find the inverse laplace transform for the question above
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The clue to the solution lies in the denominator, because the factors are recognisable as denominators of Laplace Transforms in a table. So the first step is to convert the expression into partial fractions.

F(s)=A/(s-1)+B/(s-2)+C/(s-4) where A, B and C are constants.

If the transform is 1/(s-a), then the inverse is eᵃᵗ, so f(t)=Aeᵗ+Be²ᵗ+Ce⁴ᵗ.

To find A, B and C we use the identity:

s²+6s+9≡A(s-2)(s-4)+B(s-1)(s-4)+C(s-1)(s-2)=

(A+B+C)s²-(6A+5B+3C)s+(8A+4B+2C).

Equating coefficients:

A+B+C=1, 6A+5B+3C=-6, 8A+4B+2C=9.

There are 3 equations and 3 unknowns in this system so we can find the unknowns.

8A+4B+2C-2(A+B+C)=9-2=7, 6A+2B=7,

6A+5B+3C-3(A+B+C)=-6-3=-9, 3A+2B=-9, 

3A=7-(-9)=16, A=16/3; 2B=7-6A=7-32=-25, B=-25/2;

16/3-25/2+C=1, 32/6-75/6+C=1, C=75/6-32/6+6/6=49/6.

So f(t)=16eᵗ/3-25e²ᵗ/2+49e⁴ᵗ/6.

 

by Top Rated User (599k points)

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