Tessa Peters
2022-10-28
Answered

Find the inverse laplace of this function: $F(s)=\frac{s+8}{{s}^{2}+4s+5}$

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Kristin Myers

Answered 2022-10-29
Author has **12** answers

Hint. You may observe that

$$\mathcal{L}({e}^{-at}\mathrm{cos}\omega t)=\frac{s+a}{(s+a{)}^{2}+{\omega}^{2}},$$

$$\mathcal{L}({e}^{-at}\mathrm{sin}\omega t)=\frac{\omega}{(s+a{)}^{2}+{\omega}^{2}}.$$

Then apply it to

$$F(s)=\frac{s+{2}}{(s+{2}{)}^{2}+{1}^{2}}+6\phantom{\rule{mediummathspace}{0ex}}\times \frac{{1}}{(s+2{)}^{2}+{{1}}^{2}}.$$

$$\mathcal{L}({e}^{-at}\mathrm{cos}\omega t)=\frac{s+a}{(s+a{)}^{2}+{\omega}^{2}},$$

$$\mathcal{L}({e}^{-at}\mathrm{sin}\omega t)=\frac{\omega}{(s+a{)}^{2}+{\omega}^{2}}.$$

Then apply it to

$$F(s)=\frac{s+{2}}{(s+{2}{)}^{2}+{1}^{2}}+6\phantom{\rule{mediummathspace}{0ex}}\times \frac{{1}}{(s+2{)}^{2}+{{1}}^{2}}.$$

Lara Cortez

Answered 2022-10-30
Author has **3** answers

Divide it as s+2 and 6.

$$\mathcal{L}({e}^{-at}\mathrm{cos}\omega t)=\frac{s+a}{(s+a{)}^{2}+{\omega}^{2}}.$$

$$\mathcal{L}({e}^{-at}\mathrm{cos}\omega t)=\frac{s+a}{(s+a{)}^{2}+{\omega}^{2}}.$$

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