 # Calculate the Laplace transform L\{f\} of the function f(t)=e^{4t+2}-8t^6+8 \sin (2t)+9 using the basic formulas OlmekinjP 2021-09-15 Answered

Calculate the Laplace transform $L\left\{f\right\}$ of the function $f\left(t\right)={e}^{4t+2}-8{t}^{6}+8\mathrm{sin}\left(2t\right)+9$ using the basic formulas and the linearity of the Laplace transform

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Step 1
$f\left(t\right)={e}^{4t+2}-8{t}^{6}+8\mathrm{sin}\left(2t\right)+9$
$F\left(t\right)={e}^{2}{e}^{4t}-8{t}^{6}+8\mathrm{sin}\left(2t\right)+9$
$L\left\{f\left(t\right)\right\}=L\left\{{e}^{2}{e}^{4t}-8{t}^{6}+8\mathrm{sin}\left(2t\right)+9\right\}$
Apply linearity
$L\left\{f\left(t\right)\right\}={e}^{2}L\left\{{e}^{4t}\right\}-8L\left\{{t}^{6}\right\}-8L\left\{\mathrm{sin}\left(2t\right)\right\}+9L\left\{1\right\}$
using the basic formulas
$L\left\{{e}^{at}\right\}=\frac{1}{s-a},L\left\{1\right\}=\frac{1}{s}$
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}}$
and $L\left\{\mathrm{sin}\left(at\right)\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
So, $L\left\{f\left(t\right)\right\}={e}^{2}\frac{1}{s-4}-8\frac{6!}{{s}^{7}}-8\frac{2}{{s}^{2}+{2}^{2}}+9\frac{1}{s}$
$L\left\{f\right\}=\frac{{e}^{2}}{s-4}-\frac{5760}{{s}^{7}}-\frac{16}{{s}^{2}+{2}^{2}}+\frac{9}{s}$