Where is my error in solving $y'' + y' + y = 0, y(0) = 1, y'(0) = 0$ with Laplace transform? Im trying to solve a laplace transoform question, but i am stuck.
The question is $y′′(t) + 2ζy′(t) + y(t) = 0, y(0) = 1, y′(0) = 0$ and $ζ = 0.5$.
I have so far done:
Laplace transform which gives
$s^2Y(s)-sy(0)-y'(0)+2ζ(sY(s)-y(0))+Y(s)$
When $ζ = 0.5$
$S^2Y(s)-Sy(0)-y'(0)+sY(s)-y(0)+Y(s)$
$S^2Y(s)-s-sY(s)-1+Y(s)$
$s^2Y(s)-sY(s)+Y(s)=s+1$
$Y(s)[s^2-s+1]=(s+1)$
$Y(s)=(s+1)/(s^2-s+1)$
I'm stuck on this bit not sure what to do after this.
 A: Since the denominator is not easily factorable, we avoid partial fractions and instead complete the square. Recall the following inverse Laplace transforms:

$$
Y(s) = \frac{b}{(s - a)^2 + b^2} \implies y(t) = e^{at}\sin bt \\
Y(s) = \frac{s - a}{(s - a)^2 + b^2} \implies y(t) = e^{at}\cos bt
$$

Indeed, observe that:
\begin{align*}
\frac{s + 1}{s^2 - s + 1}
&= \frac{s + 1}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{\frac{3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{3}{\sqrt 3} \cdot \frac{\frac{\sqrt 3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2}
\end{align*}
Can you take it from here?
A: EDIT: As @georg stated in his comment there is a slight mistake, I've fixed the following accordingly: $$Y(s)=\frac{s+\dfrac 1 2  + \dfrac 1 2}{(s+\dfrac 1 2)^2+\frac 3 4}$$
$$Y(s)=\frac{s+\dfrac 1 2}{(s+\dfrac 1 2)^2+\frac 3 4} +\frac 1 {\sqrt 3} \frac{\dfrac {\sqrt 3} 2}{(s+\dfrac 1 2)^2+\frac 3 4}$$
And keep in mind:
$$\mathcal L (\cos(\omega t))=\frac s {s^2+\omega^2}$$
$$\mathcal L (\sin(\omega t))=\frac \omega {s^2+\omega^2}$$
$$\mathcal L (e^{at}f(t))=F(s-a)$$
PLUS: If you're good at complex analysis, use the Bromwich inversion formula
$$f(t)=\sum \mathcal{Res}(e^{st}Y(s))$$
(where $s$, not $t$, is the variable).
