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

[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

 

[Pyrrhus]

What have you done?, on what are you getting stuck?

 

[winner2]

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.

 

[Pyrrhus]

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.

 

[supersix2]

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.

 

[saltydog]

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