<s>[INST] 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.
 [/INST] 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?
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