The instruction is to solve this IVP using the method of Laplace transforms: $$y''-y'-2y=-8cost-2sint\;\;\;\;\;\;\;\; y\left({\pi\over 2}\right)=1,\;\;\;\;y'\left({\pi\over 2}\right)=0$$ I needed to change the variables, so I did: $n=t-{\pi\over 2}\;\;\;\;t=n+{\pi\over 2}\;\;\;$
$u(n)=y(t+{\pi\over 2});\;\;\;\;\;u'(n)=y'(y+{\pi\over 2});\;\;\;\;u''(n)=y''(t+{\pi\over 2})$
Then I rewrote the original equation: $$u''+u'+2u=-8cos\left(n+{\pi\over 2}\right)-2sin\left(n+{\pi\over 2}\right)$$ I took the Laplace transform of every term: $$[s^2U(s)-su(0)-u'(0)]-[sU(s)-u(0)]-2[U(s)]={8\over s^2+1}-{2s\over s^2+1}$$ Isolated $U(s)$ $$U(s)(s^2-s-2)-s+1={{8-2s}\over s^2+1}$$ $$U(s)={{8-2s}\over (s^2+1)(s-2)(s+1)}+{{s-1}\over (s-2)(s+1)}$$ Did a bunch of partial fractions and ended up with this equation: $$U(s)={7\over 5}{s\over s^2+1}-{11\over 5}{1\over s^2+1}+{4\over 15}{1\over s-2}-{5\over 3}{1\over s+1}+{1\over 3}{1\over s-2}+{2\over 3}{1\over s+1}$$ Then inverse Laplace gave me: $$u(n)={7\over 5}sin(n)-{11\over 5}sin(n)+{4\over 15}e^{2n}-{5\over 3}e^{-n}+{1\over 3}e^{2n}+{2\over 3}e^{-n}$$ $$=-{4\over 5}sin(n)+{9\over 15}e^{2n}-e^{-n}$$ I then replaced $n$ with $t-{\pi\over 2}$: $$-{4\over 5}sin\left(t-{\pi\over 2}\right)+{9\over 15}e^{2\left(t-{\pi\over 2}\right)}-e^{\left({\pi\over 2}-t\right)}$$ I really tried my best. There is a lot more work to this that I have on my paper but it's mostly just simplifying things (and solving the partial fractions) and I didn't think it would be necessary to put all of that down. I've checked my work over and over and I keep getting the same answer. What did I do wrong?
HINT: Take a careful look at the differences between the first two terms in your final expression for $U(s)$, right before you perform the inverse Laplace transform.