Can Initial Conditions y(0)=0 and y(2pi)=1 Be Satisfied for y''(x)+y(x)=0?

[rjscott1]

Homework Statement

Solve the differential equation
y''(x)+y(x)=0
y(0) = 0
y(2pi) = 1
y(pi)=?
2. The attempt at a solution
The solution seemed really obvious to me at first
Solving the characteristic equation
r^2+1=0
r = +/- sqrt(-1)
r = +/- i
so the solution should be given by:
y(x) = Asin(x) + Bcos(x)
replacing initial condition
y(0) = 0 we get B=0 , B=0
y(2pi) = Asin(2pi) = 1, -> this is never true
so to me this means that there is no solution. Did I make any mistakes or would that be the solution?
and as a consequence y(pi) has no solution..

Also sorry if this is poor formatting, this is my first post :D

 

[Mark44]

Homework Statement

Solve the differential equation
y''(x)+y(x)=0
y(0) = 0
y(2pi) = 1
y(pi)=?
2. The attempt at a solution
The solution seemed really obvious to me at first
Solving the characteristic equation
r^2+1=0
r = +/- sqrt(-1)
r = +/- i
so the solution should be given by:
y(x) = Asin(x) + Bcos(x)
replacing initial condition
y(0) = 0 we get B=0 , B=0
y(2pi) = Asin(2pi) = 1, -> this is never true
so to me this means that there is no solution. Did I make any mistakes or would that be the solution?
and as a consequence y(pi) has no solution..

Also sorry if this is poor formatting, this is my first post :D

Welcome to Physics Forums!

I agree that there is no solution to your boundary value problem. The solution to the differential equation is y = Asin(x) + Bcos(x), but there are no constants A and B so that y(0) = 0 and y(2π) = 1.