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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?

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

How did you get zero for the inverse Laplace transform?

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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

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