Find the particular solution of the differential equation y''+36y=0 satisfying the conditions y(0)=−4 and y(pi/12)=3.
Your answer should be a function of x.
The Attempt at a Solution
I think I know how to do this kind of question, and I can't see where I'm going wrong. I set up the auxiliary equation, r^2+36 = 0, and get the roots to be + or - 6i. So then I get y = c1cos(6x)+c2sin(6t). I differentiate, and get y' = -6c1sin(6x)+6c2cos(6x). Then I plug in the initial values given. The first equation gives me c1 = -4, and the second gives me 3 = -6c1, which is clearly incorrect (and in both equations c2 is cancelled out?).
I've looked over it over and over again but I can't see where I'm going wrong.
Thanks for any help