Question

find the inverse laplace transform of L^-1(s^2-3s+4/s^3)​

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)