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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Taking the Laplace transform of the equation $$x'(t)=x(t)-t,$$ we get $$sx(s)-x(0)=x(s)-\frac{1}{s^2},$$ right? So if $x(0)=1$, don't you get $$x(s)=\frac{1-\frac{1}{s^2}}{s-1}?$$ When I take the inverse laplace of this I get 2pi*i, how do I know this works?

share|cite|improve this question
up vote 0 down vote accepted

$$ \frac{1-\frac{1}{s^2}}{s-1}=\frac{s^2-1}{s^2(s-1)}=\frac{s+1}{s^2}=\frac{1}{s}+\frac{1}{s^2}. $$

How did you get zero for the inverse Laplace transform?

share|cite|improve this answer
Okay now I get 2*pi*i for the inverse, but how does that work with the original equation x'(t) = x(t) - t? – Marcus Dec 2 '12 at 2:01
@Marcus How do you get this particular inverse? The answer is $1+t$ as you can simply check. – Artem Dec 2 '12 at 2:05
I got this, residue's not the way to go! – Marcus Dec 2 '12 at 2:22
Good for you :) Considering your deleted comment -- you should take a small break from math. – Artem Dec 2 '12 at 2:24
By the way, could you tell me why the residue way doesn't work? – Marcus Dec 2 '12 at 2:27

Your Answer


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.