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Boundary Value Problem

  1. Jul 23, 2011 #1
    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. jcsd
  3. Jul 23, 2011 #2


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    Homework Helper

    I would agree with that since both do not yield the same value for c2.
  4. Jul 23, 2011 #3
    Maybe one of the initial conditions was missing a prime after the y?
  5. Jul 24, 2011 #4
    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[itex]\prod[/itex]?
    Technically, as given it's an invalid problem.
  6. Jul 24, 2011 #5
    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).
  7. Jul 24, 2011 #6


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    Science Advisor

    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, [itex]y(2\pi)= 0[/itex] would have [itex]y(x)= Csin(x)[/itex] 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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