How do I solve this initial value problem?

Question

y"+9y=0; y(pi/3)=1, y'(pi/3)=-2

Answer 1

The general solution of this equation is
y = A cos (w x-t) where on this case w = sqrt9=3
(note that the caracteristic equation is r^2+9=0)
y(pi/3)= A cos(pi-t) = -Acos t =1
y´= -3A sin (wx-t)
y´(pi/3) = -3A sin(pi-t) =-3A sin t= -2
squaring and summing
A^2 = 1+4/9 = 13/9 so A (amplitude =1/3*sqrt(13)
tant = 2/3 so t= arctan -2/3 =0.5880

y= 1/3*sqrt(13)* cos(3x-0.5880)
In physics w=2pi f where f is the frequency