How to compute the inverse laplace transform of this term? $\frac{-\frac{3}{10}s-\frac{1}{5}}{(s+1{)}^{2}+1}$

Francis Oliver
2022-10-22
Answered

How to compute the inverse laplace transform of this term? $\frac{-\frac{3}{10}s-\frac{1}{5}}{(s+1{)}^{2}+1}$

You can still ask an expert for help

indivisast7

Answered 2022-10-23
Author has **13** answers

$\mathcal{L}(y)=\frac{-\frac{3}{10}s-\frac{1}{5}}{(s+1{)}^{2}+1}$

you want to get this into the form. $\frac{a(s+1)+b}{(s+1{)}^{2}+1}$

At which point:

$y=a{e}^{-t}\mathrm{cos}t+b{e}^{-t}\mathrm{sin}t$

$\frac{-\frac{3}{10}(s+1)+\frac{1}{10}}{(s+1{)}^{2}+1}$

you want to get this into the form. $\frac{a(s+1)+b}{(s+1{)}^{2}+1}$

At which point:

$y=a{e}^{-t}\mathrm{cos}t+b{e}^{-t}\mathrm{sin}t$

$\frac{-\frac{3}{10}(s+1)+\frac{1}{10}}{(s+1{)}^{2}+1}$

caritatsjq

Answered 2022-10-24
Author has **3** answers

Hint. One may write

$$\frac{-\frac{3}{10}s-\frac{1}{5}}{(s+1{)}^{2}+1}=-\frac{3}{10}\frac{(s+1)}{(s+1{)}^{2}+1}+\frac{1}{10}\frac{1}{(s+1{)}^{2}+1}$$

then one may apply the properties of the inverse Laplace transform to each term

$$\frac{-\frac{3}{10}s-\frac{1}{5}}{(s+1{)}^{2}+1}=-\frac{3}{10}\frac{(s+1)}{(s+1{)}^{2}+1}+\frac{1}{10}\frac{1}{(s+1{)}^{2}+1}$$

then one may apply the properties of the inverse Laplace transform to each term

asked 2022-10-01

How can we take the inverse Laplace transform of

$${f}_{1}(s)X(s)+{f}_{2}(s)({e}^{i\varphi}X(s-i{\alpha}_{1})+{e}^{-i\varphi}X(s+i{\alpha}_{1}))={f}_{0}(s)$$

Where ${f}_{1}(s)$ is in the form of $\frac{{f}^{2}(s)}{{f}^{3}(s)}$, ${f}_{2}(s)={k}_{0}s$, and ${f}_{0}(s)$ is in the form of ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$

$${f}_{1}(s)X(s)+{f}_{2}(s)({e}^{i\varphi}X(s-i{\alpha}_{1})+{e}^{-i\varphi}X(s+i{\alpha}_{1}))={f}_{0}(s)$$

Where ${f}_{1}(s)$ is in the form of $\frac{{f}^{2}(s)}{{f}^{3}(s)}$, ${f}_{2}(s)={k}_{0}s$, and ${f}_{0}(s)$ is in the form of ${k}_{1}+\frac{{f}^{4}(s)}{s({s}^{2}+{\alpha}_{0}^{2}){f}^{2}(s)}$

asked 2022-09-05

I know that the Laplace transform of a function f(t) on $[0,\mathrm{\infty})$ is defined as

$${f}^{\ast}(\theta )={\int}_{0}^{\mathrm{\infty}}{e}^{-\theta t}f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t.$$

But if I only have ${f}^{\ast}(\theta )={\displaystyle \frac{3+10\theta}{24{\theta}^{2}+22\theta +3}}$, how to evaluate f(t)?

$${f}^{\ast}(\theta )={\int}_{0}^{\mathrm{\infty}}{e}^{-\theta t}f(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t.$$

But if I only have ${f}^{\ast}(\theta )={\displaystyle \frac{3+10\theta}{24{\theta}^{2}+22\theta +3}}$, how to evaluate f(t)?

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Find $\frac{dy}{dx}$ and $\frac{{d}^{2}y}{{dx}^{2}}.\text{}x={e}^{t},\text{}y=t{e}^{-t}$ For which values of t is the curve concave upward?

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Find the Laplace transform of $3{e}^{-7t}$

asked 2022-11-16

Let $f(t)=\alpha {e}^{-\beta t}$, where $\alpha ,\beta $ are constants

Let $g(t)=y(t)$

Then the resulting convolution $f\ast g$ is:

$f\ast g={\int}_{0}^{t}\alpha {e}^{-\beta (t-\tau )}y(\tau )d\tau $

Does anyone know how one would take the derivative of this expression?

In general, are there rules for taking derivative of $f\ast g$, for some given $f,g$?

Let $g(t)=y(t)$

Then the resulting convolution $f\ast g$ is:

$f\ast g={\int}_{0}^{t}\alpha {e}^{-\beta (t-\tau )}y(\tau )d\tau $

Does anyone know how one would take the derivative of this expression?

In general, are there rules for taking derivative of $f\ast g$, for some given $f,g$?

asked 2021-09-14

Find f(t).

asked 2021-05-17

Compute $\mathrm{\u25b3}y$ and dy for the given values of x and $dx=\mathrm{\u25b3}x$

$y={x}^{2}-4x,x=3,\mathrm{\u25b3}x=0,5$

$\mathrm{\u25b3}y=???$

dy=?

dy=?