Inverse Laplace: Find 2exp(t)cos(t)+3exp(t)sin(t)

[footballxpaul]

find the inverse laplace transform

F(s)=(2s+1)/(s^2 -2s+2)

I have broken it down too 2s/((s-1)^2 +1) + 1/((s-1)^2 +1) , then i got...

2cos(t) + exp(t)sin(t)

the book states that

2exp(t)cos(t) + 3exp(t)sin(t) is the answer? Where did I go wrong?

 

[cipher42]

I'm not sure where you went wrong; I can't find a simple inverse for your first term - which is the one that doesn't match the answers - so I'm guessing that's where your problem is.

However, look at #16 on this table of transforms:

http://www.swarthmore.edu/NatSci/ec...nMethods/LaplaceZTable/LaplaceZFuncTable.html

It's exactly the form for your problem, and you don't even have to break it into two fractions.

 

[footballxpaul]

thanks that transform is not on the chart my book uses.