Recent questions in Laplace transform

Differential EquationsAnswered question

Emmanuel Giles 2022-11-20

Question is find the Laplace transform of this equation:

$x\frac{\mathrm{\partial}(w)}{\mathrm{\partial}(x)}+\frac{\mathrm{\partial}(w)}{\mathrm{\partial}(t)}=xt$

Boundary conditions :

$w(x,0)=0\phantom{\rule{2em}{0ex}}x\ge 0$

$w(0,t)=0\phantom{\rule{2em}{0ex}}t\ge 0$

I've got as far as

$\frac{d(W)}{d(x)}+\frac{s}{x}W=\frac{1}{{s}^{2}}$

$x\frac{\mathrm{\partial}(w)}{\mathrm{\partial}(x)}+\frac{\mathrm{\partial}(w)}{\mathrm{\partial}(t)}=xt$

Boundary conditions :

$w(x,0)=0\phantom{\rule{2em}{0ex}}x\ge 0$

$w(0,t)=0\phantom{\rule{2em}{0ex}}t\ge 0$

I've got as far as

$\frac{d(W)}{d(x)}+\frac{s}{x}W=\frac{1}{{s}^{2}}$

Differential EquationsAnswered question

Nola Aguilar 2022-11-20

Find the inverse Laplace transform for the following:

a. $F(s)={\displaystyle \frac{2s-3}{{s}^{2}-3s+2}}$

I used partial fractions and I ended up getting

$y(t)={e}^{t}+{e}^{2t}.$

b. $F(s)={\displaystyle \frac{{e}^{-2s}}{s-9}}$

I ended up getting ${e}^{9t-9}h(t-2)$ and 0 being $\ne 2$

a. $F(s)={\displaystyle \frac{2s-3}{{s}^{2}-3s+2}}$

I used partial fractions and I ended up getting

$y(t)={e}^{t}+{e}^{2t}.$

b. $F(s)={\displaystyle \frac{{e}^{-2s}}{s-9}}$

I ended up getting ${e}^{9t-9}h(t-2)$ and 0 being $\ne 2$

Differential EquationsAnswered question

Abdiel Mays 2022-11-19

Find the Laplace transform for

$f(t)={e}^{-4t}{t}^{\frac{3}{2}}$

$f(t)={e}^{-4t}{t}^{\frac{3}{2}}$

Differential EquationsAnswered question

Frances Pham 2022-11-19

Evaluate the following integral:

$f(t)=\frac{1}{\pi}{\int}_{0}^{\pi}\mathrm{cos}(t\mathrm{sin}\theta )\phantom{\rule{thinmathspace}{0ex}}d\theta $

I started with the 'standard' form of the Laplace Transform:

$\mathcal{L}(\mathrm{cos}\omega t)=\frac{s}{{s}^{2}+{\omega}^{2}}$

Now I can substitute to obtain:

$\mathcal{L}[f(t)]=\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:

$\mathcal{L}[f(t)]=\frac{s}{\pi}{\int}_{0}^{\pi}\frac{d\theta}{{s}^{2}+{\mathrm{sin}}^{2}\theta}$

But how to proceed now?

$f(t)=\frac{1}{\pi}{\int}_{0}^{\pi}\mathrm{cos}(t\mathrm{sin}\theta )\phantom{\rule{thinmathspace}{0ex}}d\theta $

I started with the 'standard' form of the Laplace Transform:

$\mathcal{L}(\mathrm{cos}\omega t)=\frac{s}{{s}^{2}+{\omega}^{2}}$

Now I can substitute to obtain:

$\mathcal{L}[f(t)]=\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:

$\mathcal{L}[f(t)]=\frac{s}{\pi}{\int}_{0}^{\pi}\frac{d\theta}{{s}^{2}+{\mathrm{sin}}^{2}\theta}$

But how to proceed now?

Differential EquationsAnswered question

vidamuhae 2022-11-19

Laplace transform problem at the moment, and I'm a little stuck.

$\mathcal{L}\{\mathrm{sin}(2x)\mathrm{cos}(5x)\}$

I don't recall any trig identity that would apply here. I know that

$\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

But I'm not sure if that applies in this situation.

$\mathcal{L}\{\mathrm{sin}(2x)\mathrm{cos}(5x)\}$

I don't recall any trig identity that would apply here. I know that

$\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$

But I'm not sure if that applies in this situation.

