How do I find the Laplace transform of the function $f(t)=3t$ if $t\le 6$ and $f(t)=18$ if $t>6$?

Question

I found the step function: $u(6-t)\times 3(t-6)+18$ where $u(t)$ is the step function, but I do not know how to find the inverse laplace of this. Do I need to change the form of the step function?

Answer 1

I guess there are two kinds of step functions. $$\begin{align}\mathscr{L}[u(t-T)f(t)](s)&=\int_0^{\infty}u(t-T)f(t)e^{-st}dt\\ &=\int_T^{\infty}f(t)e^{-st}dt\\ &=\int_0^{\infty}f(v+T)w^{-s(v+T)}dv\\ &=e^{-sT}\mathscr{L}[f(t+T)](s)\end{align}$$ And then there is $$\begin{align}\mathscr{L}[u(T-t)f(t)](s)&=\int_0^{\infty}u(T-t)f(t)e^{-st}dt\\ &=\int_0^{\infty}f(t)e^{-st}dt-\int_T^{\infty}f(t)e^{-st}dt\\ &=\mathscr{L}[f(t)](s)-e^{-sT}\mathscr{L}[f(t+T)](s)\end{align}$$ So $$\begin{align}\mathscr{L}[3(t-6)u(6-t)+18](s)&=3\mathscr{L}[[(t-6)u(6-t)](s)+18\mathscr{L}[1](s)\\ &=3\left\{\frac1{s^2}-\frac6s-e^{-6s}\cdot\frac{1}{s^2}\right\}+18\cdot\frac1s\\ &=\frac3{s^2}\left(1-e^{-6s}\right)\end{align}$$