Laplace Transforms of Step Functions

Question

The problem asks to find the Laplace transform of the given function:

$$ f(t) = \begin{cases} 0, & t<2 \\ (t-2)^2, & t \ge 2 \end{cases} $$

Here's how I worked out the solution:

$$\mathcal{L}[f(t)]=\mathcal{L}[0]+\mathcal{L}[u_2(t)(t-2)^2]=0+e^{-2s}\mathcal{L}[(t-2)^2]=e^{-2s}\mathcal{L}[t^2-4t+4]=2e^{-2s}\left(\frac{1}{s^3}-\frac{2}{s^2}+\frac{2}{s} \right)$$

However, the solution in the back of the book is simply:

$ 2e^{-s}s^{-3} $

What did I do wrong?

Answer 1

$$F(s) = \int_2^\infty (t-2)^2 e^{-st} \, dt \to F(s) =\int_0^\infty u^2e^{-s(u+2)}\, du=e^{-2s}\int_0^\infty t^2e^{-st}\, dt=\frac{2}{s^3}e^{-2s}. $$

Answer 2

This seems to be the thing that students in DE have more trouble with than anything else. That formula in the book reads $$\mathcal L[u_c(t)f(t-c)] =e^{-cs}\mathcal L[f(t)].$$ You're trying to use this to find $\mathcal L[u_2(t)(t-2)^2]$. To fit this into that formula you must have $$u_2(t)f(t-2)^2=u_c(t)f(t-2).$$

But in the next step you seem to think that $f(t)=(t-2)^2$. Not so. You need to ask yourself, if $f(t-2) = (t-2)^2$ what is $f(t)$? Answer, $f(t)=t^2$. So your transform should be $e^{-2s}\mathcal L[t^2].$