Question

Find inverse laplace transform for x(s) = (2)/(s(s+1)^2)

Answer 1

Answer:

x(t)=$2-2e^{-t}-2te^{-t}$

Step-by-step explanation:

The function x(s) = (2)/(s(s+1)^2) can be expressed as partial fractions:

X(s)=$\frac{2}{s(s+1)^2}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+1)^2}$

2=A·(s+1)²+B·s·(s+1)+C·s

2=A·(s²+2s+1)+B·(s²+s)+C·s

2=A·s²+2sA+A+B·s²+Bs+C·s

2=s²(A+B)+s(2A+B+C)+A

So we can find the values of A, B and C by solving these equations:

A+B=0 ⇒ 2+B=0 ⇒ B= -2

2A+B+C=0 ⇒ 2·2+(-2)+C=0 ⇒ C= -2

A=2

So X(s) is expressed as:

X(s)=$\frac{2}{s}+\frac{-2}{s+1}+\frac{-2}{(s+1)^2}$

Using the inverse laplace transform tables we obtain x(t):

ℒ⁻¹{X(s)}=ℒ⁻¹{$\frac{2}{s}+\frac{-2}{s+1}+\frac{-2}{(s+1)^2}$}

ℒ⁻¹{X(s)}=ℒ⁻¹{$\frac{2}{s}$}+ℒ⁻¹{$\frac{-2}{s+1}$}+ℒ⁻¹{$\frac{-2}{(s+1)^2}$}

x(t)=$2-2e^{-t}-2te^{-t}$