Hi I have this question and I am horribly stuck at one part and I cant seem to figure out if i did something wrong so any advice or help would be greatly apprecaited. Here is the question:
$$y''-y'-6y=\begin{cases}0 &\text{if } 0 \le t < \pi,\\ cos(t)&\text{if } t \ge \pi.\end{cases} $$
Which Is: $$ y''-y'-6y= u(t-\pi)cos(t)$$ Take Laplace transform of entire DE and get: Where Y(s)=L{y(t)} $$s^2Y(s)-sy(0)-y'(0)-sY(s)+y(0)-6Y(s)=e^{-\pi s}(\frac{s}{s^2+1})$$ Plug in initial conditions and simplify and get: $$Y(s)=e^{-\pi s}(\frac{s}{(s^2+1)(s^2-s-6)}) + \frac{3s+1}{s^2-s-6}$$ $$Y(s)=e^{-\pi s}(\frac{s}{(s^2+1)(s+2)(s-3)}) + \frac{3s+1}{(s+2)(s-3)}$$ Now this is where I am stuck because I have no idea how to solve this and when I tried to use partial fractions it got every ugly. Is there a better way to solve this than using partial fractions? Or have I made a mistake somewhere in the steps above that caused this to be very messy?
Thanks, any advice or help would be greatly appreciated.
The Finale Answer after doing partial fractions ended up being (and hopefully correct): $$y(t)=u(t-\pi)[\frac{1}{50}(-sin(t-\pi)-7cos(t-\pi))+\frac{2}{25}e^{-2t-2\pi} + \frac{3}{50}e^{3t+3\pi}] + e^{-2t} + 2e^{3t}$$