what is the inverse Laplace transform of -8e^(-9s)(s^2+6s-9)/s(s^2-9)

Question

I'm not sure how to solve this, any help would be appreciated, thank you in advance

what is the inverse Laplace transform of -8e^(-9s)(s^2+6s-9)/s(s^2-9)

Answer 1

Consider the expression (s²+6s-9)/(s(s²-9)).

This can be expressed in partial fractions:

(s²+6s-9)/(s(s²-9))=(s²+6s-9)/(s(s-3)(s+3))=A/s+B/(s-3)+C/(s+3).

A, B and C are constants.

A(s²-9)+B(s²+3s)+C(s²-3s)≡s²+6s-9,

So A+B+C=1 (s² coefficients)

3B-3C=6 (s coefficients)

-9A=-9 (constant)

Therefore A=1, C=-B, 6B=6, B=1, C=-1.

Partial fractions: 1/s+1/(s-3)-1/(s+3).

The complete expression is -8e^-9s(1/s+1/(s-3)-1/(s+3)).

Now ℒ^-1 can be applied to each term:

ℒ^-1{-8e^-9s/s}=-8u(t-9) where u is the unit-step function defined as:

u(t)=

⎛ 0 | t<0

⎝ 1 | t⩾0

In this case t-9⇒u is zero when t<9 and 1 when t⩾9. If t represents time, then this is known as time displacement or shift. The unit step (function=1) takes place after a delay of 9 time units. Before this lapse the function is zero.

ℒ^-1{-8e^-9s/(s-3)}=-8e^3(t-9).u(t-9)

ℒ^-1{8e^-9s/(s+3)}=8e^-3(t-9).u(t-9)

So, putting these together:

ℒ^-1{-8e^-9s(s²+6s-9)/(s(s²-9))}=ℒ^-1{-8e^-9s(1/s+1/(s-3)-1/(s+3))}=

-8u(t-9)-8e^3(t-9).u(t-9)+8e^-3(t-9).u(t-9)