# Learn Laplace Transform Equations with Plainmath

Recent questions in Laplace transform
Kayden Mills 2022-11-17

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

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 $v=\alpha u+w$

InjegoIrrenia1mk 2022-11-16

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

vidamuhae 2022-11-16

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

django0a6 2022-11-16

## Try to find the two sided Laplace transform of${\int }_{-\mathrm{\infty }}^{t}{e}^{-\left(t-\tau \right)-{\tau }^{2}}d\tau ={\int }_{-\mathrm{\infty }}^{t}{e}^{-\left(t-\tau \right)}{e}^{-{\tau }^{2}}d\tau$which seems to be some kind of convolution integral. I figured that I could apply the theorem$\mathcal{L}\left\{f\ast g\phantom{\rule{thinmathspace}{0ex}}\left(t\right)\right\}=\mathcal{L}\left\{f\left(t\right)\right\}\phantom{\rule{thinmathspace}{0ex}}\cdot \mathcal{L}\left\{g\left(t\right)\right\}$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${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\left(t-\tau \right)}{e}^{-{\tau }^{2}}\left(1-H\left(\tau -t\right)\right)\phantom{\rule{thinmathspace}{0ex}}d\tau .$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 $H\left(t-\tau \right)$ instead of $H\left(\tau -t\right)$

drogaid1d8 2022-11-16

## Let $f\left(t\right)=\alpha {e}^{-\beta t}$, where $\alpha ,\beta$ are constantsLet $g\left(t\right)=y\left(t\right)$Then the resulting convolution $f\ast g$ is:$f\ast g={\int }_{0}^{t}\alpha {e}^{-\beta \left(t-\tau \right)}y\left(\tau \right)d\tau$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 $f,g$?

Aryanna Fisher 2022-11-16

## Express your answer in terms of the error function:${L}^{-1}\left[\frac{1}{\sqrt{{s}^{3}+a{s}^{2}}}\right]$

Nola Aguilar 2022-11-16

## Can someone give me a clue on how to compute this Laplace transform?$\mathcal{L}\left[\frac{\mathrm{cosh}\left(at\right)}{at}\right]$

Clara Dennis 2022-11-16

## I have a very easy inverse Laplace transform.I can not figure out why${L}^{-1}\left(\frac{1}{{s}^{2}}\frac{s-a}{s+a}\right)=-t+\frac{2}{a}-\frac{2}{a}{e}^{-at}$Is there a conversion I don't see?

Humberto Campbell 2022-11-16

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

Leanna Jennings 2022-11-16

## Inverse Laplace Transform of $1/{s}^{a+1}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{x}^{a}$ by contour integrationMy try:We get Laplace transform of$g\left(t\right)={t}^{a}$is:$\stackrel{^}{g}\left(t\right)=1/{s}^{a+1}{\int }_{0}^{\mathrm{\infty }}{e}^{-t}{x}^{a}$then I was stuck at question 2) which asks me to evaluate the inverse laplace transform of $\stackrel{^}{g}\left(p\right)$ which is$1/2\pi i{\int }_{0}^{\mathrm{\infty }}{e}^{pt}\stackrel{^}{g}\left(p\right)dp$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.

Kayley Dickson 2022-11-15

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

Hanna Webster 2022-11-15

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

linnibell17591 2022-11-15

## If there is a way to calculate the Laplace transform for a given f(t):$\frac{f\left(t\right)}{1-{e}^{-at}}$

Siena Erickson 2022-11-14

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

klasyvea 2022-11-13

## Try to do this Laplace Transform${L}_{t}\left(u\left(t-2\right)\left(2{t}^{2}-6t+5\right)\right)$What I tried was:${L}_{t}\left(u\left(t-2\right)\left(2{t}^{2}-6t+5\right)\right)$$={e}^{-2s}{L}_{t}\left(2{t}^{3}-10{t}^{2}+17t-10\right)$$={e}^{-2s}\left(\frac{12}{{s}^{4}}-\frac{20}{{s}^{3}}+\frac{17}{{s}^{2}}-\frac{10}{s}\right)$But this is wrong. Can anyone tell what I have done wrong and help me figure it out.?

Howard Nelson 2022-11-13