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?
Trying to find laplace transform $x(s)$ that satisfies $x'(t)=x(t)-t$?
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1 Answers
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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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0I 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