Laplace transform of (2s^2 +10s) / ((s^2 -2s +5)(s+1))

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SUMMARY

The discussion centers on the inverse Laplace transform of the function (2s^2 + 10s) / ((s^2 - 2s + 5)(s + 1)). The correct inverse Laplace transform is confirmed to be (e^t)[(3cos(2t) + 2.5sin(2t))] - (e^-t), while the user's calculation yielded (e^t)[(3cos(2t) + 4sin(2t))] - (e^-t). The partial fraction decomposition was verified as correct, leading to the conclusion that the user's answer is incorrect despite their confidence in the calculation.

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foo9008
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Homework Statement


(2s^2) +10s / (s^2 -2s +5 )(s+1) , I have checked the partial fraction , it's correct , but according to the ans it's (e^t)[(3cos2t + 2.5sin2t)] - (e^-t), but my ans is (e^t)[(3cos2t + 4sin2t)] - (e^-t)

Homework Equations

The Attempt at a Solution


UsCyv8m.jpg
 
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foo9008 said:

Homework Statement


(2s^2) +10s / (s^2 -2s +5 )(s+1) , I have checked the partial fraction , it's correct , but according to the ans it's (e^t)[(3cos2t + 2.5sin2t)] - (e^-t), but my ans is (e^t)[(3cos2t + 4sin2t)] - (e^-t)

Homework Equations

The Attempt at a Solution


UsCyv8m.jpg

You typed
2s^2 +\frac{10s}{(s^2 -2s +5 )(s+1)}
If you mean
\frac{2s^2 + 10s}{(s^2-2s+5)(s+1)},
you must either use LaTeX (as I did just now) or else use parentheses, like this:
(2s^2+ 10s)/[(s^2-2s+5)(s+1)]

Anyway, that form gives an inverse Laplace that agrees with your answer.
 
Ray Vickson said:
You typed
2s^2 +\frac{10s}{(s^2 -2s +5 )(s+1)}
If you mean
\frac{2s^2 + 10s}{(s^2-2s+5)(s+1)},
you must either use LaTeX (as I did just now) or else use parentheses, like this:
(2s^2+ 10s)/[(s^2-2s+5)(s+1)]

Anyway, that form gives an inverse Laplace that agrees with your answer.
Sorry, I mean the second one. You mean my answer is correct??
 
foo9008 said:
Sorry, I mean the second one. You mean my answer is correct??

Isn't that what I said?
 

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