Question

Question:

How do we find the inverse laplace transformation of

s/[ ((s^2) + 9)^2 ]

My attempt at the question:

I noted that s/[ ((s^2) + 9)^2 ] can be written as follows:
s/[ ((s^2) + 9)^2 ] ={ s/[(s^2) + 9] }* { 1/[(s^2) + 9] }

By further simplying, I got,

s/[ ((s^2) + 9)^2 ] =(1/3) * { s/[(s^2) + 9] }* { 3/[(s^2) + 9] }

The RHS now seems to be a multiplication of two functions F(S) and G(S) such that

F(S) = s/[ ((s^2) + 9)^2 ]

and

G(S) = 1/[(s^2) + 9], which gives,

s/[ ((s^2) + 9)^2 ] = (1/3)* F(S) * G(S)


Now ILT(Inverse Laplace Transform) of s/[ ((s^2) + 9)^2 ] = ILT[ (1/3)* F(S) * G(S) ] = (1/3) * ILT [ (F(S)*G(S) ]

From the Convolution theorem, we know that,

ILT( F(S) * G(S) ) = integration from 0-t { f(u)*g(t-u) du } , where f(t) = ILT(F(S)) and g(t) = ILT(G(S))

Hence,

ILT of s/[ ((s^2) + 9)^2 ] = -cos(6t) | limits from 0 - t,

which gives the answer as, 1 - cos(6t)

Is this approach correct?

Answer 1

clearly not, since ℒ(1 - cos(6t)) = 36/(s(s^2+6^2))
and the s^2+9 should have been a clue to look for sin/cos (3t), not 6t.
I get a result of 1/6 t sin(3t)

see whether you can find where you went astray.