Find the particular solution to y ' = 2sin(x) given the general solution is y = C - 2cos(x) and the initial condition y(pi/2) = 1

Answer 1

y' = 2sinx
y = c-2cosx
use the point they provided to find c:
1 = c - 2sin(π/2) = c-2
c = 3

so, y=3-2cosx

Answer 2

Steve is wrong, C=1
because sin(pi/2) is zero and anything times zero is zero, so the two goes away
1=C-2sin(pi/2)
1=C-2(0)
1=C-0
C=1

y=1-2cos(x)

Answer 3

idk but both are wrong.

Answer 4

We know that the general solution is y = C - 2cos(x), so now just plug the initial condition in!

The initial condition is y (pi/3) = 1. In other words, when you plug pi/3 into the equation, you’ll get 1. Therefore, the y is 1.

1 = C - 2cos(pi/3)

I’ll give you a hint: 2cos(pi/3) = 1

I’m going to make you do some math, but at this point, it’s just solving for C based on the formula I gave you (should take under 5 seconds). Then, plug your C back into y = C - 2cos(x)!