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# determine the inverse Laplace transform of the given function. F(s)=\frac{2s+1}{s^2+16}

Laplace transform
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asked 2021-09-18
determine the inverse Laplace transform of the given function.
$$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}+{1}}}{{{s}^{{2}}+{16}}}}$$

## Expert Answers (1)

2021-09-19

Step 1 Find the inverse Laplace transform of $$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}+{1}}}{{{s}^{{2}}+{16}}}}$$
Step 2
$$L^{-1}\left\{\frac{2s+1}{s^2+16}\right\}=L^{-1}\left\{\frac{2s}{s^2+16}\right\}+L^{-1}\left\{\frac{1}{s^2+16}\right\}$$
$$=2L^{-1}\left\{\frac{s}{s^2+16}\right\}+L^{-1}\left\{\frac{1}{s^2+16}\right\}$$
$$=2(\cos 4t)+\frac{1}{4}\sin 4t \left[ \because L^{-1}\left\{\frac{s}{s^2+a^2}\right\}=\cos at \text{ and } L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at \right]$$
$$\displaystyle={2}{\cos{{4}}}{t}+{\frac{{{1}}}{{{4}}}}{\sin{{4}}}{t}$$

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