Given IVP is \(\displaystyle{y}{''}+{4}{y}'+{3}{y}={e}^{{-{t}}}.\) To find the auxiliary equation, let \(\displaystyle{y}={e}^{{{m}}}\times{x}.\) The auxiliary equation of is

\(\displaystyle{\left({m}^{{{2}}}\right)}+{4}{m}+{3}={0}\)

\(\displaystyle\Rightarrow{\left({m}+{3}\right)}{\left({m}+{1}\right)}={0}\)

\(\displaystyle\Rightarrow{m}=-{3},-{1}\)

So, the complimentary function is given by

\(\displaystyle{y}{c}{f}={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{e}^{{-{x}}}.\)

The particular integral is given by

PSKypi=1/((D^2)+4d+3)e^-t =e^-t((1/(D-1)^2)+4(D-1)+3)*1 =e^-t(1/(D^2)-2D+1+4D-4+3)*1 =e^-t((1/2D)(D/2)+1)*^-1*1 =e^-t((1/2D)(1-(D/2)+...)*^-1*1 =e^-t(1/2D)*1 =t(e^-t)/2ZSK

Therefore the solition is

\(\displaystyle{y}={y}{c}{f}+{y}\pi={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{\left({e}^{{-{x}}}\right)}+\frac{{{t}{e}^{{-{t}}}}}{{2}}\)

\(\displaystyle{\left({m}^{{{2}}}\right)}+{4}{m}+{3}={0}\)

\(\displaystyle\Rightarrow{\left({m}+{3}\right)}{\left({m}+{1}\right)}={0}\)

\(\displaystyle\Rightarrow{m}=-{3},-{1}\)

So, the complimentary function is given by

\(\displaystyle{y}{c}{f}={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{e}^{{-{x}}}.\)

The particular integral is given by

PSKypi=1/((D^2)+4d+3)e^-t =e^-t((1/(D-1)^2)+4(D-1)+3)*1 =e^-t(1/(D^2)-2D+1+4D-4+3)*1 =e^-t((1/2D)(D/2)+1)*^-1*1 =e^-t((1/2D)(1-(D/2)+...)*^-1*1 =e^-t(1/2D)*1 =t(e^-t)/2ZSK

Therefore the solition is

\(\displaystyle{y}={y}{c}{f}+{y}\pi={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{\left({e}^{{-{x}}}\right)}+\frac{{{t}{e}^{{-{t}}}}}{{2}}\)