From the given linear differential equation, \({P}{\left({x}\right)}=-{3}{\quad\text{and}\quad}{Q}{\left({x}\right)}={e}^{{{2}{x}}}\).

Therefore, the integrating factor is given by

\(A{\left({x}\right)}={\int_{{0}}^{{x}}}-{3}{t}{\left.{d}{t}\right.}=-{3}\)

So that, the solution is given by

\(y={0}{e}^{{{3}{x}}}+{e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{{-{3}{t}}}{e}^{{{2}{t}}}{\left.{d}{t}\right.}\)

\(={e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{ -{{t}}}{\left.{d}{t}\right.}\)

\(={e}^{{{3}{x}}}{{\left[\frac{{e}^{ -{{t}}}}{ -{{1}}}\right]}_{{0}}^{{x}}}\)

\(={e}^{{{3}{x}}}{\left({1}-{e}^{ -{{x}}}\right)}\)

\(={e}^{{{3}{x}}}-{e}^{{{2}{x}}}\)

We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

Therefore, the integrating factor is given by

\(A{\left({x}\right)}={\int_{{0}}^{{x}}}-{3}{t}{\left.{d}{t}\right.}=-{3}\)

So that, the solution is given by

\(y={0}{e}^{{{3}{x}}}+{e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{{-{3}{t}}}{e}^{{{2}{t}}}{\left.{d}{t}\right.}\)

\(={e}^{{{3}{x}}}{\int_{{0}}^{{x}}}{e}^{ -{{t}}}{\left.{d}{t}\right.}\)

\(={e}^{{{3}{x}}}{{\left[\frac{{e}^{ -{{t}}}}{ -{{1}}}\right]}_{{0}}^{{x}}}\)

\(={e}^{{{3}{x}}}{\left({1}-{e}^{ -{{x}}}\right)}\)

\(={e}^{{{3}{x}}}-{e}^{{{2}{x}}}\)

We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.