# 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]}

Question
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) $$\displaystyle{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}\cdot{e}^{{-{3}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{\infty}}{e}^{{-{\left({s}+{3}\right)}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={{\left[\frac{{-{e}^{{-{\left({s}+{3}\right)}{t}}}}}{{{s}+{3}}}\right]}_{{0}}^{\infty}}={\left[\frac{{{e}^{{-{\left({s}+{3}\right)}\cdot\infty}}-{e}^{{-{\left({s}+{3}\right)}\cdot{0}}}}}{{{s}+{3}}}\right]}={\left[\frac{{{0}+{1}}}{{{s}+{3}}}\right]}$$
$$\displaystyle{L}{\left\lbrace{e}^{{-{3}{t}}}\right\rbrace}=\frac{1}{{{s}+{3}}}$$

### Relevant Questions

Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
Use the definition of Laplace Transforms to show that:
\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},\ldots\)
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
Given that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}$$
a) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ using the theorem for differentiation
c) Repeat a) and b) for the case that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
d) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Find the Laplace transforms of the following
$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}$$
$$2y'+y=0 , y(0)=-3$$
a) $$f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}$$
b)$$f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}$$
c)$$f{{\left({t}\right)}}={6}{e}^{{{2}{t}}} d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$