Inverse Laplace transform ,L^{-1}\left\{\frac{2s}{(s-3)^2}\right\}

Nannie Mack 2021-09-12 Answered

Inverse Laplace transform
L1{2s(s3)2}

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Expert Answer

davonliefI
Answered 2021-09-13 Author has 79 answers

Step 1
Here,
f(s)=2s(s3)2
Step 2
Take the partial fraction of f(s)
L1{2s(s3)2}
2s(s3)2=as3+b(s3)2
=a(s3)+b(s3)2
=as3a+b(s3)2
2s=as3a+b
Step 3
Comparing coefficients of s
a=2
Comparing constant terms.
3a+b=0
3(2)+b=0
b=6
So, function will be,
f(s)=2s3+6(s3)2
L1f(s)=L1(2s3)+L1(6(s3)2)
Step 4
Solving first part.
L1(2s3)=2L1(1s3)
=2e3t
Step 5
Solving second part.
L1(6(s3)2)=e3tL1{6s2}
=6e3tt21
=6te3t
Step 6
Determine inverse laplace of the function.
L1{f(s)}=2e3t+6te3t

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