Having trouble with the Laplace Transform

[shamieh]

Solve by Laplace Transforms.

$y'' + 4y' + 4y = e^t$ $y(0) = 1$, $y'(0) = 0$So I've got

$s^2Y - s + 4sY - 1 + 4Y = \frac{1}{s+1}$

then I got:
$ Y = \frac{s^2+2s+2}{(s+2)(s+2)}$

Now here is where I am getting lost on the partial fraction decomposition..

I've got $s^2+2s+2 = A(s+2) + B$ I got $A =1$ but can't remember what to do to get $B$ .. is $B=0$?

 

[shamieh]

Nevermind, I solved it using polynomial long division which really sucked.

But I ended up with $1 - 2e^{-2t} + 2te^{-2t}$ but why does wolphram alpha replace the 1 with a $\delta$?

 

[chisigma]

Nevermind, I solved it using polynomial long division which really sucked.

But I ended up with $1 - 2e^{-2t} + 2te^{-2t}$ but why does wolphram alpha replace the 1 with a $\delta$?

... because is ...

$\displaystyle \mathcal {L} \{ \delta (t) \} = \int_{0}^{\infty} \delta (t)\ e^{- s\ t}\ d t = e^{0} = 1 \implies \mathcal{L}^{-1} \{ 1\} = \delta(t)\ (1)$

Kind regards

$\chi$ $\sigma$