Find the Laplace transform of $f(t) = \begin{cases} 0, & \text{if $t<5$} \\ t^2−10t+31, & \text{if $t\ge 5$} \\ \end{cases} $

Question

Find the Laplace transform of

$$f(t) = \begin{cases} 0, & \text{if $t<5$} \\ t^2−10t+31, & \text{if $t\ge 5$} \\ \end{cases} $$

$F(s)=$ __________?

Here is my work. I went wrong somewhere. Can someone tell me the correct answer

First, rewrite $t^2 - 10t + 31$ in powers of $t - 5$: $$\begin{split} t^2 - 10t + 31 &= [(t - 5) + 5]^2 - 10 [(t - 5) + 5] + 31 \\ &= [(t - 5)^2 + 10(t - 5) + 25] - [10(t - 5) + 50] + 31 \\ &= (t - 5)^2 + 1. \end{split} $$

Hence, $$\begin{split} f(t) &= ((t - 5)^2 + 1) u(t - 5), \text{which implies}\\ F(s) &= \frac{2!}{s^3} + \frac{\exp(-5s)}{s} \end{split}$$

by the shifting theorem.

Answer 1

Your simplification is wrong. Note that $(t-5)^2+1 = t^2 -10t + 26$ and you should have $t^2-10t+31$, so you need to use $(t-5)^2+6$ instead.