Find the inverse Laplace transform of (any two) i) frac{(s^2+3)}{s(s^2+9)} ii) logleft(frac{(s+1)}{(s-1)}right)

Joni Kenny 2021-02-20 Answered
Find the inverse Laplace transform of (any two)
i) (s2+3)s(s2+9)
ii) log((s+1)(s1))
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Expert Answer

Clelioo
Answered 2021-02-21 Author has 88 answers
(i) The given Laplace transformation is,(s2+3)s(s2+9)
We can write it,
s2+3s(s2+9)=s2s(s2+9)+3s(s2+9)
=s(s2+9)+s2(s2+9)s(s2+9))(39)
=s(s2+9)+13ss2+9+131s
=23ss2+9+131s
=23ss2+32+131s
Now taking inverse Laplace of both sides we have,
L1[s2+3s(s2+9)]=23L1[s(s2+32)]+13L1[1s]
=23cos3t+13
ANSWER:L1[s2+3s(s2+9)]=23cos3t+13
Step 3
(ii)
The given Laplace transformation is,
log(s+1s1)
We can write it,
log(s+1s1)=log(s+1)log(s1)
Now,the property of Laplace transformation is,
tf(t)dds[F(s)]
Therefore, 
tf(t)dds[log(s+1)log(s1)]
tf(t)1s+1+1s1
tf(t)et+et
f(t)etett=2(etet2t)=2(sinhtt)
Therefore,
L1[logs+1s1]=2(sinhtt)
Answer: L1[logs+1s1]=2(sinhtt)
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