 # Learn Laplace Transform Equations with Plainmath

Recent questions in Laplace transform Emmanuel Giles 2022-11-20

## Question is find the Laplace transform of this equation:$x\frac{\mathrm{\partial }\left(w\right)}{\mathrm{\partial }\left(x\right)}+\frac{\mathrm{\partial }\left(w\right)}{\mathrm{\partial }\left(t\right)}=xt$Boundary conditions :$w\left(x,0\right)=0\phantom{\rule{2em}{0ex}}x\ge 0$$w\left(0,t\right)=0\phantom{\rule{2em}{0ex}}t\ge 0$I've got as far as$\frac{d\left(W\right)}{d\left(x\right)}+\frac{s}{x}W=\frac{1}{{s}^{2}}$ Nola Aguilar 2022-11-20

## Find the inverse Laplace transform for the following:a. $F\left(s\right)=\frac{2s-3}{{s}^{2}-3s+2}$I used partial fractions and I ended up getting$y\left(t\right)={e}^{t}+{e}^{2t}.$b. $F\left(s\right)=\frac{{e}^{-2s}}{s-9}$I ended up getting ${e}^{9t-9}h\left(t-2\right)$ and 0 being $\ne 2$ Abdiel Mays 2022-11-19

## Find the Laplace transform for$f\left(t\right)={e}^{-4t}{t}^{\frac{3}{2}}$ Frances Pham 2022-11-19

## Evaluate the following integral:$f\left(t\right)=\frac{1}{\pi }{\int }_{0}^{\pi }\mathrm{cos}\left(t\mathrm{sin}\theta \right)\phantom{\rule{thinmathspace}{0ex}}d\theta$I started with the 'standard' form of the Laplace Transform:$\mathcal{L}\left(\mathrm{cos}\omega t\right)=\frac{s}{{s}^{2}+{\omega }^{2}}$Now I can substitute to obtain:$\mathcal{L}\left[f\left(t\right)\right]=\frac{1}{\pi }{\int }_{0}^{\pi }\frac{s}{{s}^{2}+{\mathrm{sin}}^{2}\theta }d\theta$But how do I get back? Taking the 's' out:$\mathcal{L}\left[f\left(t\right)\right]=\frac{s}{\pi }{\int }_{0}^{\pi }\frac{d\theta }{{s}^{2}+{\mathrm{sin}}^{2}\theta }$But how to proceed now? vidamuhae 2022-11-19

## Laplace transform problem at the moment, and I'm a little stuck.$\mathcal{L}\left\{\mathrm{sin}\left(2x\right)\mathrm{cos}\left(5x\right)\right\}$I don't recall any trig identity that would apply here. I know that$\mathrm{sin}\left(2x\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$But I'm not sure if that applies in this situation. Zackary Diaz 2022-11-19

## How do you evaluate the inverse transform below using convolution ? ${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]$I tried$\begin{array}{rl}{\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]\left(t\right)& =\underset{0}{\overset{t}{\int }}\mathrm{sin}t\cdot \mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau =\mathrm{sin}t\cdot \underset{0}{\overset{t}{\int }}\mathrm{cos}\left(at-a\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau \\ & =\mathrm{sin}t\cdot {\left[-\frac{1}{a}\mathrm{sin}\left(at-a\tau \right)\right]}_{0}^{t}=\mathrm{sin}t\cdot \left[0-\left(-\frac{1}{a}\mathrm{sin}\left(at\right)\right)\right]\\ & =\frac{1}{a}{\mathrm{sin}}^{2}\left(at\right)\end{array}$What have I done wrong? Noe Cowan 2022-11-19

## The Laplace transform of $\mathcal{L}\left(t{e}^{t}\mathrm{cos}t\right)$How do I find it? apopiw83 2022-11-18

## Find the inverse Laplace transform of the giveb function by using the convolution theorem.$F\left(x\right)=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}$If I use partial fractions I get:$\frac{s+4}{5\left({s}^{2}+4\right)}-\frac{1}{5\left(x+1\right)}$which gives me Laplace inverses:$\frac{1}{5}\left(\mathrm{cos}2t+\mathrm{sin}2t\right)-\frac{1}{5}{e}^{-t}$But the answer is:$f\left(t\right)={\int }_{0}^{t}{e}^{-\left(t-\tau \right)}\mathrm{cos}\left(2\tau \right)d\tau$How did they get that? inurbandojoa 2022-11-18

## How to find the Laplace inversion of $\frac{p}{{p}^{4}+4}$? Alice Chen 2022-11-18

## Find Inverse Laplace Transform of $\frac{{s}^{2}+2s+2}{s+1}$ Zackary Diaz 2022-11-18

## Evaluate${L}^{-1}\left(\frac{1}{{p}^{3}+1}\right)$Any clue on this? undergoe8m 2022-11-18

## Laplace transform property for convolutions of the type${\int }_{0}^{x}f\left(x-y\right)g\left(y\right)\phantom{\rule{thinmathspace}{0ex}}dy$is very well known. There exists a Laplace transform property to calculate functions of the type below${\int }_{a}^{x}f\left(x-y\right)g\left(y\right)\phantom{\rule{thinmathspace}{0ex}}dy,$when a>0? dannigurl21ck2 2022-11-18

## Inverse Laplace transform of $\frac{1}{s-\lambda }$ is ${e}^{\lambda t}$? figoveck38 2022-11-18

## Why this equality?$\mathcal{L}\left({\int }_{t}^{+\mathrm{\infty }}\phantom{\rule{thinmathspace}{0ex}}\psi \left(\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau \right)=\frac{1-\psi \left(s\right)}{s}$Can you help me with the steps? Davirnoilc 2022-11-18

## Solve the differential equation using Laplace transformationWrite your answer as an integral. The hint suggests using convolution, but I'm not sure how to write ${e}^{-{t}^{2}}$ in terms of that. I'm only confused on this part. I can do the rest of the problem. Simone Watts 2022-11-18

## Use Laplace to solve IVP: ${y}^{″}-2{y}^{\prime }3y=3{e}^{t};y\left(0\right)=-1,{y}^{\prime }\left(0\right)=2$ drogaid1d8 2022-11-18

## How can I demonstrate this? If F(t) is a periodic function with a period of T>0, then$\mathcal{L}\left\{F\left(t\right)\right\}=\frac{\underset{0}{\overset{T}{\int }}{e}^{-st}F\left(t\right)\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t}{1-{e}^{-sT}}\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t$ Ty Moore 2022-11-18

## There is a timely unchanged continuous function :$H\left(s\right)=\frac{s-1}{s+1}$At the entry of the system exists a x(t) which Laplace's transformation is:$X\left(s\right)=\frac{\left(5{s}^{2}-15s+7\right)}{\left(s-2{\right)}^{3}\left(s-1\right)}$Which is the impulse response?and which is the exit signal of the system? Owen Mathis 2022-11-17
## I am trying to work out the Laplace inverse of $\frac{1}{s}{e}^{-s{t}_{0}-sb}$, I was trying to use the second shift theorem and rewrote it as ${e}^{-s{t}_{0}}\frac{1}{s}{e}^{-sb}$.I ended up with $H\left(t-{t}_{0}\right)H\left(t-b-{t}_{0}\right)$, but have been told the solution is $H\left(t-b-{T}_{0}\right)$, what happens to the other heaviside unit function? Is there a general rule as to why this happens or have I just worked out the transform wrong? Siemensueqw 2022-11-17