Example 8.
\(\ \ \)
\begin{equation*}
\lap{15 + 6e^{7t} - 11t}.
\end{equation*}
Solution.
Let’s start by applying the definition of the Laplace transform:
\begin{align*}
\lap{15\ +\amp\ 6e^{7t} - 11t} \\
=\amp\ \int_0^{\infty} e^{-st} \cdot \left(15 + 6e^{7t} - 11t\right)\ dt\\
=\amp\ \int_0^{\infty} \left( 15e^{-st} + 6e^{-st} \cdot e^{7t} - 11e^{-st} \cdot t \right)\ dt
\end{align*}
Here, we’ve expanded the expression inside the integral. Notice that each term is something we’ve already encountered in previous sections. We can now use the linearity property to separate the integral:
\begin{align*}
=\amp\ 15\int_0^{\infty} e^{-st}\cdot 1\ dt + 6\int_0^{\infty} e^{-st} \cdot e^{7t}\ dt - 11\int_0^{\infty} e^{-st} \cdot t\ dt\\
=\amp\ 15\lap{1} + 6\lap{ e^{7t} } - 11\lap{ t }.
\end{align*}
We already know the Laplace transforms of these individual functions:
\begin{align*}
=\amp\ \ub{\frac{15}{s}}_{s \gt 0} + 6\ub{\frac{1}{s - 7}}_{s \gt 7} - 11\ub{\frac{1}{s^2}}_{s \gt 0}.
\end{align*}
To satisfy all the conditions on \(s\text{,}\) we must have \(s \gt 7\) since that ensures all terms are defined. Therefore:
\begin{equation*}
\lap{15 + 6e^{7t} - 11t} = \frac{15}{s} + 6\cdot \frac{1}{s-7} - \frac{11}{s^2}, \quad s \gt 7.
\end{equation*}
