Where is my error in solving $y'' + y' + y = 0, y(0) = 1, y'(0) = 0$ with Laplace transform?

Question

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.

Answer 1

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).

Answer 2

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?