# 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
$$L\left\{3e^{-4t}-t^{2}+6t-9\right\}$$
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$$3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}$$
$$4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}$$
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$$a) \frac{3}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
$$B) \frac{(3!)}{(s^3)}+\frac{s}{(s^2+4)}+\frac{(2!)}{(s-1)^3}$$
$$c) \frac{3!}{(s^4)}+\frac{2}{(s^2+2)}+\frac{1}{(s-1)^3}$$
$$d) \frac{3!}{(s^4)}+\frac{4}{(s^2+4)}+\frac{2}{(s^3)} \cdot \frac{1}{(s-1)}$$
$$e) \frac{3!}{(s^4)}+\frac{2}{(s^2+4)}+\frac{2}{(s-1)^3}$$
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) - ?
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$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
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