Find the inverse laplace trans. displaystyle{F}{left({s}right)}=frac{10}{{{s}{left({s}^{2}+{9}right)}}}

tricotasu

tricotasu

Answered question

2021-02-19

Find the inverse laplace trans.
F(s)=10s(s2+9)

Answer & Explanation

avortarF

avortarF

Skilled2021-02-20Added 113 answers

Step 1 
Given, 
F(s)=10s(s2+9) 
The Laplace transform of this function's inverse is needed.
Step 2 
F(s)=10s(s2+9) 
=109s10s9(s2+9) 
Using the reverse Both sides' Laplace transform
L1[F(s)]=L1[109s10s9(s2+9)] 
Then, f(t)=L1[109s]L1[10s9(s2+9)] 
=109L1[1s]109L1[ss2+9] 
=109L1[1s]109L1[ss2+32] 
Step 3 
Apply the formula in a way that
L1[1s]=t 
L1[ss2+a2]=cos(at) 
f(t)=109t109cos(3t) 
=109[tcos(3t)] 
Step 4 
Therefore, 
f(t)=109[tcos(3t)]

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