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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?2012-12-02
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    @Marcus How do you get this particular inverse? The answer is $1+t$ as you can simply check.2012-12-02
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    I got this, residue's not the way to go!2012-12-02
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    Good for you :) Considering your deleted comment -- you should take a small break from math.2012-12-02
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    By the way, could you tell me why the residue way doesn't work?2012-12-02
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    I am not sure how to answer your question. The simplest answer is "because it is not the way to find the inverse Laplace transform". A better answer is "Look at the formula which you use to find the inverse Laplace transform and look at the integrals you can evaluate by using the residues".2012-12-02