## Solution:

The function can be written as follows;

f(t)=3e^{-2t}\cos(6t)-5e^{-2t}\sin(6t)

## 1. Use linearity property

As \mathscr{L}\{\cdot\} is a linear application, we have:

\begin{aligned}& \mathscr{L}\{f(t)\}=\mathscr{L}\{3e^{-2t}\cos(6t)-5e^{-2t}\sin(6t)\} \\ & \mathscr{L}\{f(t)\}=3\mathscr{L}\{e^{-2t}\cos(6t)\}-5\mathscr{L}\{e^{-2t}\sin(6t)\} \end{aligned}

## 2. Use table of laplace transforms

Line 8 from the transform table tells us:

\displaystyle \mathscr{L}\{e^{at}\cos(bt)\}=\frac{s-a}{(s-a)^{2}+b^{2}}

Comparing to the first term in \mathscr{L}\{f(t)\}, we have a=-2 and b=6 and then:

\displaystyle\mathscr{L}\{e^{-2t}\cos(6t)\}=\frac{s-(-2)}{(s-(-2))^{2}+6^{2}}

i.e.,

\displaystyle \mathscr{L}\{e^{-2t}\cos(6t)\}=\frac{s+2}{(s+2)^{2}+36}

## 3. Use table of laplace transforms for the second term

As we did above, from the line 7 of the table, a=-2 and b=6:

\displaystyle \mathscr{L}\{e^{-2t}\sin(6t)\}=\frac{6}{(s+2)^{2}+36}

Therefore,

\displaystyle \mathscr{L}\{f(t)\}=3\cdot\frac{s+2}{(s+2)^{2}+36}-5\cdot\frac{6}{(s+2)^{2}+36}

## 4. Result

which may be rewritten as follows:

\begin{aligned} & \mathscr{L}\{f(t)\}=\frac{3s+6}{(s+2)^{2}+36}-\frac{30}{(s+2)^{2}+36} \\ & \mathscr{L}\{f(t)\}=\frac{3s-24}{(s+2)^{2}+36} \end{aligned}

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