# Find the Laplace transform Lleft{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}

Question
Laplace transform
Find the Laplace transform $$L\left\{u_3(t)(t^2-5t+6)\right\}$$
$$a) F(s)=e^{-3s}\left(\frac{2}{s^4}-\frac{5}{s^3}+\frac{6}{s^2}\right)$$
$$b) F(s)=e^{-3s}\left(\frac{2}{s^3}-\frac{5}{s^2}+\frac{6}{s}\right)$$
$$c) F(s)=e^{-3s}\frac{2+s}{s^4}$$
$$d) F(s)=e^{-3s}\frac{2+s}{s^3}$$
$$e) F(s)=e^{-3s}\frac{2-11s+30s^2}{s^3}$$

2020-11-03
Step 1
According to the given information, it is required to determine the laplace transform of the given function.
$$L\left\{u_3(t)(t^2-5t+6)\right\}$$
Step 2
Use the below formula to find the laplace transform.
$$L(u_c(t)g(t))=e^{-cs}L(g(t+c))$$
$$L(u_c(t))=\frac{e^{-cs}}{s}$$
Step 3
Solving further to get:
$$L(u_3(t)(t^2-5t+6))=L(u_3(t)t^2)-5L(u_3(t)t)+6L(u_3(t))$$
$$=e^{-3s}L((t+3)^2)-5e^{-3s}L(t+3)+\frac{6e^{-3s}}{s}$$
$$=e^{-3s}\left[\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}\right]-5e^{-3s}\left[\frac{1}{s^2}+\frac{3}{s}\right]+\frac{6e^{-3s}}{s}$$
$$=e^{-3s}\left[\frac{2}{s^3}+\frac{6}{s^2}+\frac{9}{s}-\frac{5}{s^2}-\frac{15}{s}+\frac{6}{s}\right]$$
$$=e^{-3s}\left[\frac{2}{s^3}+\frac{1}{s^2}\right]$$
$$L(u_3(t)(t^2-5t+6))=e^{-3s}\left[\frac{2+s}{s^3}\right]$$
Step 4
Therefore, the correct option is (D).

### Relevant Questions

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 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) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Let $$y(t)=\int_0^tf(t)dt$$ If the Laplace transform of y(t) is given $$Y(s)=\frac{19}{(s^2+25)}$$ , find f(t)
a) $$f(t)=19 \sin(5t)$$
b) none
c) $$f(t)=6 \sin(2t)$$
d) $$f(t)=20 \cos(6t)$$
e) $$f(t)=19 \cos(5t)$$
Using Laplace Transform , solve the following differential equation
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
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}}}}$$
The Laplace transform of the function $${\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)}$$
Given that $$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)}$$
If the Laplace Transforms of fimetions $$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}$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\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}-?$$
b) Obtain the solution y(t).
y(t) - ?
Find the inverse Laplace transform of $$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)$$