Question

Inverse Laplace Transform (2s^2+4s+8)/(s^2+4)(s^2+2s+1)

Answer 1

@AccessDenied
can u tall me the way how to do this equation ?

Answer 2

partial fraction...then find inverse

Answer 3

can u show me it plz

Answer 4

Let\[\frac{As+B}{s^2+4}+\frac{Cs+D}{s^2+2s+1}=\frac{2s^2+4s+8}{(s^2+4)(s^2+2s+1)}\]
So, we get
\[(As+B)(s^2+2s+1)+(Cs+D)(s^2+4) \equiv 2s^2+4s+8 \]
\[(A+C)s^3+(2A+B+D)s^2+(A+2B+4C)s+(B+4D) \equiv 2s^2+4s+8 \]Hence, we have four equations in four unknown. They are:
A+C =0
2A+B+D = 2
A+2B+4C = 4
B+4D = 8

Solve A,B,C,D, then you'll be done with the partial fractions part.