# Find the Laplace transform for displaystyle f{{left({t}right)}}={2}{u}{left({t}-{3}right)}{{cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}delta{left({t}+{3}right)}

Question
Laplace transform
Find the Laplace transform for
$$\displaystyle f{{\left({t}\right)}}={2}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}$$

2021-01-06
Step 1
Calculate the Laplace transform:
$$\displaystyle{\int_{ -{\infty}}^{\infty}} f{{\left({t}\right)}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{ -{\infty}}^{\infty}}{\left({2}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{\int_{ -{\infty}}^{\infty}}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}-{6}{\int_{ -{\infty}}^{\infty}}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{{\cos}^{2}{2}}{t}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{\left( \cos{{4}}{t}+{1}\right)}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}} \cos{{4}}{t}{\left.{d}{t}\right.}+{\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}} \cos{{4}}{t}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}\frac{{{e}^{{{4}{j}{t}}}+{e}^{{-{4}{j}{t}}}}}{{2}}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle={\int_{{3}}^{\infty}}\frac{{{e}^{{-{s}{t}+{4}{j}{t}}}+{e}^{{-{s}{t}-{4}{j}{t}}}}}{{2}}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle=\frac{{{1}-{e}^{{-{3}{s}+{12}{j}}}}}{{{2}{s}-{8}{j}}}+\frac{{{1}-{e}^{{-{3}{s}-{12}{j}}}}}{{{2}{s}+{8}{j}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle=\frac{{{2}{s}+{8}{j}-{\left({2}{s}+{8}{j}\right)}{e}^{{-{3}{s}+{12}{j}}}+{\left({2}{s}-{8}{j}\right)}{e}^{{-{3}{s}-{12}{j}}}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
$$\displaystyle=\frac{{{4}{s}-{2}{e}^{{-{3}{s}}}{\left({2}{s} \cos{{12}}-{8} \sin{{12}}\right)}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$
Step 2
Thus, Laplace transform is $$\displaystyle\frac{{{4}{s}-{2}{e}^{{-{3}{s}}}{\left({2}{s} \cos{{12}}-{8} \sin{{12}}\right)}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}$$

### Relevant Questions

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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
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 $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
Use Laplace transform to find the solution of the IVP
$$2y'+y=0 , y(0)=-3$$
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}$$
Find the Laplace transform of $$\displaystyle f{{\left({t}\right)}}={t}{e}^{{-{t}}} \sin{{\left({2}{t}\right)}}$$
Then you obtain $$\displaystyle{F}{\left({s}\right)}=\frac{{{4}{s}+{a}}}{{\left({\left({s}+{1}\right)}^{2}+{4}\right)}^{2}}$$
Please type in a = ?
Use the Laplace transform to solve the given initial-value problem.
$$\displaystyle{y}{''}+{8}{y}'+{41}{y}=\delta{\left({t}-\pi\right)}+\delta{\left({t}-{3}\pi\right)},\ \ \ \ {y}{\left({0}\right)}={1},\ \ \ \ {y}'{\left({0}\right)}={0}\ \ {y}{\left({t}\right)}=?$$
Find the laplace transform by definition.
a) $$\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]}$$
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
What is the Laplace transform of $$\displaystyle{x}{\left({t}\right)}={e}^{{-{t}}} \cos{{\left(\pi\right)}}{t}$$?
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}$$