 # The Laplace transform of the function {L}{leftlbrace{6}{e}^{{-{3}{t}}}-{t}^{2}+{2}{t}-{8}rightrbrace} Wierzycaz 2020-10-21 Answered
The Laplace transform of the function $L\left\{6{e}^{-3t}-{t}^{2}+2t-8\right\}$
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Calculation:
Consider,
$L\left\{6{e}^{-3t}-{t}^{2}+2t-8\right\}$
The objective is to determine the Laplace Transform
Consider and c be a constant
Then, for
$L\left\{{f}_{1}+{f}_{2}\right\}=L\left\{{f}_{1}\right\}+L\left\{{f}_{2}\right\}$
Also,
$L\left\{cf\right\}=cL\left\{f\right\}$
Using above properties:
$L\left\{6{e}^{-3t}-{t}^{2}+2t-8\right\}=L\left\{\left\{6{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}+L\left\{2t\right\}-8L\left\{1\right\}$
$=6L\left\{{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}-L\left\{{t}^{2}\right\}+2L\left\{t\right\}-8L\left\{1\right\}$
Since the Laplace Transform from the Laplace Transform table:
$L\left\{{e}^{0\cdot t}=\frac{1}{s},\mathrm{\forall }s>0$
$L\left\{{e}^{a}\cdot t\right\}=\frac{1}{s-a},\mathrm{\forall }s>a$
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},\mathrm{\forall }s>0,n=1,2,3,\dots$
Using above formula:
$6L\left\{{e}^{-3t}\right\}-L\left\{{t}^{2}\right\}+2L\left\{t\right\}-8L\left\{1\right\}=\frac{6}{s-\left(-3\right)}-\frac{2!}{{s}^{2+1}}+2×\frac{1!}{{s}^{1+1}}-8×\frac{1}{s},$
$\mathrm{\forall }s>0$
$=\frac{6}{s+3}-\frac{2}{{s}^{3}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall }s>0$
Laplace transforms into
$F\left(s\right)=\frac{6}{s+3}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall }s>0$
Conclusion:
The required Laplace transform for this function is
$F\left(s\right)=\frac{6}{s+3}-\frac{2}{{s}^{2}}+\frac{2}{{s}^{2}}-\frac{8}{s},\mathrm{\forall }s>0$