# Determine L^{-1}left[frac{(s-4)e^{-3s}}{s^2-4s+5}right]

Determine ${L}^{-1}\left[\frac{\left(s-4\right){e}^{-3s}}{{s}^{2}-4s+5}\right]$
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yunitsiL
Concept used:
The inverse laplace transform rule.
If ${L}^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$ then
1. ${L}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right)$, Here, H(t) is Heaviside step function.
2.${L}^{-1}\left\{F\left(s-a\right)\right\}=eatf\left(t\right)$
3.${L}^{-1}\left\{sF\left(s\right)\right\}=f\prime \left(t\right)+f\left(0\right)$
The linearity property of the inverse laplace transform is
${L}^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot {L}^{-1}\left\{f\left(s\right)\right\}+b\cdot {L}^{-1}\left\{g\left(s\right)\right\}$
From the Inverse Laplace transform table.
${L}^{-1}\left\{\frac{a}{{s}^{2}+{a}^{2}}\right\}=\mathrm{sin}\left(at\right)$
Calculation
Apply the inverse transform rule "If ${L}^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$
then ${L}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right)$".
For the function $\frac{\left(s-4\right){e}^{-3s}}{{s}^{2}-4s+5},$
$F\left(s\right)=\frac{s-4}{{s}^{2}-4s+5}, a=3$
${L}^{-1}\left\{\frac{{e}^{-3s}s-4}{{s}^{2}-4s+5}\right\}=H\left(t-3\right){L}^{-1}\left\{F\left(s-3\right)\right\}\dots \left(1\right)$

${L}^{-1}\left\{\frac{s-4}{{s}^{2}-4s+5}\right\}={L}^{-1}\left\{\frac{s-2}{\left(s-2{\right)}^{2}+1}-2\cdot \frac{1}{\left(s-2{\right)}^{2}+1}\right\}$
Step 3
Now apply the linearity property of inverse Laplace transform.
${L}^{-1}\left\{\frac{s-4}{{s}^{2}-4s+5}\right\}={L}^{-1}\left\{\frac{s-2}{\left(s-2{\right)}^{2}+1}-2\cdot \frac{1}{\left(s-2{\right)}^{2}+1}\right\}$
${L}^{-1}\left\{\frac{s-2}{\left(s-2{\right)}^{2}+1}\right\}-2{L}^{-1}\left\{\frac{1}{\left(s-2{\right)}^{2}+1}\right\}$

${L}^{-1}\left\{\frac{s-2}{\left(s-2{\right)}^{2}+1}\right\}-2{L}^{-1}\left\{\frac{1}{\left(s-2{\right)}^{2}+1}\right\}={e}^{2t}{L}^{-1}\left\{\frac{s}{{s}^{2}+1}\right\}-2{e}^{2t}{L}^{-1}\left\{\frac{1}{{s}^{2}+1}\right\}\dots \left(2\right)$
Step 4
Now solve ${L}^{-1}\left\{\frac{1}{{s}^{2}+1}\right\}$ by the use of