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
idk but both are wrong.
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)!
To find the particular solution to the given differential equation, we need to substitute the general solution into the differential equation and then solve for the constant C using the initial condition.
Given general solution: y = C - 2cos(x)
Differential equation: y' = 2sin(x)
Substituting the general solution into the differential equation:
(C - 2cos(x))' = 2sin(x)
Now, differentiate C and -2cos(x) separately:
0 - (-2sin(x)) = 2sin(x)
Simplifying the equation:
2sin(x) = 2sin(x)
This equation is true for all values of x, so there are no restrictions on C.
Now, we need to find the value of C that satisfies the initial condition y(pi/2) = 1.
Substitute x = pi/2 and y = 1 into the general solution:
1 = C - 2cos(pi/2)
1 = C - 2(0)
1 = C
Therefore, the constant C = 1.
Substitute this value of C back into the general solution:
y = 1 - 2cos(x)
This is the particular solution to the given differential equation with the initial condition y(pi/2) = 1.
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)
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