Inverse Laplace Transform of s2-5/s3+4s2+3s

[drsmoothe2004]

Homework Statement

s²-5 / s³+4s²+3s

Homework Equations

find the inverse laplace transform

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

 

[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. Homework Statement
3s / (s+1)⁴


2. Homework 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 don't know how to find A, B or C (sad i know) maybe its just a late night and my brain isn't working well because i can't seem to figure it out

 

[HallsofIvy]

another problem

1. Homework Statement
3s / (s+1)⁴


2. Homework 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 don't know how to find A, B or C (sad i know) maybe its just a late night and my brain isn't working well because i can't 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.