# text{Find the Laplace transform } F(s)=Lleft{f(t)right} text{of the function } f(t)=6+sin(3t) text{defined on the interval } tgeq0

Question
Laplace transform
$$\text{Find the Laplace transform }\ F(s)=L\left\{f(t)\right\}\ \text{of the function }\ f(t)=6+\sin(3t) \ \text{defined on the interval }\ t\geq0$$

2021-02-17
$$\text{Step 1}$$
$$\text{From the given statement, the function is }\ f(t)=6+\sin(3t)$$
$$\text{Step 2}$$
$$\text{To find the Laplace transform of the function as follows.}$$
$$L(f(t))=L(6+\sin(3t))$$
$$=L(6)+L(\sin(3t))$$
$$\text{Known fact: }$$
$$L(1)=\frac{1}{s}$$
$$L(\sin(\omega t))=\frac{\omega}{s^2+\omega^2}$$
$$\text{Therefore, }$$
$$L(6)+L(\sin(3t))=6L(1)+L(\sin(3t))$$
$$=6\bigg(\frac{1}{s}\bigg)+\frac{3}{s^2+3^2}$$
$$=\frac{6}{s}+\frac{3}{s^2+9}$$
$$=\frac{6s^2+3s+54}{s^3+9s}$$
$$\text{Thus, the Laplace transform of the function is } \frac{6s^2+3s+54}{s^3+9s}$$

### Relevant Questions

Find the laplace transform of the following:
Change of Scale
$$\text{If } L\left\{f(t)\right\}=\frac{s^2-s+1}{(2s+1)^2(s-2)} \text{ , find } L\left\{f(2t)\right\}$$
$$f(t)=3e^{2t}$$
Determine L[f]
Let f be a function defined on an interval $$[0,\infty)$$
The Laplace transform of f is the function F(s) defined by
$$F(s) =\int_0^\infty e^{-st}f(t)dt$$
provided that the improper integral converges. We will usually denote the Laplace transform of f by L[f].
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.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$
$$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$
$$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$
$$\text{ Verify the following Laplace transforms, where u is a real number. }$$
$$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$
Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
$$L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\}$$
$$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}$$
Use properties of the Laplace transform to answer the following
(a) If $$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)$$
Find the inverse Laplace transform of $$F(s)=\frac{(s+4)}{(s^2+9)}$$
a)$$\cos(t)+\frac{4}{3}\sin(t)$$
c) $$\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)$$
$$\begin{cases}t & 0,4\leq t<\infty \\0 & 4\leq t<\infty \end{cases}$$
$$L\left\{f(t)\right\} - ?$$