Use the table of laplace transform to find the inverse Laplace transform of F(s)=\frac{2+s(s+1)}{s(s^2-s-6)}

sibuzwaW

sibuzwaW

Answered question

2021-09-08

Use the table of laplace transform to find the inverse Laplace transform of F(s)=2+s(s+1)s(s2s6)

Answer & Explanation

likvau

likvau

Skilled2021-09-09Added 75 answers

Step 1
solution: We need to find the inverse laplace transform of F(s)=2+s(s+1)s(s2s6)
Step 2
For this
F(s)=2+s(s+1)s(s2s6)
=2s(s2s6)+s+1(s2s6)
F(s)=2s(s3)(s+2)+s+1(s3)(s+2)
using partial fraction
F(s)=13s+2151s3+151(s+2)+451(s3)+15(s+2)
Taking inverse laplace transform is
f(t)=L1{F(s)}=13L1{1s}+215L1{1s3}+15L1{1s+2}+45L1{1s3}+15L1{1s+2}
f(t)=13+2e3t15+e2t5+45e3t+e2t5
f(t)=1415e3t+25e2t13
Hence, inverse laplace transform of F(s)
f(t)=L1{F(s)}=1415e3t+25e2t13
L1{F(s)}=1415e3t+25e2t13

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