Differential EquationsAnswered question

Zackary Diaz 2022-11-19

How do you evaluate the inverse transform below using convolution ? ${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{({s}^{2}+{a}^{2}{)}^{2}}\right]$

I tried

$\begin{array}{rl}{\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{({s}^{2}+{a}^{2}{)}^{2}}\right](t)& =\underset{0}{\overset{t}{\int}}\mathrm{sin}t\cdot \mathrm{cos}(at-a\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau =\mathrm{sin}t\cdot \underset{0}{\overset{t}{\int}}\mathrm{cos}(at-a\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau \\ & =\mathrm{sin}t\cdot {[-\frac{1}{a}\mathrm{sin}(at-a\tau )]}_{0}^{t}=\mathrm{sin}t\cdot [0-(-\frac{1}{a}\mathrm{sin}(at))]\\ & =\frac{1}{a}{\mathrm{sin}}^{2}(at)\end{array}$

What have I done wrong?

I tried

$\begin{array}{rl}{\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left[\frac{s}{({s}^{2}+{a}^{2}{)}^{2}}\right](t)& =\underset{0}{\overset{t}{\int}}\mathrm{sin}t\cdot \mathrm{cos}(at-a\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau =\mathrm{sin}t\cdot \underset{0}{\overset{t}{\int}}\mathrm{cos}(at-a\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau \\ & =\mathrm{sin}t\cdot {[-\frac{1}{a}\mathrm{sin}(at-a\tau )]}_{0}^{t}=\mathrm{sin}t\cdot [0-(-\frac{1}{a}\mathrm{sin}(at))]\\ & =\frac{1}{a}{\mathrm{sin}}^{2}(at)\end{array}$

What have I done wrong?

Differential EquationsAnswered question

Noe Cowan 2022-11-19

The Laplace transform of $\mathcal{L}(t{e}^{t}\mathrm{cos}t)$

How do I find it?

How do I find it?

Differential EquationsAnswered question

apopiw83 2022-11-18

Find the inverse Laplace transform of the giveb function by using the convolution theorem.

$F(x)=\frac{s}{(s+1)({s}^{2}+4)}$

If I use partial fractions I get:

$\frac{s+4}{5({s}^{2}+4)}-\frac{1}{5(x+1)}$

which gives me Laplace inverses:

$\frac{1}{5}(\mathrm{cos}2t+\mathrm{sin}2t)-\frac{1}{5}{e}^{-t}$

But the answer is:

$f(t)={\int}_{0}^{t}{e}^{-(t-\tau )}\mathrm{cos}(2\tau )d\tau $

How did they get that?

$F(x)=\frac{s}{(s+1)({s}^{2}+4)}$

If I use partial fractions I get:

$\frac{s+4}{5({s}^{2}+4)}-\frac{1}{5(x+1)}$

which gives me Laplace inverses:

$\frac{1}{5}(\mathrm{cos}2t+\mathrm{sin}2t)-\frac{1}{5}{e}^{-t}$

But the answer is:

$f(t)={\int}_{0}^{t}{e}^{-(t-\tau )}\mathrm{cos}(2\tau )d\tau $

How did they get that?

Differential EquationsAnswered question

inurbandojoa 2022-11-18

How to find the Laplace inversion of $\frac{p}{{p}^{4}+4}$?

Differential EquationsAnswered question

Alice Chen 2022-11-18

Find Inverse Laplace Transform of $\frac{{s}^{2}+2s+2}{s+1}$

Differential EquationsAnswered question

Zackary Diaz 2022-11-18

Evaluate

${L}^{-1}(\frac{1}{{p}^{3}+1})$

Any clue on this?

${L}^{-1}(\frac{1}{{p}^{3}+1})$

Any clue on this?

Differential EquationsAnswered question

undergoe8m 2022-11-18

Laplace transform property for convolutions of the type

${\int}_{0}^{x}f(x-y)g(y)\phantom{\rule{thinmathspace}{0ex}}dy$

is very well known. There exists a Laplace transform property to calculate functions of the type below

${\int}_{a}^{x}f(x-y)g(y)\phantom{\rule{thinmathspace}{0ex}}dy,$

when a>0?

${\int}_{0}^{x}f(x-y)g(y)\phantom{\rule{thinmathspace}{0ex}}dy$

is very well known. There exists a Laplace transform property to calculate functions of the type below

${\int}_{a}^{x}f(x-y)g(y)\phantom{\rule{thinmathspace}{0ex}}dy,$

when a>0?

Differential EquationsAnswered question

dannigurl21ck2 2022-11-18

Inverse Laplace transform of $\frac{1}{s-\lambda}$ is ${e}^{\lambda t}$?

Differential EquationsAnswered question

figoveck38 2022-11-18

Why this equality?

$\mathcal{L}({\int}_{t}^{+\mathrm{\infty}}\phantom{\rule{thinmathspace}{0ex}}\psi (\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau )=\frac{1-\psi (s)}{s}$

Can you help me with the steps?

$\mathcal{L}({\int}_{t}^{+\mathrm{\infty}}\phantom{\rule{thinmathspace}{0ex}}\psi (\tau )\phantom{\rule{thinmathspace}{0ex}}d\tau )=\frac{1-\psi (s)}{s}$

Can you help me with the steps?

