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One of my homework problems last week was to find the inverse Laplace transform of the following:

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

The answer is $f(t)= 2e^t \cos t + 3e^t \sin t$.

Obviously once you have the decomposed fraction the remainder of the problem is simple but I can't seem to get to that point. I already turned in the assignment (they are ungraded) but I have a test coming up and want to make sure I'm ready. Could someone please lay out the steps to decompose $F(s)$?

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You want to set things up to use the formulas $$ \cal L [ e^{at}\sin (kt)]={k\over (s-a)^2+k^2},\quad \cal L [ e^{at}\cos (kt)]={s-a\over (s-a)^2+k^2} $$Towards this end, write $$ {2s+1\over s^2-2s+2}= {2s+1\over (s-1)^2+1}={2s-2+2+1\over(s-1)^2+1} ={2(s-1)\over(s-1)^2+1}+{3\over(s-1)^2+1}. $$

Note that in the third equality above, we had to do something somewhat sneaky in order to use our formulas. We needed a term of the form $a(s-1)$ or a constant term upstairs (just $2s+1$ won't do). Writing $2s+1=2(s-1)+3$ allowed us to appeal to the formulas.

The salient thing to note in this problem is that the denominator of the original expression does not factor into linear terms. So, complete the square to obtain $(s-1)^2+1$; and then the formulas above should spring to mind.

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Can you explain the step to attain the far right side of that equality? This is the step that I could not arrive at. –  rmh52 Jul 17 '12 at 3:05
    
@rmh52 I added more detail. I hope it clears things up for you. –  David Mitra Jul 17 '12 at 3:12

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