Use the Laplace transformation table and the linearity of the Laplace transform to determine the following transforms.

\(L\left\{ {{{\bf{t}}^{\bf{3}}}{\bf{ - t}}{{\bf{e}}^{\bf{t}}}{\bf{ + }}{{\bf{e}}^{{\bf{4t}}}}{\bf{cost}}} \right\}\)

The given Laplace transform is \(L\left\{ {{{\bf{t}}^{\bf{3}}}{\bf{ - t}}{{\bf{e}}^{\bf{t}}}{\bf{ + }}{{\bf{e}}^{{\bf{4t}}}}{\bf{cost}}} \right\}\).

The objective is to find the Laplace transformation.

Some important formulas of Laplace transform:

\(\begin{array}{l}L\left\{ {{t^n}} \right\} = \frac{{n!}}{{{s^{n + 1}}}},\;\;\;\;\;n = 1,2, \ldots \\L\left\{ {{t^n}{e^{at}}} \right\} = \frac{{n!}}{{{{\left( {s - a} \right)}^{n + 1}}}},\;\;\;\;\;s > a\\L\left\{ {{e^{at}}\cos bt} \right\} = \frac{{s - a}}{{{{\left( {s - a} \right)}^2} + {b^2}}},\;\;\;\;\;s > a\end{array}\)

Recall the linearity of Laplace transform:

\(L\left( {{f_1} + {f_2}} \right) = L\left( {{f_1}} \right) + L\left( {{f_2}} \right)\)and \(L\left( {{\rm{c}}f} \right) = {\rm{c}}L\left( f \right)\)

Now, simplify the given function as follows:

\(\begin{array}{c}L\left\{ {{t^3} - t{e^t} + {e^{4t}}\cos t} \right\} = L\left\{ {{t^3}} \right\} - L\left\{ {t{e^t}} \right\} + L\left\{ {{e^{4t}}\cos t} \right\}\\ = \frac{{3!}}{{{s^{3 + 1}}}} - \frac{{1!}}{{{{\left( {s - 1} \right)}^{1 + 1}}}} + \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + {1^2}}},\;\;\;\;\;s > 4\\ = \frac{6}{{{s^4}}} - \frac{1}{{{{\left( {s - 1} \right)}^2}}} + \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + 1}},\;\;\;\;\;s > 4\end{array}\)

Hence, the Laplace transform is \(\frac{6}{{{s^4}}} - \frac{1}{{{{\left( {s - 1} \right)}^2}}} + \frac{{s - 4}}{{{{\left( {s - 4} \right)}^2} + 1}},\;\;\;\;\;s > 4\).