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 $x'(t)=x(t)-t$?
0
$\begingroup$
laplace-transform
1 Answers
0
$$ \frac{1-\frac{1}{s^2}}{s-1}=\frac{s^2-1}{s^2(s-1)}=\frac{s+1}{s^2}=\frac{1}{s}+\frac{1}{s^2}. $$
How did you get zero for the inverse Laplace transform?
-
0Okay 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
-
0@Marcus How do you get this particular inverse? The answer is $1+t$ as you can simply check. – 2012-12-02
-
0I got this, residue's not the way to go! – 2012-12-02
-
0Good for you :) Considering your deleted comment -- you should take a small break from math. – 2012-12-02
-
0By the way, could you tell me why the residue way doesn't work? – 2012-12-02
-
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