Parker Pitts

2022-10-10

find the inverse laplace transformation of
$\frac{13{s}^{2}+3s+6}{\left(s-2\right)\left({s}^{2}+9\right)}.$
Here is my work:
Using partial factions:
$\begin{array}{rl}Y\left(s\right)& =\frac{13{s}^{2}+3s+6}{\left(s-2\right)\left({s}^{2}+9\right)}\\ & =\frac{64}{13\left(-2+s\right)}+\frac{3\left(83+35s\right)}{13\left(9+{s}^{2}\right)}\end{array}$
And:
$\frac{3\left(83+35s\right)}{13\left(9+{s}^{2}\right)}=\frac{83}{9+{s}^{2}}+\frac{35s}{9+{s}^{2}}$

Jayleen Copeland

The partial fraction expansion yields:
$\frac{3\left(35s+83\right)}{13\left({s}^{2}+9\right)}+\frac{64}{13\left(s-2\right)}=\frac{3\left(35s\right)}{13\left({s}^{2}+9\right)}+\frac{3\left(83\right)}{13\left({s}^{2}+9\right)}+\frac{64}{13\left(s-2\right)}$
Now, we put that result into the desired forms:
$\frac{3\left(35s\right)}{13\left({s}^{2}+{3}^{2}\right)}+\frac{3\left(83\right)}{13\left({s}^{2}+{3}^{2}\right)}+\frac{64}{13\left(s-2\right)}$
From this, we can see the forms we need.
This yields:
$y\left(t\right)=\frac{1}{13}\left(105\mathrm{cos}3t+83\mathrm{sin}3t+64{e}^{2t}\right)$

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