Differential Equations L^(-1) { (3s-4) / (s(s-4)) }

In summary, for the first question, decompose the expression using partial fractions to simplify the inverse Laplace transform. For the second question, use the property of Laplace transforms of derivatives to solve the IVP problem. A good reference for using Laplace transforms is the provided link.
  • #1
winner2
Need some help here:

Find each Laplace transform or Inverse as indicated:

1. L^(-1) { (3s-4) / (s(s-4)) }

2. Solve the following IVP problem using the method of Laplace transforms:

y'' - 3y' + 2y = 0 y(0)=0 y'(0)=1
 
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  • #2
What have you done?, on what are you getting stuck?
 
  • #3
I haven't been able to figure out how to solve either type. Essentially, I'm stuck at step 1. The examples in our books are not very good and the teachers here don't really do a good job at explaining things.
 
  • #4
I see, Well if i do the problem i doubt it will help you in anyway, because you are not stuck on trivial algebra or anything like that, but on the procedure of how to do it.

Here go to this webpage

http://www.sosmath.com/diffeq/diffeq.html

If you still have any questions, ask them.
 
  • #5
The thing about Laplace transforms is that you basically need to just know their identities and properties. Thats the only way to do them really. Especially their inverse. Go to your book and find the table that shows the Laplace Transform identities. The inverse Laplace just changes the function of s back to the original function of t. For example The Laplace of 1 = 1/s therefore the inverse Laplace of 1/s = 1. Its quite simple.

As for your second problem you need to find the property of Laplace Transforms of derivatives. And apply that to the IVP you have.
 
  • #6
winner2 said:
Need some help here:

Find each Laplace transform or Inverse as indicated:

1. L^(-1) { (3s-4) / (s(s-4)) }

2. Solve the following IVP problem using the method of Laplace transforms:

y'' - 3y' + 2y = 0 y(0)=0 y'(0)=1

The first thing to think of when "inverting" a transform is "partial fractions". For the first question, decompose the expression using partial fractions. This produces elementary transforms which are easily inverted.

Here's a link about using Laplace transforms we worked on earlier:

Click here
 

Related to Differential Equations L^(-1) { (3s-4) / (s(s-4)) }

What is a differential equation?

A differential equation is a type of mathematical equation that involves an unknown function and its derivatives. It describes the relationship between the function and its derivatives, and is often used to model real-world phenomena such as motion, growth, and decay.

What does the notation "L^(-1)" mean in "Differential Equations L^(-1) { (3s-4) / (s(s-4)) }"?

The notation "L^(-1)" represents the inverse Laplace transform. It is a mathematical operation that takes a function in the Laplace domain and transforms it back to the time domain.

What is the purpose of the Laplace transform in differential equations?

The Laplace transform is used to convert a differential equation from the time domain to the Laplace domain, where it can be solved more easily using algebraic methods. The inverse Laplace transform is then applied to obtain the solution in the time domain.

How do I solve "Differential Equations L^(-1) { (3s-4) / (s(s-4)) }"?

To solve this differential equation, you would first need to apply the inverse Laplace transform to the given function. This would give you the solution in the time domain. From there, you could use various techniques such as separation of variables or integrating factors to obtain a more explicit solution.

What are some practical applications of differential equations?

Differential equations are used in a wide range of fields, including physics, engineering, economics, and biology. They are used to model and understand various natural phenomena, from the motion of objects to the spread of diseases. They are also used in the design and analysis of systems such as circuits, chemical reactions, and control systems.

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