Solve no.4 inverse laplace L^{-1}left{s lnleft(frac{s}{sqrt{s^2+1}}right)+cot^{-1s}right}

babeeb0oL 2021-02-25 Answered
Solve no.4 inverse laplace
L1{sln(ss2+1)+cot1s}
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Expert Answer

okomgcae
Answered 2021-02-26 Author has 93 answers
Step 1
Given:
L1{sln(ss2+1)+cot1s} Step 2
Use the linear property of inverse Laplace transformation, we get
L1{sln(ss2+1)+cot1s}=L1{sln(ss2+1)}+L1{cot1s}
Now first find the inverse Laplace transformation of the first term
We know that the Laplace transformation formula:
If L{f(t)}=F(s) then L{tf(t)}=dds(F(s))
f(t)=1tL1{ddsF(s)}
L1{sln(ss2+1)}=1tL1{ds(sln(ss2+1))}
Step 3
compute the derivative, we get
1tL1lnss2+1+1s2+1
1tL1{lnss2+1}1tL1{1s2+1}
Now we know that L1{as2+a2}=sin(at)
1tL1{ln(ss2+1)}sin(t)t
Use the quotient rule of logarithmic, we get
1tL1{ln(s)12ln(s2+1)}sin(t)t
Step 4
Use the inverse Laplace transformation formula in the first term:
f(t)=1tL1{ddsF(s)}
1t2L1{dds[ln(s)12ln(s2+1)]}sin(t)t
1t2L1{1s}1t2L1{ss2+1}sin(t)t
1t2(L1{1s}L1{ss2+1})sin(t)t
L1{sln(ss2+1)}=1t2(1cos(t))sin(t)t
Step 5
Now find the inverse Laplace transformation of
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