00:01
Okay, we're given that y -equal c1 cosine 2x plus c2 sine 2x is a solution to a differential equation.
00:08
And they want us to find out what c1 and c2 are by using these initial conditions.
00:14
So first, let's draw picture at pi, which is right here.
00:21
This point's name is negative 1 -0.
00:25
So that's the cosine, that's the sign.
00:28
And then at pi for 6, this is 1, 2, square root of 3.
00:40
All right, so here you go.
00:42
First we're going to plug in pi for x, 9 for y.
00:45
So 9 equals c1, cosine 2 pi plus c2 sine 2 pi.
00:53
Oh, i forgot.
00:54
I'm going to have to multiply by 2.
00:56
Okay, so i'm interested in 2 pi, whose name is 1 ,0...