Transform the given initial value problem into an initial value problem for the first two first order equations.
u'' + .25u' + 4u = 2cos(3t), u(0)=1, u'(0)=-2
The Attempt at a Solution
x1=u , x2=u' => x2' = -.25x2 -4x1 + 2cos(3t); x1'=x2
There's the system. I don't understand the initial value part, though; and my professor didn't do any examples.
I know x1 and x2 are functions of t, so the second equation is saying that the derivative of x1 is x2, and x1'(0)=x2(0)=-2; x1(0)=1. Where do I go from here?