Question

Laplace inverse transform of S/(S2+1)(s2-4)

Answer 1

First we need to express this in partial fractions:

F(s)=s/((s²+1)(s²-4))≡(As+B)/(s²+1)+C/(s+2)+D/(s-2).

Therefore, s≡(As+B)(s²-4)+C(s²+1)(s-2)+D(s²+1)(s+2).

s≡As³-4As+Bs²-4B+Cs³-2Cs²+Cs-2C+Ds³+2Ds²+Ds+2D,

s≡(A+C+D)s³+(B-2C+2D)s²+(-4A+C+D)s+(-4B-2C+2D).

Equating coefficients:

s³: A+C+D=0, so C+D=-A;

s²: B-2C+2D=0, -2C+2D=-B

s: -4A+C+D=1, -4A-A=1, A=-⅕;

constant: -4B-2C+2D=0, -4B-B=0, B=0, C=D.

C+D=⅕, C=D=1/10.

F(s)=-1/(5(s²+1))+1/(10(s+2))+1/(10(s-2)).

We can look up each transform from a table:

f(t)=(-⅕)sin(t)+(e²ᵗ+e⁻²ᵗ)/10, without using hyperbolic trigonometry.