Differential equations - inverse laplace transforms

In summary, the conversation discusses using LaPlace Transformations to find g(t) and h(t). The suggested method involves using the summation formula for convergent geometric series and the time shift property of LaPlace transforms. This approach is expected to be more successful in finding the solutions.
  • #1
sugarplum31
7
0

Homework Statement



We are given the following LaPlace Transformations and we are to find g(t) and h(t).

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

L(h)(t) = (e^-s)/s(1-e^-s)

Homework Equations



By looking at the Table of LaPlace Transformations, the only one I found that could possibly work for both of these is

f(t+T) = f(t) ---> (integral from 0 to T of e^(-st)f(t) dt)/ 1-e^(-sT)

The Attempt at a Solution



For the first one, I broke it into two separate equations, 1/(1-e^-2s) * 1/(s-1) and for the second part it would be e^t. However, I am very confused on the equation I found in the Table and I'm not sure how to use it on either of these.
 
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  • #2
sugarplum31 said:

Homework Equations



By looking at the Table of LaPlace Transformations, the only one I found that could possibly work for both of these is

f(t+T) = f(t) ---> (integral from 0 to T of e^(-st)f(t) dt)/ 1-e^(-sT)

That formula makes it easy to compute the Laplace transform of a periodic function, but I don't see how it's going to make it easy to do the inverse transform. I think the following is a better approach.

* Use the summation formula for convergent geometric series:

[tex]\sum_{n=0}^\infty x^n = \frac{1}{1-x}[/tex], [tex]|x|<1[/tex].

Here, [itex]x=e^-Ts[/itex]. Since [itex]T>0[/itex], you can apply the above sum formula.

Once you have done that, then...

* Make repeated use the time shift property of Laplace transforms:

[tex]e^{-Ts}F(s) \longrightarrow f(t-T)U(t-T)[/tex]

I think you'll have more success taking that route.
 

Related to Differential equations - inverse laplace transforms

What is the purpose of using inverse Laplace transforms in differential equations?

The inverse Laplace transform is used to convert a function from the Laplace domain back to the time domain. In differential equations, this is useful for finding the solution to a differential equation in terms of time.

How do you perform an inverse Laplace transform?

To perform an inverse Laplace transform, you need to use a table of Laplace transforms and their corresponding inverse transforms. You can also use algebraic manipulation, partial fractions, and other techniques to simplify the function before applying the inverse transform.

What is the difference between a one-sided and two-sided inverse Laplace transform?

A one-sided inverse Laplace transform is used when the function being transformed only exists for positive values of time. A two-sided inverse Laplace transform is used when the function exists for both positive and negative values of time.

Can an inverse Laplace transform be used for any function?

No, not every function has a Laplace transform or an inverse Laplace transform. The function must meet certain criteria, such as being piecewise continuous and having a finite number of discontinuities.

What are some real-world applications of inverse Laplace transforms in differential equations?

Inverse Laplace transforms are used in a variety of fields, including physics, engineering, and economics. They can be used to model and analyze systems such as electrical circuits, mechanical systems, and population growth.

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