Question

$L\left\{\right(3t+1\left)U\right(t-1\left)\right\}$

Short answer

The Laplace transform is$\frac{3}{s^2}{e}^{-s}+\frac{4}{s}{e}^{-s}$

Step-by-step solution

Unit step function

The unit step function $U(t-a)$is defined to be

$$\begin{gathered} U(t-a)=\left\{0,0⩽t<a \\ 1,t⩾a\right. \end{gathered}$$

Apply Laplace transform

Try to make the function look like the unit step function whenever you can.

$$\begin{gathered} \left\{\right(3t+1\left)U\right(t-1\left)\right\}=L\left\{\right(3(t-1)+4\left)U\right(t-1\left)\right\} \\ =3L\left\{\right(t-1\left)U\right(t-1)+4L\{U(t-1)\left\} \\ Y\right(s)=\frac{3}{s^2}{e}^{-s}+\frac{4}{s}{e}^{-s} \end{gathered}$$

Hence, the solution is $Y\left(s\right)=\frac{3}{s^2}{e}^{-s}+\frac{4}{s}{e}^{-s}$