Find the inverse Laplace transform?

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SUMMARY

The inverse Laplace transform of the function F(s)=(8s^2-4s+12)/(s(s^2+4)) is calculated to be f(t)=3-2 sin(2t)+5 cos(2t). The solution involves decomposing the function into partial fractions, identifying constants A, B, and C, and applying known Laplace transforms. The constants are determined as A=3, B=5, and C=-4, leading to the final expression for f(t).

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Homework Statement


Find the inverse Laplace transform of F(s)=(8s^2-4s+12)/(s(s^2+4)).


Homework Equations


A/s+(Bs+C)/(s^2+4)
8s^2-4s+12=A(s^2+4)+(Bs+C)(s)=As^2+4A+Bs^2+Cs=s^2(A+B)+Cs+4A
8=A+B
C=-4
A=3
B=5
L^-1 (3/s)+L^-1 ((5s-4)/(s^2+4))
=3+ (Now I'm stucked.)



The Attempt at a Solution


The answer is f(t)=3-2 sin 2t+5 cos 2t.
 
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[tex] \frac{5s - 4}{s^2 + 4} = 5\frac{s}{s^2 + 4} - 2\frac{2}{s^2 + 4}[/tex]

Hopefully you should recognise the two terms on the right hand side as being Laplace transforms.
 
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Yes, I did. Thank you.
 

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