Differential EquationsAnswered question

Davirnoilc 2022-11-18

Solve the differential equation using Laplace transformation

${x}^{\u2033}+4x={e}^{-{t}^{2}},\text{}t\ge 0,\text{}\text{}x(0)=0,{x}^{\prime}(0)=0$

Write 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.

${x}^{\u2033}+4x={e}^{-{t}^{2}},\text{}t\ge 0,\text{}\text{}x(0)=0,{x}^{\prime}(0)=0$

Write 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.

Differential EquationsAnswered question

Simone Watts 2022-11-18

Use Laplace to solve IVP:

${y}^{\u2033}-2{y}^{\prime}3y=3{e}^{t};y(0)=-1,{y}^{\prime}(0)=2$

${y}^{\u2033}-2{y}^{\prime}3y=3{e}^{t};y(0)=-1,{y}^{\prime}(0)=2$

Differential EquationsAnswered question

drogaid1d8 2022-11-18

How can I demonstrate this? If F(t) is a periodic function with a period of T>0, then

$\mathcal{L}\{F(t)\}=\frac{\underset{0}{\overset{T}{\int}}{e}^{-st}F(t)\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t}{1-{e}^{-sT}}\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t$

$\mathcal{L}\{F(t)\}=\frac{\underset{0}{\overset{T}{\int}}{e}^{-st}F(t)\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t}{1-{e}^{-sT}}\mathrm{d}\phantom{\rule{negativethinmathspace}{0ex}}t$

Differential EquationsAnswered question

Ty Moore 2022-11-18

There is a timely unchanged continuous function :

$H(s)=\frac{s-1}{s+1}$

At the entry of the system exists a x(t) which Laplace's transformation is:

$X(s)=\frac{(5{s}^{2}-15s+7)}{(s-2{)}^{3}(s-1)}$

Which is the impulse response?and which is the exit signal of the system?

$H(s)=\frac{s-1}{s+1}$

At the entry of the system exists a x(t) which Laplace's transformation is:

$X(s)=\frac{(5{s}^{2}-15s+7)}{(s-2{)}^{3}(s-1)}$

Which is the impulse response?and which is the exit signal of the system?

Differential EquationsAnswered question

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 $H(t-{t}_{0})H(t-b-{t}_{0})$, but have been told the solution is $H(t-b-{T}_{0})$, 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?

I ended up with $H(t-{t}_{0})H(t-b-{t}_{0})$, but have been told the solution is $H(t-b-{T}_{0})$, 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?

Differential EquationsAnswered question

Siemensueqw 2022-11-17

$x={x}^{\prime}\mathrm{cos}\theta -{y}^{\prime}\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}y={x}^{\prime}\mathrm{sin}\theta +{y}^{\prime}\mathrm{cos}\theta $

$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{cos}\theta +\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\mathrm{sin}\theta $

are given. I don't understand how the partial derivative with respect to x′ is obtained. Can you explain?

$\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{x}^{\prime 2}}=\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{x}^{2}}{\mathrm{cos}}^{2}\theta +\frac{\mathrm{\partial}}{\mathrm{\partial}x}(\frac{\mathrm{\partial}f}{\mathrm{\partial}y})\mathrm{sin}\theta \mathrm{cos}\theta +\frac{\mathrm{\partial}}{\mathrm{\partial}y}(\frac{\mathrm{\partial}f}{\mathrm{\partial}x})\mathrm{cos}\theta \mathrm{sin}\theta +\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{y}^{2}}{\mathrm{sin}}^{2}\theta $

$\frac{\mathrm{\partial}f}{\mathrm{\partial}{x}^{\prime}}=\frac{\mathrm{\partial}f}{\mathrm{\partial}x}\mathrm{cos}\theta +\frac{\mathrm{\partial}f}{\mathrm{\partial}y}\mathrm{sin}\theta $

are given. I don't understand how the partial derivative with respect to x′ is obtained. Can you explain?

$\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{x}^{\prime 2}}=\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{x}^{2}}{\mathrm{cos}}^{2}\theta +\frac{\mathrm{\partial}}{\mathrm{\partial}x}(\frac{\mathrm{\partial}f}{\mathrm{\partial}y})\mathrm{sin}\theta \mathrm{cos}\theta +\frac{\mathrm{\partial}}{\mathrm{\partial}y}(\frac{\mathrm{\partial}f}{\mathrm{\partial}x})\mathrm{cos}\theta \mathrm{sin}\theta +\frac{{\mathrm{\partial}}^{2}f}{\mathrm{\partial}{y}^{2}}{\mathrm{sin}}^{2}\theta $

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.