What is the invers laplas funsion of (3s-s)/(2s^2-6s-2) and (s+7)/(s^2-3s-10)

Question

This is a math of laplas. Answer will be shown on invers laplas funsion.

Answer 1

(1) 2s²-6s-2=2(s²-3s-1)=2(s²-3s+9/4-9/4-1)=2((s-3/2)²-13/4).

This can be written 2(s-3/2-√13/2)(s-3/2+√13/2). 3s-s=2s.

s/((s-3/2-√13/2)(s-3/2+√13/2))=A/(s-3/2-√13/2)+B/(s-3/2+√13/2), where A and B are constants to be found.

A(s-3/2+√13/2)+B(s-3/2-√13/2)=s, so equating coefficients:

A+B=1, A(-3/2+√13/2)+B(-3/2-√13/2)=0=(A+B)(-3/2)+A√13/2-B√13/2,

(A-B)√13/2=3/2, A-B=3/√13. Therefore 2A=1+3/√13, A=(1+3/√13)/2=(13+3√13)/26 and B=(13-3√13)/26.

So (3s-s)/(2s²-6s-2)=(1/26)((13+3√13)/(s-3/2-√13/2)+(13-3√13)/(s-3/2+√13/2)).

ℒ^-1{1/(s-a)}=e^at, so if we let a=3/2+√13/2, b=3/2-√13/2, p=13+3√13 and q=13-3√13:

ℒ^-1{(3s-s)/(2s²-6s-2)}=(pe^at+qe^bt)/26=

[(13+3√13)e^{(3/2+√13/2)t}+(13-3√13)e^{(3/2-√13/2)t}]/26=(e^3/2/26)((13+3√13)e^t√13/2+(13-3√13)e^-t√13/2).

(2) s²-3s-10=(s-5)(s+2),

(s+7)/(s²-3s-10)=A/(s-5)+B/(s+2),

A(s+2)+B(s-5)=s+7,

A+B=1, 2A-5B=7, 2A-5(1-A)=7, 2A-5+5A=7, 7A=12, A=12/7, B=-5/7.

(s+7)/(s²-3s-10)=(1/7)(12/(s-5)-5/(s+2)).

ℒ^-1{(s+7)/(s²-3s-10)}=(1/7)(12e^5t-5e^-2t).

I suspect that the first given Laplace expression (1) has been wrongly presented because of the complexity that resulted in finding the inverse. The numerator 3s-s is too simplistic to be correct, and the quadratic denominator should have had rational factors. The Laplace expression in (2) was a lot simpler and uncomplicated.