Based on the linearity property and the table of common Laplace transforms, calculate the Laplace transform of the function: f(t)=e^{-2t}(3\cos(6t)-5\sin(6t))
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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