What is the Inverse Laplace Transform of 2s+7-e^-2s/(s+1)^2?

Click For Summary
SUMMARY

The inverse Laplace transform of the expression $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$ can be computed by separating the terms into $\displaystyle \frac{2s+7}{(s+1)^2}$ and $\displaystyle -\frac{e^{-2s}}{(s+1)^2}$. The first term can be solved using standard inverse Laplace techniques, while the second term utilizes the shifting theorem for Laplace transforms. The relevant formula for the second piece is $\displaystyle \mathfrak{L}^{-1}\left[\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}\right]$, which applies to the exponential decay component.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with the shifting theorem in Laplace transforms
  • Knowledge of basic calculus and differential equations
  • Ability to manipulate algebraic expressions and fractions
NEXT STEPS
  • Study the properties of the Laplace transform, focusing on linearity and shifting
  • Learn about the inverse Laplace transform techniques for rational functions
  • Explore the application of the shifting theorem in solving differential equations
  • Practice solving inverse Laplace transforms with exponential terms
USEFUL FOR

Students and professionals in engineering, mathematics, and physics who are working with differential equations and require a solid understanding of Laplace transforms for problem-solving.

alexmahone
Messages
303
Reaction score
0
Find the inverse Laplace transform of $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$.
 
Last edited:
Physics news on Phys.org
Alexmahone said:
Find the inverse Laplace transform of $\displaystyle\frac{2s+5-e^{-2s}}{s^2+s+1}$.

$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.
 
dwsmith said:
$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.

I changed the question. (Sorry about that.)
 
Alexmahone said:
I changed the question. (Sorry about that.)

Then look at

$$
\frac{2s+7}{(s+1)^2} - \frac{e^{-2s}}{(s+1)^2}
$$

The formula for the second piece is

$$
\frac{(t-\tau)^n}{n!}e^{-\alpha(t-\tau)}u(t-\tau) = \mathfrak{L}^{-1}\left[\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}\right]
$$

The other one shouldn't be too bad. Just ask if you need help with that one.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K