sjeikdom0

2021-02-08

Find the inverse Laplace transform of $F\left(s\right)=\frac{\left(s+4\right)}{\left({s}^{2}+9\right)}$
a)$\mathrm{cos}\left(t\right)+\frac{4}{3}\mathrm{sin}\left(t\right)$
b)non of the above
c) $\mathrm{cos}\left(3t\right)+\mathrm{sin}\left(3t\right)$
d) $\mathrm{cos}\left(3t\right)+\frac{4}{3}\mathrm{sin}\left(3t\right)$
e)$\mathrm{cos}\left(3t\right)+\frac{2}{3}\mathrm{sin}\left(3t\right)$
f) $\mathrm{cos}\left(t\right)+4\mathrm{sin}\left(t\right)$

d2saint0

Step 1
The given function is $F\left(s\right)=\frac{\left(s+4\right)}{\left({s}^{2}+9\right)}$
The following formulae are used in finding the inverse laplace transform of the given function.
${L}^{-1}\left\{f\left(s\right)+g\left(s\right)\right\}={L}^{-1}\left\{f\left(t\right)\right\}+{L}^{-1}\left\{g\left(t\right)\right\}$
${L}^{-1}\left\{a\cdot f\left(s\right)\right\}=a\cdot {L}^{-1}\left\{f\left(t\right)\right\}$
${L}^{-1}\left\{\frac{s}{\left({s}^{2}+{a}^{2}\right)}\right\}=\mathrm{cos}at$
${L}^{-1}\left\{\frac{1}{\left({s}^{2}+{a}^{2}\right)}\right\}=\mathrm{sin}at$
Step 2
Evaluate the inverse laplace transform of the given function as follows.
${L}^{-1}\left\{s+4s2+9\right\}={L}^{-1}\left\{\frac{s}{\left({s}^{2}+9\right)}\right\}+{L}^{-1}\left\{\frac{4}{\left({s}^{2}+9\right)}\right\}$
$={L}^{-1}\left\{\frac{s}{\left({s}^{2}+{3}^{2}\right)}\right\}+4{L}^{-1}\left\{\frac{1}{\left({s}^{2}+{3}^{2}\right)}\right\}$
$=\mathrm{cos}3t+4\left(\frac{1}{3}\mathrm{sin}3t\right)$
$=\mathrm{cos}3t+\frac{4}{3}\mathrm{sin}3t$
Therefore, the inverse Laplace transform of the given function is $\mathrm{cos}3t+\frac{4}{3}\mathrm{sin}3t$
Thus, the correct option is (d).

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