How do I solve the Laplace transform of te^t without using a table?

In summary: So if you set e^[t(1-s)] = 0 and integrate by parts, you'll get something likete^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2
  • #1
Pi Face
76
0

Homework Statement



f(t)=te^t, find laplace

Homework Equations





The Attempt at a Solution



I started doing integration by parts and after doing it three times I wasn't sure if I was going in the right direction/making any progress. I'm not supposed to use a table to solve this (I have to do the integral out) so could anyone give me a hint as to how to start?

I can manipulate the original equation to
(0 to inf for all integrals)
∫ te^(t(1-s)) dt and then i set u=t, du=dt, dv=e^(t(1-s)), v=e^(t(1-s))/(1-s)
and then I end up with a longer expression and I need to integrate by parts again. is this is right direction?
 
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  • #2
It's the right direction. After all, your t-factor in the integrand goes away, so you just have to integrate e^(t*something).
 
  • #3
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

EDIT: nevermind, I had a brainfart, I got the answer
 
Last edited:
  • #4
Pi Face said:
I think I'm missing a step.

after doing interation by parts the first time, i get

te^[t(1-s)] / (1-s) - ∫ e^[t(1-s)] / (1-s) dt

=te^[t(1-s)] / (1-s) - e^[t(1-s)] / (1-s)^2

i then need to evaluate fro 1 to inf but what does that do to te^[t(1-s)], assuming s>1? t is inf and e^[t(1-s)] becomes 0?

Laplace transform generally has a region of convergence to the right of some axis parallel to the y-axis on the complex s plane, in this case this axis happens to be x=1.
 

Related to How do I solve the Laplace transform of te^t without using a table?

What is the Laplace transform of te^t?

The Laplace transform of te^t is given by the formula:
L{te^t} = 1/(s-1)^2, where s is the Laplace variable.

What is the purpose of using Laplace transform on te^t?

The Laplace transform is a mathematical tool used to solve differential equations and analyze linear systems. It transforms a function of time into a function of a complex variable, making it easier to solve certain problems.

What is the inverse Laplace transform of 1/(s-1)^2?

The inverse Laplace transform of 1/(s-1)^2 is te^t. This can be found using the formula for the inverse Laplace transform, or by using tables of Laplace transform pairs.

What is the region of convergence for the Laplace transform of te^t?

The Laplace transform of te^t has a region of convergence of Re(s) > -1, which means that it will converge for all values of s with a real part greater than -1.

How is the Laplace transform of te^t used in real-life applications?

The Laplace transform of te^t has many applications in engineering, physics, and other scientific fields. It can be used to analyze and solve problems related to electrical circuits, control systems, and mechanical systems. It is also used in signal processing and image processing to understand and manipulate data.

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