Deragz

2021-12-10

I have a Laplace Transform problem which Im

MoxboasteBots5h

Beginner2021-12-11Added 35 answers

The definition of the Laplace transform is $L\left(f\right)\left(s\right){\textstyle \phantom{\rule{0.222em}{0ex}}}={\int}_{0}^{\mathrm{\infty}}{e}^{-st}f\left(t\right)dtask{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{sinh}thefollow\in gholds:PSK\mathrm{sin}hx=\frac{1}{2}({e}^{x}-{e}^{-x})$ . Now you put these two things together and compute

$\frac{1}{2}{\int}_{0}^{\mathrm{\infty}}x{e}^{4x}{e}^{-xs}dx-\frac{1}{2}{\int}_{0}^{\mathrm{\infty}}x{e}^{-4x}{e}^{-xs}dx$

Hope this helps

Hope this helps

levurdondishav4

Beginner2021-12-12Added 38 answers

You got $t\mathrm{sin}h4t=t\frac{{e}^{4t}-{e}^{-4t}}{2}$ ,so:

$L\left\{t\mathrm{sin}h4t\right\}=\frac{1}{2}(L\left\{t{e}^{4t}\right\}+L\left\{t{e}^{-4t}\right\})$

We know that$L\left\{{e}^{at}f\left(t\right)\right\}=F(s-a)$ . Furthermore $L\left\{t\right\}=\frac{1}{{s}^{2}}$ .So in our case $L\left\{t{e}^{4t}\right\}=F(s-4)=\frac{1}{{(s-4)}^{2}}$ . Finally:

$L\left\{t\mathrm{sin}h4t\right\}=\frac{0.5}{{(s-4)}^{2}}+\frac{0.5}{{(s+4)}^{2}}$

We know that

- A body falls from rest against resistance proportional to the square root of the speed at any instant. If the body attains speed V1 and V2 feet per second, after 1 and 2 seconds in motion, respectively, find an expression for the limiting velocity.

The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True or False

The Laplace transform of

is$u(t-2)$

(a)$\frac{1}{s}+2$

(b)$\frac{1}{s}-2$

(c) ??$e}^{2}\frac{s}{s}\left(d\right)\frac{{e}^{-2s}}{s$ 1 degree on celsius scale is equal to

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scaleThe Laplace transform of $t{e}^{t}$ is A. $\frac{s}{(s+1{)}^{2}}$ B. $\frac{1}{(s-1{)}^{2}}$ C. $\frac{s}{(s+1{)}^{2}}$ D. $\frac{s}{(s-1)}$

What is the Laplace transform of

into the s domain?$t\mathrm{cos}t$ Find the general solution of the given differential equation:

${y}^{\u2033}-2{y}^{\prime}+y=0$The rate at which a body cools is proportional to the difference in

temperature between the body and its surroundings. If a body in air

at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more

minutes will it take the body to cool from 100℃ 𝑡𝑜 50℃ ?A body falls from rest against a resistance proportional to the velocity at any instant. If the limiting velocity is 60fps and the body attains half that velocity in 1 second, find the initial velocity.

What's the correct way to go about computing the Inverse Laplace transform of this?

$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

$\frac{-2s+1}{(s+1{)}^{2}+4}$How to find inverse Laplace transform of the following function?

$X(s)=\frac{s}{{s}^{4}+1}$

I tried to use the definition: $f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi i}\underset{T\to \mathrm{\infty}}{lim}{\int}_{\gamma -iT}^{\gamma +iT}{e}^{st}F(s)\phantom{\rule{thinmathspace}{0ex}}ds$or the partial fraction expansion but I have not achieved results.How do i find the lapalace transorm of this intergral using the convolution theorem? ${\int}_{0}^{t}{e}^{-x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx$

How can I solve this differential equation? : $xydx-({x}^{2}+1)dy=0$

Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$