Question

In problems, $37-48$find either $F\left(s\right)$or$f\left(t\right)$, as indicated $L^{-1}\left\{\frac{e^{-s}}{s(s+1)}\right\}$.

Short answer

The Laplace transform is$y\left(t\right)=U(t-1)-e^{-(t-1)}U(t-1)$

Step-by-step solution

Formula to be used 

${\text{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}=f(t-a)U(t-a)$

Substituting in the equation of step 1

$$\begin{gathered} L^{-1}\left\{\frac{e^{-s}}{s(s+1)}\right\} \\ Herea=1 \end{gathered}$$

We need to decompose the fraction to obtain f(t)

Decomposing,

$\frac{1}{s(s+1)}=\frac{a}{s}+\frac{b}{s+1}=\frac{1}{s}-\frac{1}{s+1}=\frac{1}{s}-\frac{1}{s-(-1)}$

Explanation

$y\left(t\right)=U(t-1)-e^{-(t-1)}U(t-1)$

Hence, the answer is$y\left(t\right)=U(t-1)-e^{-(t-1)}U(t-1)$

$\frac{1}{s(s+1)}=\frac{a}{s}+\frac{b}{s+1}=\frac{1}{s}-\frac{1}{s+1}=\frac{1}{s}-\frac{1}{s-(-1)}$