# Determine L^-1[F(s)] for the given F. F(s)=(s-2)/(s^2+2s+3)

Determine ${L}^{-1}\left[F\left(s\right)\right]$ for the given F.
$F\left(s\right)=\frac{s-2}{{s}^{2}+2s+3}$

You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Malena

Step 1
Given information:
$F\left(s\right)=\frac{s-2}{{s}^{2}+2s+3}$
Linearity property of inverse Laplace Transform, for function f(s) and g(s) and constants a and b.
${L}^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot {L}^{-1}\left\{f\left(s\right)\right\}+b\cdot {L}^{-1}\left\{g\left(s\right)\right\}$
Step 2
First can be written as,
${L}^{-1}\left\{\frac{s-2}{{s}^{2}+2s+3}\right\}={L}^{-1}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+2}-3\cdot \frac{1}{{\left(s+1\right)}^{2}+2}\right\}$
Apply the linearity property and inverse Laplace transform,
${L}^{-1}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+2}-3\cdot \frac{1}{{\left(s+1\right)}^{2}+2}\right\}={L}^{-1}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+2}\right\}-3{L}^{-1}\left\{\frac{1}{{\left(s+1\right)}^{2}+2}\right\}$
$={e}^{-t}\mathrm{cos}\left(\sqrt{2}t\right)-3{e}^{-t}\frac{1}{\sqrt{2}}\mathrm{sin}\left(\sqrt{2}t\right)$
$={e}^{-t}\mathrm{cos}\left(\sqrt{2}t\right)-\frac{3}{\sqrt{2}}{e}^{-t}\mathrm{sin}\left(\sqrt{2}t\right)$