Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Isee. so what would the correct answer be now that we have the 6 –  Michael Rametta Apr 3 '13 at 22:07
1  
(2!/s^3 + 6/s) e^(-5s) this is the correct anwer. Thanks so much –  Michael Rametta Apr 3 '13 at 22:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.