# Boundary Value Problem

1. Jul 23, 2011

### Precursor

The problem statement, all variables and given/known data
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?

2. Jul 23, 2011

### rock.freak667

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

3. Jul 23, 2011

### Bohrok

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

4. Jul 24, 2011

### 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.

5. Jul 24, 2011

### Precursor

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

6. Jul 24, 2011

### 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 "existance 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.