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

[foo9008]

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

 

[Ray Vickson]

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

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.

 

[foo9008]

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??

 

[Ray Vickson]

Sorry, I mean the second one. You mean my answer is correct??

Isn't that what I said?