Question

$inverse\:laplace\:\frac{s}{s^2+4s+5}$

solve it

Answer 1

$\mathrm{Laplace\:Inverse\:Transform\:of\:}\frac{s}{s^2+4s+5}:\quad e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$

$L^{-1}\left\{\frac{s}{s^2+4s+5}\right\}$

$\mathrm{Expand\:}\frac{s}{s^2+4s+5}:\quad \frac{s+2}{\left(s+2\right)^2+1}-2\cdot \frac{1}{\left(s+2\right)^2+1}$

$=L^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+1}-2\cdot \frac{1}{\left(s+2\right)^2+1}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}$

$\mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}$

$=L^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+1}\right\}-2L^{-1}\left\{\frac{1}{\left(s+2\right)^2+1}\right\}$

$L^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+1}\right\}:\quad e^{-2t}\cos \left(t\right)$

$L^{-1}\left\{\frac{1}{\left(s+2\right)^2+1}\right\}:\quad e^{-2t}\sin \left(t\right)$

$=e^{-2t}\cos \left(t\right)-2e^{-2t}\sin \left(t\right)$

Answer 2

e

−2t

cos(t)−2e

−2t

sin(t)

L^{-1}\{\frac{s}{s^2+4s+5}\}L

−1

{

s

2

+4s+5

s

}

\mathrm{Expand\:}\frac{s}{s^2+4s+5}:\quad \frac{s+2}{(s+2)^2+1}-2\cdot \frac{1}{(s+2)^2+1}Expand

s

2

+4s+5

s

:

(s+2)

2

+1

s+2

−2⋅

(s+2)

2

+1

1

=L^{-1}\{\frac{s+2}{(s+2)^2+1}-2\cdot \frac{1}{(s+2)^2+1}\}=L

−1

{

(s+2)

2

+1

s+2

−2⋅

(s+2)

2

+1

1

}

\mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}UsethelinearitypropertyofInverseLaplaceTransform:

\mathrm{For\:functions\:}f(s),\:g(s)\mathrm{\:and\:constants\:}a,\:b:\quad L^{-1}\{a\cdot f(s)+b\cdot g(s)\}=a\cdot L^{-1}\{f(s)\}+b\cdot L^{-1}\{g(s)\}Forfunctionsf(s),g(s)andconstantsa,b:L

−1

{a⋅f(s)+b⋅g(s)}=a⋅L

−1

{f(s)}+b⋅L

−1

{g(s)}

=L^{-1}\{\frac{s+2}{(s+2)^2+1}\}-2L^{-1}\{\frac{1}{(s+2)^2+1}\}=L

−1

{

(s+2)

2

+1

s+2

}−2L

−1

{

(s+2)

2

+1

1

}

L^{-1}\{\frac{s+2}{(s+2)^2+1}\}:\quad e^{-2t}\cos (t)L

−1

{

(s+2)

2

+1

s+2

}:e

−2t

cos(t)

L^{-1}\{\frac{1}{(s+2)^2+1}\}:\quad e^{-2t}\sin (t)L

−1

{

(s+2)

2

+1

1

}:e

−2t

sin(t)

=e^{-2t}\cos (t)-2e^{-2t}\sin (t)=e

−2t

cos(t)−2e

−2t

sin(t)