First find solution of homogenenous problem.

\(\displaystyle{r}^{{2}}-{2}{r}-{3}={0}\to{r}_{{{1},{2}}}=-{1},{3}\to\) using quadratic formula

Homogeneous solution:

\(\displaystyle{y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}\)

Let \(\displaystyle{Y}={A}{e}^{{{2}{t}}}\) because \(\displaystyle{g{{\left({t}\right)}}}={3}{e}^{{{2}{t}}}\)

Plug Y into starting equation to find particular solution:

\(\displaystyle{\left({A}{e}^{{{2}{t}}}\right)}{''}-{2}{\left({A}{e}^{{{2}{t}}}\right)}'-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle{4}{A}{e}^{{{2}{t}}}-{4}{A}{e}^{{{2}{t}}}-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle{A}=-{1}\)

\(\displaystyle{Y}=-{e}^{{{2}{t}}}\)

Solution: \(\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}\)

Result: \(\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}\)

\(\displaystyle{r}^{{2}}-{2}{r}-{3}={0}\to{r}_{{{1},{2}}}=-{1},{3}\to\) using quadratic formula

Homogeneous solution:

\(\displaystyle{y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}\)

Let \(\displaystyle{Y}={A}{e}^{{{2}{t}}}\) because \(\displaystyle{g{{\left({t}\right)}}}={3}{e}^{{{2}{t}}}\)

Plug Y into starting equation to find particular solution:

\(\displaystyle{\left({A}{e}^{{{2}{t}}}\right)}{''}-{2}{\left({A}{e}^{{{2}{t}}}\right)}'-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle{4}{A}{e}^{{{2}{t}}}-{4}{A}{e}^{{{2}{t}}}-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle-{3}{A}{e}^{{{2}{t}}}={3}{e}^{{{2}{t}}}\)

\(\displaystyle{A}=-{1}\)

\(\displaystyle{Y}=-{e}^{{{2}{t}}}\)

Solution: \(\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}\)

Result: \(\displaystyle{y}={y}_{{c}}={c}_{{1}}{e}^{{-{t}}}+{c}_{{2}}{e}^{{{3}{t}}}-{e}^{{{2}{t}}}\)