Recent questions in Laplace transform

Differential EquationsAnswered question

Kayden Mills 2022-11-17

Finding the inverse Laplace transform of $\mathrm{ln}\phantom{\rule{negativethinmathspace}{0ex}}(1+\frac{1}{{s}^{2}})$

Differential EquationsAnswered question

Annie French 2022-11-17

Determine the constant $\alpha \in \mathbb{R}$ and the vector $w\in {\mathbb{R}}^{3}$ that is perpendicular to the vector u such that $v=\alpha u+w$

Differential EquationsAnswered question

InjegoIrrenia1mk 2022-11-16

Try to calculate

${\mathcal{L}}^{-1}\left(\frac{3{s}^{3}-3{s}^{2}+3s-5}{{s}^{2}({s}^{2}+2s+5)}\right)$

${\mathcal{L}}^{-1}\left(\frac{3{s}^{3}-3{s}^{2}+3s-5}{{s}^{2}({s}^{2}+2s+5)}\right)$

Differential EquationsAnswered question

vidamuhae 2022-11-16

Find the inverse Laplace Transform of: ${\mathcal{L}}^{-1}\left\{\frac{{e}^{-2s}}{{s}^{3}}\right\}$

Differential EquationsAnswered question

django0a6 2022-11-16

Try to find the two sided Laplace transform of

${\int}_{-\mathrm{\infty}}^{t}{e}^{-(t-\tau )-{\tau}^{2}}d\tau ={\int}_{-\mathrm{\infty}}^{t}{e}^{-(t-\tau )}{e}^{-{\tau}^{2}}d\tau $

which seems to be some kind of convolution integral. I figured that I could apply the theorem

$\mathcal{L}\{f\ast g\phantom{\rule{thinmathspace}{0ex}}(t)\}=\mathcal{L}\{f(t)\}\phantom{\rule{thinmathspace}{0ex}}\cdot \mathcal{L}\{g(t)\}$

if I could just change the upper limit to ∞ (because of two sided laplace transform). The way I tried to do this was by subtracting the part of the integrand >t using a step function, which gives me

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-(t-\tau )}{e}^{-{\tau}^{2}}{\textstyle (}1-H(\tau -t){\textstyle )}\phantom{\rule{thinmathspace}{0ex}}d\tau .$

This is were I am stuck as I can't see how to correctly apply the theorem here. It seems like it would work if I had $H(t-\tau )$ instead of $H(\tau -t)$

${\int}_{-\mathrm{\infty}}^{t}{e}^{-(t-\tau )-{\tau}^{2}}d\tau ={\int}_{-\mathrm{\infty}}^{t}{e}^{-(t-\tau )}{e}^{-{\tau}^{2}}d\tau $

which seems to be some kind of convolution integral. I figured that I could apply the theorem

$\mathcal{L}\{f\ast g\phantom{\rule{thinmathspace}{0ex}}(t)\}=\mathcal{L}\{f(t)\}\phantom{\rule{thinmathspace}{0ex}}\cdot \mathcal{L}\{g(t)\}$

if I could just change the upper limit to ∞ (because of two sided laplace transform). The way I tried to do this was by subtracting the part of the integrand >t using a step function, which gives me

${\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}{e}^{-(t-\tau )}{e}^{-{\tau}^{2}}{\textstyle (}1-H(\tau -t){\textstyle )}\phantom{\rule{thinmathspace}{0ex}}d\tau .$

This is were I am stuck as I can't see how to correctly apply the theorem here. It seems like it would work if I had $H(t-\tau )$ instead of $H(\tau -t)$

Differential EquationsAnswered question

drogaid1d8 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 $f,g$?

Differential EquationsAnswered question

Aryanna Fisher 2022-11-16

Express your answer in terms of the error function:

${L}^{-1}\left[\frac{1}{\sqrt{{s}^{3}+a{s}^{2}}}\right]$

${L}^{-1}\left[\frac{1}{\sqrt{{s}^{3}+a{s}^{2}}}\right]$

Differential EquationsAnswered question

Nola Aguilar 2022-11-16

Can someone give me a clue on how to compute this Laplace transform?

$\mathcal{L}\left[\frac{\mathrm{cosh}(at)}{at}\right]$

$\mathcal{L}\left[\frac{\mathrm{cosh}(at)}{at}\right]$

Differential EquationsAnswered question

Clara Dennis 2022-11-16

To find Inverse Laplace of

$\phantom{\rule{thinmathspace}{0ex}}F(s)=\mathrm{log}{\displaystyle \frac{s+1}{(s+2)(s+3)}}$

$\phantom{\rule{thinmathspace}{0ex}}F(s)=\mathrm{log}{\displaystyle \frac{s+1}{(s+2)(s+3)}}$

Differential EquationsAnswered question

Adrian Brown 2022-11-16

I have a very easy inverse Laplace transform.

I can not figure out why

${L}^{-1}(\frac{1}{{s}^{2}}\frac{s-a}{s+a})=-t+\frac{2}{a}-\frac{2}{a}{e}^{-at}$

Is there a conversion I don't see?

I can not figure out why

${L}^{-1}(\frac{1}{{s}^{2}}\frac{s-a}{s+a})=-t+\frac{2}{a}-\frac{2}{a}{e}^{-at}$

Is there a conversion I don't see?

