Question

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

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

2021-03-07
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:
$$\displaystyle{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}\cdot f{{\left({t}\right)}}{\left.{d}{t}\right.}$$
a) $$f(t)=2$$
$$\displaystyle\Rightarrow{L}{\left\lbrace{2}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}\cdot{2}{\left.{d}{t}\right.}={2}{\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{{\left[\frac{{-{e}^{{-{s}{t}}}}}{{s}}\right]}_{{0}}^{\infty}}={2}{\left[\frac{{-{e}^{{-{s}\cdot\infty}}+{e}^{{{0}\cdot{s}}}}}{{s}}\right]}={2}{\left[\frac{{{0}+{1}}}{{s}}\right]}={2}\cdot\frac{1}{{s}}$$
$$\displaystyle\Rightarrow{L}{\left\lbrace{2}\right\rbrace}=\frac{2}{{s}}$$
b) $$\displaystyle{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}\cdot{e}^{{{2}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{\infty}}{e}^{{{\left({2}-{s}\right)}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={{\left[\frac{{{e}^{{{\left({2}-{s}\right)}{t}}}}}{{{2}-{s}}}\right]}_{{0}}^{\infty}}={\left[\frac{{{e}^{{{\left({2}-{s}\right)}\cdot\infty}}-{e}^{{{\left({2}-{s}\right)}\cdot{0}}}}}{{{2}-{s}}}\right]}={\left[\frac{{{0}-{1}}}{{{2}-{s}}}\right]}$$
$$\displaystyle=\frac{1}{{{2}-{s}}}=\frac{1}{{{s}-{2}}}\Rightarrow{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}=\frac{1}{{{s}-{2}}}$$
c) $$\displaystyle{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}\cdot{e}^{{-{3}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{\infty}}{e}^{{-{\left({s}+{3}\right)}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={{\left[\frac{{-{e}^{{-{\left({s}+{3}\right)}{t}}}}}{{{s}+{3}}}\right]}_{{0}}^{\infty}}={\left[\frac{{{e}^{{-{\left({s}+{3}\right)}\cdot\infty}}-{e}^{{-{\left({s}+{3}\right)}\cdot{0}}}}}{{{s}+{3}}}\right]}={\left[\frac{{{0}+{1}}}{{{s}+{3}}}\right]}$$
$$\displaystyle{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}=\frac{1}{{{s}+{3}}}$$