I know when trying to solve an initial value problem using Laplace transforms I need to put them in this form $f(0)=a$ and $f'(0)=b$. So when I'm given this problem : $$y''-y'-2y=-8cost-2sint\\ y({\pi\over 2})=1, \\ y'({\pi\over 2})=0$$ I have to change the initial values $t=0$. This website told me I needed to do a change of variable to make this work. It said to define $n=t-{\pi\over 2}$ and $t=n+{\pi\over 2}$
Then it said "to simplify life a little let's define" $$u(n)=y(n+3)$$ then $u(0)=y(0+{\pi\over 2})=y({\pi\over 2})=1$ and $u'(0)=y'(0+{\pi\over 2})=y'({\pi\over 2})=0$
At this point I'm a little lost, but I'll figure it out at some point. The point of the post was to ask a different question that has to do with the following. My friend did the problem using this method: $$t_i=z(0)=y({\pi\over 2})$$
His way obviously is much less work, but I'm confused as to why the website (which is normally very good) would tell me to change variables. Is there a preferred method?
Edit: I realize that my friend did change variables, what I meant was why I had to do so much more work than he did.