Differential EquationsAnswered question

Humberto Campbell 2022-11-16

How do I take inverse Laplace transform of $\frac{-2s+3}{{s}^{2}-2s+2}$?

Differential EquationsAnswered question

Leanna Jennings 2022-11-16

Inverse Laplace Transform of $1/{s}^{a+1}{\int}_{0}^{\mathrm{\infty}}{e}^{-t}{x}^{a}$ by contour integration

My try:

We get Laplace transform of

$g(t)={t}^{a}$

is:

$\hat{g}(t)=1/{s}^{a+1}{\int}_{0}^{\mathrm{\infty}}{e}^{-t}{x}^{a}$

then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $\hat{g}(p)$ which is

$1/2\pi i{\int}_{0}^{\mathrm{\infty}}{e}^{pt}\hat{g}(p)dp$

I know the answer should be ${t}^{a}$ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $1/2\pi i$ but was stuck then.

My try:

We get Laplace transform of

$g(t)={t}^{a}$

is:

$\hat{g}(t)=1/{s}^{a+1}{\int}_{0}^{\mathrm{\infty}}{e}^{-t}{x}^{a}$

then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $\hat{g}(p)$ which is

$1/2\pi i{\int}_{0}^{\mathrm{\infty}}{e}^{pt}\hat{g}(p)dp$

I know the answer should be ${t}^{a}$ as the inverse transform comes back to itself, but I cannot figure out how to make the contour integration. I tried to apply Cauchy's residue theorem to eliminate the $1/2\pi i$ but was stuck then.

Differential EquationsAnswered question

Kayley Dickson 2022-11-15

Proof $L[\frac{x(t)}{t}]={\int}_{s}^{\mathrm{\infty}}X(u)du$

Differential EquationsAnswered question

Hanna Webster 2022-11-15

How to find the Laplace Transformation of $H(t-\pi )$, but what about if the t is negative.

Differential EquationsAnswered question

linnibell17591 2022-11-15

If there is a way to calculate the Laplace transform for a given f(t):

$\frac{f(t)}{1-{e}^{-at}}$

$\frac{f(t)}{1-{e}^{-at}}$

Differential EquationsAnswered question

Siena Erickson 2022-11-14

How to find ${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{F(s)}{s+a}\right]$ where F(s) is the Laplace transform of f(t).

Differential EquationsAnswered question

klasyvea 2022-11-13

Try to do this Laplace Transform

${L}_{t}(u(t-2)(2{t}^{2}-6t+5))$

What I tried was:

${L}_{t}(u(t-2)(2{t}^{2}-6t+5))$

$={e}^{-2s}{L}_{t}(2{t}^{3}-10{t}^{2}+17t-10)$

$={e}^{-2s}(\frac{{\textstyle 12}}{{\textstyle {s}^{4}}}-\frac{{\textstyle 20}}{{\textstyle {s}^{3}}}+\frac{{\textstyle 17}}{{\textstyle {s}^{2}}}-\frac{{\textstyle 10}}{{\textstyle s}})$

But this is wrong. Can anyone tell what I have done wrong and help me figure it out.?

${L}_{t}(u(t-2)(2{t}^{2}-6t+5))$

What I tried was:

${L}_{t}(u(t-2)(2{t}^{2}-6t+5))$

$={e}^{-2s}{L}_{t}(2{t}^{3}-10{t}^{2}+17t-10)$

$={e}^{-2s}(\frac{{\textstyle 12}}{{\textstyle {s}^{4}}}-\frac{{\textstyle 20}}{{\textstyle {s}^{3}}}+\frac{{\textstyle 17}}{{\textstyle {s}^{2}}}-\frac{{\textstyle 10}}{{\textstyle s}})$

But this is wrong. Can anyone tell what I have done wrong and help me figure it out.?

Differential EquationsAnswered question

Howard Nelson 2022-11-13

If $f(t)=1$ when $0\le t<1$, and 0 when $1\le t<2$, where $f(t+2)=f(t)$, find the Laplace transform.

Differential EquationsAnswered question

szklanovqq 2022-11-12

How to find the inverse Laplace transform and solve for a?

$\frac{Y(s)}{{s}^{2}}}+{\displaystyle \frac{{Y}^{\prime}(s)}{s}}={\displaystyle \frac{-a}{{s}^{4}}$

is in the Laplace transform. How can I take the inverse i.e transform back to time domain and solve for a?

$\frac{Y(s)}{{s}^{2}}}+{\displaystyle \frac{{Y}^{\prime}(s)}{s}}={\displaystyle \frac{-a}{{s}^{4}}$

is in the Laplace transform. How can I take the inverse i.e transform back to time domain and solve for a?

Differential EquationsAnswered question

Jefferson Booth 2022-11-11

Calculate the inverse Laplace transform of $\frac{1}{1-{e}^{-s}}$

If you came across the necessity of Laplace transform, it is most likely that you are coming from a mechanical engineering or electrical background. The concept is used to solve differential equations, which is why it is vital to consider Laplace transform examples as you are looking through the questions and connect the dots as the equations are being approached. Remember to look through our list of answers as these will help you to address various Laplace transform problems and find solutions to complex Laplace transform questions as you are dealing with your Laplace transform equation homework.