I was wondering if one was easier than the other or if it is situational.
Original problem was: y'' + 4y' + 5y = 39e^(t)sint and y(0) =-1 and y'(0) = -1
I took the LT resulting: . I was wondering about the denominator, since I could set it as . And then I would have to multiply that by . Would it be easier to do it as [(s-1)^2 +1]*[ - s -5 ]
You'll end up with the same answer either way. I think it's easier to just remember the ILT for a quadratic in s in the denominator.
Tables prefer a form of
$\dfrac {\omega_0} {(s+\alpha)^2+\omega_0^2}$
corresponding to
$e^{-at}\sin(\omega_0 t)u(t)$
there are a couple similar forms.
Assuming we're going with what you've got at the bottom.
$L[y]=\dfrac{\alpha_1 s + \beta_1}{s^2+4s+5}+\dfrac{\alpha_2 s + \beta_2}{s^2-2s+2}$
I would rewrite it as
$\dfrac{\alpha_1 s + \beta_1}{(s+2)^2+1}+\dfrac{\alpha_2 s+\beta_2}{(s-1)^2+1}$
you can see with a bit of algebra these map to sinusoids times an exponential in the time domain as
$\dfrac {\omega_0}{(s+\gamma)^2+\omega_0^2}\rightarrow e^{-\gamma t}\sin(\omega_0 t)u(t)$
and
$\dfrac{s + \gamma}{(s+\gamma)^2+\omega_0^2}\rightarrow e^{-\gamma t}\cos(\omega_0 t)u(t)$
For example the first term
$\dfrac {\alpha_1 s + \beta_1}{(s+2)^2+1} = \alpha_1\left(\dfrac {s + \frac {\beta_1}{\alpha_1}}{(s+2)^2+1}\right)=$
$\alpha_1\left(\dfrac {s + 2+\left(\frac {\beta_1}{\alpha_1}-2\right)}{(s+2)^2+1}\right)=$
$\alpha_1\left(\dfrac {s + 2}{(s+2)^2+1}\right)+\dfrac {\alpha_1}{\frac {\beta_1}{\alpha_1}-2}\left(\dfrac {1}{(s+2)^2+1}\right) \Rightarrow$
$\alpha_1 e^{-2t}\cos(t)u(t) + \dfrac {\alpha_1}{\frac {\beta_1}{\alpha_1} - 2} e^{-2t}\sin(t)u(t)$
see how it works?
Just do the exact same thing for the 2nd term.
it shows in mine. It depends how you define the lower limit of the Laplace Transform as to whether you need the u(t).
I didn't check your work. You can go through it as well as I can, but if you just substitute your numbers into what I did below you should end up with the correct answers.