# 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

Staff Emeritus
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.

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