Find the Laplace transform for

midtlinjeg
2021-01-05
Answered

Find the Laplace transform for

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Benedict

Answered 2021-01-06
Author has **108** answers

Step 1

Calculate the Laplace transform:

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}f\left(t\right){e}^{-st}dt={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}(2u(t-3){\mathrm{cos}}^{2}2t-6{e}^{2t+7}\delta (t+3)){e}^{-st}dt$

$=2{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}u(t-3){\mathrm{cos}}^{2}2t\cdot {e}^{-st}dt-6{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{2t+7}\delta (t+3){e}^{-st}dt$

$=2{\int}_{3}^{\mathrm{\infty}}{e}^{-st}{\mathrm{cos}}^{2}2tdt-6{e}^{3s+1}$

$={\int}_{3}^{\mathrm{\infty}}{e}^{-st}(\mathrm{cos}4t+1)dt-6{e}^{3s+1}$

$={\int}_{3}^{\mathrm{\infty}}{e}^{-st}\mathrm{cos}4tdt+{\int}_{3}^{\mathrm{\infty}}{e}^{-st}dt-6{e}^{3s+1}$

$={\int}_{3}^{\mathrm{\infty}}{e}^{-st}\mathrm{cos}4tdt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$

$={\int}_{3}^{\mathrm{\infty}}{e}^{-st}\frac{{e}^{4jt}+{e}^{-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$

$={\int}_{3}^{\mathrm{\infty}}\frac{{e}^{-st+4jt}+{e}^{-st-4jt}}{2}dt+\frac{1-{e}^{-3s}}{s}-6{e}^{3s+1}$

Calculate the Laplace transform:

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