# Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform. Lleft{e^{3t}sin(4t)-t^{4}+e^{t}right} Question
Laplace transform Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform.
$$L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\}$$ 2021-02-09
$$\text{Step 1}$$
$$L\left\{e^{3t}\sin(4t)-t^{4}+e^{t}\right\} \(=L\left\{e^{3t}\sin(4t)\right\}-L\left\{t^{4}\right\}+L\left\{e^{t}\right\}\ \ \ \ \ \ \text{1)} \(\text{we know that } L\left\{t^{n}\right\}=\frac{n!}{s^{n+1}} \(L\left\{e^{at}\right\}=\frac{1}{s-a} \(L\left\{\sin(at)\right\}=\frac{a}{s^{2}+a^{2}} \(L\left\{\sin(4t)\right\}=\frac{4}{s^{2}+16}: \(\text{by first shifting } L\left\{e^{at}\sin(at)\right\}=\frac{b}{(s-a)^{2}+b^{2}} \(L\left\{e^{3t}\sin(4t)\right\}=\frac{4}{(s-3)^{2}+16} , \(L\left\{t^{4}\right\}=\frac{4!}{s^{4+1}}=\frac{24}{s^{5}} \(L\left\{e^{t}\right\}=\frac{1}{s-1} \(\text{Step 2} \(\text{Then 1) will becomes as } \(L\left\{e^{3t}\sin(t)-t^{4}+e^{t}\right\}=\frac{4}{(s-3)^{2}+16}-\frac{24}{s^{5}}+\frac{1}{s-1}$$

### Relevant Questions Use the Laplace transform table and the linearity of the Laplace transform to determine the following transform
$$L\left\{3e^{-4t}-t^{2}+6t-9\right\}$$ 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, $$L^{-1}[F(s)] = f(t)$$ use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $$f(t)=e^{3t}\cos5t-e^{-t}\sin2t$$ $$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}$$ $$\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$$ Obtain the Laplace Transform of
$$L\left\{t^3-t^2+4t\right\}$$
$$L\left\{3e^{4t}-e^{-2t}\right\}$$ Use the Laplace transform to solve the following initial value problem:
$$2y"+4y'+17y=3\cos(2t)$$
$$y(0)=y'(0)=0$$
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for $$L\left\{y(t)\right\}$$ b) Express the solution $$y(t)$$ in terms of a convolution integral If the Laplace Transforms of fimetions $$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}-?$$
b) Obtain the solution y(t).
y(t) - ? Find the inverse Laplace transform $$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}}}$$
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$