laplace transform

[drsmoothe2004]

1. The problem statement, all variables and given/known data

s2-5 / s3+4s2+3s


2. Relevant equations

find the inverse laplace transform


3. The attempt at a solution
for the denominator, it can be factored out to s(s+3)(s+1) or one could complete the square and thus the denominator would be s(s+2)2-1. neither of this help in finding the laplace transform
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[rock.freak667]

Factor it into s(s+3)(s+1), then split into partial fractions.

[drsmoothe2004]

haha, thanks. just as i received your post, i actually figured it out. but thank you for taking the time to look at my problem

[drsmoothe2004]

another problem

1. The problem statement, all variables and given/known data
3s / (s+1)4


2. Relevant equations
find the inverse laplace


3. The attempt at a solution
i used partial fractions to split it up into A/(s+1) + B/(s+1)2 +c/(s+1)3 +D/(s+1)4

which in turn gives me A(s+1)3 + B(s+1)2 + C(s+1) + D = 3s
i plugged in s=-1 to get D=-3, i dunno how to find A, B or C (sad i know) maybe its just a late night and my brain isnt working well because i cant seem to figure it out

[HallsofIvy]

another problem

1. The problem statement, all variables and given/known data
3s / (s+1)4


2. Relevant equations
find the inverse laplace


3. The attempt at a solution
i used partial fractions to split it up into A/(s+1) + B/(s+1)2 +c/(s+1)3 +D/(s+1)4

which in turn gives me A(s+1)3 + B(s+1)2 + C(s+1) + D = 3s
i plugged in s=-1 to get D=-3, i dunno how to find A, B or C (sad i know) maybe its just a late night and my brain isnt working well because i cant seem to figure it out

As long as you have distinct first order factors, you can put in a single value for x and immediately reduce to one coefficient. With powers or irreducible quadratics, its not so trival but still not hard.

Probably easiest: put in 3 more values for s, say s= 0, s= 1, and s= 2, to get 3 linear equations in A, B, C. They won't reduce to three separated equations but still you can solve them.

Harder: go ahead and multiply everything out: A(s^3+ 3s^2+ 3s+ 1)+ B(s^2+ 2s+ 1)+ C(s+ 1)- 3= As^3+ 3As^2+ 3As+ A+ Bs^2+ 2Bs+ B+ Cs+ C- 3= 3s.

Now combine "like terms" to get 3 equations for A, B, and C.

[Count Iblis]

3s = 3(s+1) - 3

[chriscolose]

3s = 3(s+1) - 3

Yes, it looks like a shifting property would be the best to tackle this problem.