Question

Find the inverse Laplace transform of f(s)=(s^2 +1)/(s^2 +3s^2 +2s)

Answer 1

if denominator is s³+3s²+2s then

f(s)= (s²+1)/(s³+3s²+2s)

L^-1(f(s))=L^-1((s²+1)/(s³+3s²+2s))    ---------eq(1)

now,

(s²+1)/(s³+3s²+2s)=(s²+1)/s(s²+3s+2)

=>(s²+1)/(s³+3s²+2s)=(s²+1)/s(s+2)(s+1)

using partial fraction , let (s²+1)/s(s+2)(s+1)=A/s   +   B/s+2    +    C/s+1

 equating numerator both sides,

(s²+1)= A(s+2)(s+1) + B(s)(s+1) +C(s)(s+2)

put s=0;

A=1/2;

put s=-2

B=5/2

put s=-1

C=-2

=>(s²+1)/s(s+2)(s+1)=1/2/s   +   (5/2)/s+2    +    (-2)/s+1-------------eq(2)

from eq(1) and eq(2)

L^-1(f(s))=L^-1 ((1/2)/s)   +  L^-1 ((5/2)/s+2)    +  L^-1  ((-2)/s+1)

            =(1/2) + (5/2) e^-2t -2e^-t        (ans)