# Find the laplace transform by definition. a) L{2}, b)L{e^(2t)}, c)L[3^(-3t)]

Find the laplace transform by definition.
a) $L\left\{2\right\}$
b) $L\left\{{e}^{2t}\right\}$
c) $L\left[{e}^{-3t}\right]$

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Step 1
Here we finding the Laplace transform of given function by the definition . We perform the definite integral as definition :
Step 2
Answer: By the definition Laplace transform of f(t) is given as:
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot f\left(t\right)dt$
a) $f\left(t\right)=2$
$⇒L\left\{2\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot 2dt=2{\int }_{0}^{\mathrm{\infty }}{e}^{-st}dt$
$=2{\left[\frac{-{e}^{-st}}{s}\right]}_{0}^{\mathrm{\infty }}=2\left[\frac{-{e}^{-s\cdot \mathrm{\infty }}+{e}^{0\cdot s}}{s}\right]=2\left[\frac{0+1}{s}\right]=2\cdot \frac{1}{s}$
$⇒L\left\{2\right\}=\frac{2}{s}$
b) $L\left\{{e}^{2t}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot {e}^{2t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{\left(2-s\right)t}dt$
$={\left[\frac{{e}^{\left(2-s\right)t}}{2-s}\right]}_{0}^{\mathrm{\infty }}=\left[\frac{{e}^{\left(2-s\right)\cdot \mathrm{\infty }}-{e}^{\left(2-s\right)\cdot 0}}{2-s}\right]=\left[\frac{0-1}{2-s}\right]$
$=\frac{1}{2-s}=\frac{1}{s-2}⇒L\left\{{e}^{2t}\right\}=\frac{1}{s-2}$
c) $L\left\{{e}^{-3t}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot {e}^{-3t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+3\right)t}dt$
$={\left[\frac{-{e}^{-\left(s+3\right)t}}{s+3}\right]}_{0}^{\mathrm{\infty }}=\left[\frac{{e}^{-\left(s+3\right)\cdot \mathrm{\infty }}-{e}^{-\left(s+3\right)\cdot 0}}{s+3}\right]=\left[\frac{0+1}{s+3}\right]$
$L\left\{{e}^{-3t}\right\}=\frac{1}{s+3}$