Solve 2nd order ODE using Laplace transform

Question

Im trying to solve a laplace transoform question, but i am stuck.

The question is $y''(t)+2\zeta y'(t)+y(t)=0$,$y(0)=1$,$y'(0)=0$ and $\zeta=2$.

I have so far done: Laplace transform which gives

\begin{align*} s^2y(s)−sy(0)−y′(0)+2\zeta(sy(s)−y(0))+y(s)& \\ s^2y(s)−sy(0)−y′(0)+4sy(s)−4y(0)+y(s)&&\text{When $\zeta=2$}\\ s^2y(s)−s+4sy(s)−4+y(s)&\\ s^2y(s)+4sy(s)+y(s)&=s+4\\ y(s)[s^2+4s+1]&=(s+4)\\ y(s)&=(s+4)/(s^2+4s+1)\\ \end{align*} I'm stuck on this bit not sure what to do after this.

Answer 1

$$y(t)=\frac{1}{2i\pi}\int_\gamma g(s,t)ds=\frac{1}{2i\pi}\int_\gamma \frac{s+4}{s^2+4s+1}e^{st}ds $$ Poles: $s=-2 \pm \sqrt 3$ $$\mathcal{Res}_s(g)=\frac{(s+4)e^{st}}{2s+4}$$ $$y(t)=\frac{((-2 + \sqrt 3)+4)e^{(-2 + \sqrt 3)t}}{2(-2 + \sqrt 3)+4}+\frac{((-2 - \sqrt 3)+4)e^{(-2 - \sqrt 3)t}}{2(-2 - \sqrt 3)+4}$$

The rest is up to you: use hyperbolic trigonometric functions

$$y(t)=\frac 1 3 e^{-2 t} (3 \cosh(\sqrt 3 t)+2 \sqrt 3 \sinh(\sqrt 3 t))$$

Answer 2

If you're not familiar with the integral approach for the inverse transform, you can refer to a table. Completing the square in the denominator is a good way to start:$$\frac{s+4}{s^2+4s+1}=\frac{s+2}{(s+2)^2-3}+\frac{2}{(s+2)^2-3}$$ Next, recall that $$\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2},\quad\mathcal{L}\{\cosh at\}=\frac{s}{s^2-a^2}$$and$$ \mathcal{L}\{e^{ct}f(t)\}=F(s-c)$$ where $F(s)$ denotes the Laplace transform of $f(t)$.