find the inverse of Laplace transform frac{3}{(s+2)^2}-frac{2s+6}{(s^2+4)}

Josalynn 2021-01-08 Answered
find the inverse of Laplace transform
3(s+2)22s+6(s2+4)
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

broliY
Answered 2021-01-09 Author has 97 answers
Step 1
Given Laplace transform is,
3(s+2)22s+6(s2+4)
Step 2
Taking inverse of Laplace transformation as,
L1{3(s+2)22s+6(s2+4)}=L1{3(s+2)2}L1{2s+6(s2+4)}
=3L1{1(s+2)2}2L1{ss2+4}6L1{1s2+4}
Step 3
Some formulas of inverse of Laplace transformation are
1) L1{F(s)}=f(t) then L1{F(sa)}=eatf(t)
L1{1(s+2)2}=e2tL1{1s2}=e2tt
2)L1{as2+a2}=sinat
L1{1s2+4}=12L1{2s2+4}=(sin2t)2
3)L1{sF(s)}=f(t)+f(0)
Then L1{ss2+4}=12L1{s2s2+4}=12((sin2t)+sin(0))=cos2t
Step 4
Then, consider equation (1),
3L1{1(s+2)2}2L1{ss2+4}6L1{1s2+4}=3L1{1(s+2)2}2L1{ss2+4}62L1{2s2+4}=3e2tt2cos2t3sin2t
Step 5
Therefore, the inverse of the given Laplace transformation is,
L1{3(s+2)22s+6s2+4}=3e2tt2cos2t3sin2t
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions