-{25}{A}_{{{1}}}-{100}{A}_{{{2}}}={3}\)

We solve this and get

\(\displaystyle{A}_{{{1}}}={0.2}\) and \(\displaystyle{A}_{{{2}}}=-{0.08}\)

Now we can write the expression for the voltage

\(\displaystyle{v}{\left({t}\right)}=-{25}{\left({0.2}\right)}{e}^{{-{500}{t}}}-{100}{\left(-{0.08}\right)}{e}^{{-{2000}{t}}}\)

\(\displaystyle{v}{\left({t}\right)}=-{5}{e}^{{-{500}{t}}}+{8}{e}^{{-{2000}{t}}}{V}\)