The Laplace transform of $tu(t-2)={e}^{-2s}\left[\frac{s+2{s}^{2}}{{s}^{3}}\right]$

True or False?

True or False?

coexpennan
2021-02-03
Answered

The Laplace transform of $tu(t-2)={e}^{-2s}\left[\frac{s+2{s}^{2}}{{s}^{3}}\right]$

True or False?

True or False?

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Faiza Fuller

Answered 2021-02-04
Author has **108** answers

Step 1

The given function is$tu(t-2)$

To find the Laplace transform proceed as follows

$L\{tu(t-2)\}$ . The function $u(t-2)$ is unit step function .

Find the Laplace of$u(t-2)$

$L\{u(t-2)\}=\frac{{e}^{-2s}}{s}$

The function$u(t-2)$ is multiplied by t, use the formula for the multiplication by t^n
use the formula $L\{f(t)\}=L\{f(t)\}=(\overline{f})(s)\text{then}L\{{t}^{n}f(t)\}=(-1{)}^{n}\frac{{d}^{n}}{d{s}^{n}}(\overline{f})(s)$

here n =1

therefore,$L\{tu(t-2)\}=(-1)\frac{d}{ds}\frac{{e}^{-2s}}{s}$

Step 2

Solve$L\{tu(t-2)\}=(-1)\frac{d}{ds}\frac{{e}^{-2s}}{s}$

$L\{tu(t-2)\}=(-1)\frac{\frac{sd}{ds}{e}^{-2s}-{e}^{-2s}\frac{d}{ds}s}{{s}^{2}}$

$=(-1)\left(\frac{s(-2){e}^{-2s}-{e}^{-2s}(1)}{{s}^{2}}\right)$

$=\left(\frac{2s{e}^{-2s}+{e}^{-2s}}{{s}^{2}}\right)$

$={e}^{-2s}\left(\frac{2s+1}{{s}^{2}}\right)\cdot \frac{s}{s}$

$={e}^{-2s}\left(\frac{2{s}^{2}+s}{{s}^{3}}\right)$

The Laplace transform of the given function is${e}^{-2s}\left(\frac{2{s}^{2}+s}{{s}^{3}}\right)$

The answer is TRUE

The given function is

To find the Laplace transform proceed as follows

Find the Laplace of

The function

here n =1

therefore,

Step 2

Solve

The Laplace transform of the given function is

The answer is TRUE

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I'm try to solve this differential equation:

${y}^{\prime}=x-1+xy-y$

After rearranging it I can see that is a linear differential equation:

${y}^{\prime}+(1-x)y=x-1$

So the integrating factor is $l(x)={e}^{\int (1-x)dx}={e}^{(1-x)x}$

That leaves me with an integral that I can't solve... I tried to solve it in Wolfram but the result is nothing I ever done before in the classes so I'm wondering if I made some mistake...

This is the integral:

$y{e}^{(1-x)x}=\int (x-1){e}^{(1-x)x}dx$

${y}^{\prime}=x-1+xy-y$

After rearranging it I can see that is a linear differential equation:

${y}^{\prime}+(1-x)y=x-1$

So the integrating factor is $l(x)={e}^{\int (1-x)dx}={e}^{(1-x)x}$

That leaves me with an integral that I can't solve... I tried to solve it in Wolfram but the result is nothing I ever done before in the classes so I'm wondering if I made some mistake...

This is the integral:

$y{e}^{(1-x)x}=\int (x-1){e}^{(1-x)x}dx$