Solving a Boundary Value Problem: y + y = 0 ; 0<x<2π, y(0)=0 , y(2π)=1

[Precursor]

Homework Statement
Determine all the solutions, if any, to the given boundary value problem by first finding a general solution to the differential equation:

y" + y = 0 ; 0<x<2π
y(0)=0 , y(2π)=1


The attempt at a solution

So the general solution is given by: y = c₁sin(x) + c₂cos(x)

Substituting in the boundary conditions we get:

y(0)=0=c₂ ==> c₂=0
y(2π)=1=c₂ ==> c₂=1

Since the above is contradictory, does it mean that there are no solutions to this boundary value problem?

 

[rock.freak667]

I would agree with that since both do not yield the same value for c₂.

 

[Bohrok]

Maybe one of the initial conditions was missing a prime after the y?

 

[nickalh]

I'm betting Bohrok is right. I always thought it was odd, but nearly every initial value problem I see specifies, y' and y. Rarely or ever, two points, y(a)=c, y(b)=d.

Minor clarification:
Should it be 0<= x <= 2\prod?
Technically, as given it's an invalid problem.

 

[Precursor]

The way I've written the problem is exactly how it's given in the textbook (Fundamentals of Differential Equations, 7^th edition, section 10.2, question 6).

 

[HallsofIvy]

The problem specifically said "Determine all the solutions, if any, to the given boundary value problem". Being given y(0) and y'(0) would be an initial value problem, not a boundary value problem. The correct "answer" here is that there is no solution.

Note, by the way that the same d.e. with boundary condition y(0)= 0, y(2\pi)= 0 would have y(x)= Csin(x) for any C.

The "existence and uniqueness" of solutions to an initial value problem depend only on the equation. For a boundary value problem, they depend on the boundary values also.