# Laplace transform

Discussion in 'Math' started by boks, Dec 14, 2008.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
1. The problem statement, all variables and given/known data

Find the laplace transform of t sin(t) and t cos(t), and the inverse transform of $\frac{1}{(1+s^2)^2}$

2. The attempt at a solution

I found the two laplace forms:

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

and

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

I guess I'm supposed to use the two laplace transforms to find the inverse of this one, but I don't know how to do that.

2. ### vvkannan Active Member

Aug 9, 2008
138
11
hi,

1/(1+s²)² can be written as

1/2[(s²+1-(s²-1))/(s²+1)²]

this can be split as
1/2[(s²+1)/(s²+1)²] - 1/2[(s²-1)/(s²+1)²]

1/2{[1/(s²+1)] - [(s²-1)/(s²+1)²]}

inverse of 1/(s²+1) is sin t and the inverese of next term is (t cost).
Just to bring the denominator in appropriate form we have rearranged as in 1st step.

3. ### boks Thread Starter Active Member

Oct 10, 2008
218
0

Did you obtain that using separation of variables?

4. ### vvkannan Active Member

Aug 9, 2008
138
11
just splitting (a - b)/c as a/c - b/c

where
a=s²+1 , b = s² - 1 and c= (s² + 1)².
is this the step you were referring to?

5. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Then I'm done with maths for this year. Thanks for helping me with all my problems!