Lleft{t-e^{-3t}right}] which of the laplace transform is 1.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}+frac{1}{s-3} 2.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}-frac{1}{s-3} 3.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}+frac{1}{s+3} 4.) Lleft{t-e^{-3t}right}=frac{1}{s^{2}}-frac{1}{s+3}

Question
Laplace transform
$$L\left\{t-e^{-3t}\right\}\] which of the laplace transform is \(1.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s-3}$$
$$2.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s-3}$$
$$3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}$$
$$4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}$$

2020-12-26
Step 1
We know the Laplace transforms $$L\left\{t^{n}\right\}=\frac{n!}{s^{n+1}},s>0$$
$$L\left\{e^{at}\right\}=\frac{1}{(s-a)},s>0$$
$$\text{Now, }L\left\{t-e^{-3t}\right\}=L\left\{t\right\}-L\left\{e^{-3t}\right\}$$
$$L\left\{t\right\}=\frac{1}{s^{2}},s>0$$
$$L\left\{t-e^{-3t}\right\}=-\frac{1}{s+3},s>-3$$
$$L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{(s+3)}$$
Step 2
Answer: $$L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}$$ option 4) is correct

Relevant Questions

If the Laplace Transforms of fimetions $$y_1(t)=\int_0^\infty e^{-st}t^3dt , y_2(t)=\int_0^\infty e^{-st} \sin 2tdt , y_3(t)=\int_0^\infty e^{-st}e^t t^2dt$$ exist , then Which of the following is the value of $$L\left\{y_1(t)+y_2(t)+y_3(t)\right\}$$
$$a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
$$B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}$$
$$c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}$$
$$d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}$$
$$e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
Part II
29.[Poles] (a) For each of the pole diagrams below:
(i) Describe common features of all functions f(t) whose Laplace transforms have the given pole diagram.
(ii) Write down two examples of such f(t) and F(s).
The diagrams are: $$(1) {1,i,-i}. (2) {-1+4i,-1-4i}. (3) {-1}. (4)$$ The empty diagram.
(b) A mechanical system is discovered during an archaeological dig in Ethiopia. Rather than break it open, the investigators subjected it to a unit impulse. It was found that the motion of the system in response to the unit impulse is given by $$w(t) = u(t)e^{-\frac{t}{2}} \sin(\frac{3t}{2})$$
(i) What is the characteristic polynomial of the system? What is the transfer function W(s)?
(ii) Sketch the pole diagram of the system.
(ii) The team wants to transport this artifact to a museum. They know that vibrations from the truck that moves it result in vibrations of the system. They hope to avoid circular frequencies to which the system response has the greatest amplitude. What frequency should they avoid?
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
$$L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\}$$
Find the inverse Laplace transform of $$F(s)=\frac{(s+4)}{(s^2+9)}$$
a)$$\cos(t)+\frac{4}{3}\sin(t)$$
b)non of the above
c) $$\cos(3t)+\sin(3t)$$
d) $$\cos(3t)+\frac{4}{3} \sin(3t)$$
e)$$\cos(3t)+\frac{2}{3} \sin(3t)$$
f) $$\cos(t)+4\sin(t)$$
Use properties of the Laplace transform to answer the following
(a) If $$f(t)=(t+5)^2+t^2e^{5t}$$, find the Laplace transform,$$L[f(t)] = F(s)$$.
(b) If $$f(t) = 2e^{-t}\cos(3t+\frac{\pi}{4})$$, find the Laplace transform, $$L[f(t)] = F(s)$$. HINT:
$$\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha) \sin(\beta)$$
(c) If $$F(s) = \frac{7s^2-37s+64}{s(s^2-8s+16)}$$ find the inverse Laplace transform, $$L^{-1}|F(s)| = f(t)$$
(d) If $$F(s) = e^{-7s}(\frac{1}{s}+\frac{s}{s^2+1})$$ , find the inverse Laplace transform, $$L^{-1}[F(s)] = f(t)$$
The Laplace transform of the function $${\left({2}{t}-{3}\right)}{e}^{{\frac{{{t}+{2}}}{{3}}}}$$ is equal to:
a) $${e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}-\frac{3}{{{s}-\frac{1}{{3}}}}\right)}$$
b) $${e}^{{\frac{2}{{3}}}}{\left(\frac{2}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\cdot\frac{3}{{{s}-\frac{1}{{3}}}}\right)}$$
c) $${e}^{{\frac{2}{{3}}}}{\left(\frac{6}{{\left({s}-\frac{1}{{3}}\right)}^{2}}\right)}$$
d) $${\left(\frac{2}{{{\left({s}-\frac{1}{{3}}\right)}^{2}+\frac{2}{{3}}}}\right)}$$
$$\text{If } L\left\{f(t)\right\}=\frac{s^2-s+1}{(2s+1)^2(s-2)} \text{ , find } L\left\{f(2t)\right\}$$
$$\text{Find the Laplace transform }\ F(s)=L\left\{f(t)\right\}\ \text{of the function }\ f(t)=6+\sin(3t) \ \text{defined on the interval }\ t\geq0$$
Find the Laplace transform of $$\displaystyle f{{\left({t}\right)}}={t}{e}^{{-{t}}} \sin{{\left({2}{t}\right)}}$$
Then you obtain $$\displaystyle{F}{\left({s}\right)}=\frac{{{4}{s}+{a}}}{{\left({\left({s}+{1}\right)}^{2}+{4}\right)}^{2}}$$