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Can't solve this ODE

  1. Oct 7, 2009 #1
    I can't seem to find anywhere how to find the exact solution to the equation when the non-homogeneous part is just 1 and not a function. I would very much appreciate it if anyone could give me some assistance. Also for the bounday counditions doesn't one of them need to be phi prime?
  2. jcsd
  3. Oct 7, 2009 #2


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    substitute your solution into

    otherwise change variable y=u+1

    and 1 is a function f(x)=1 so you could just solve
  4. Oct 7, 2009 #3


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    If you're given phi and the derivative of phi at 0, that's called initial conditions, since it describes the position and velocity of your particle (historically you're interested in this for physics reasons). If you're given phi at two points, that's called boundary conditions because you know the position at the beginning and at the end.

    Either way you get two equation with which to solve your unknowns, so either way is fine (although for certain cases boundary conditions are a little bit weaker, but that shouldn't be a problem here)
  5. Oct 8, 2009 #4


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    There is a crucial difference between "initial value" problems and "boundary value" problems. The existence and uniqueness of a solution to an initial value problem depends entirely upon the equation, not the initial conditions but the existence and uniqueness of a solution to a boundary value problem may depend upon the boundary conditions as well.

    For example, y"+ y= 0 y(0)= A, y'(0)= B has a unique solution for all A and B. But y"+ y= 0, y(0)= 0, y([itex]\pi[/itex])= 1 has no solution while y"+ y= 0, y(0)= 0, y([itex]\pi[/itex])= 0 has an infinite number of solutions.

    rugabug, the simplest way to solve your problem is to first find the solution to the associated homogeneous equation (I imagine you have already done that), then "try" an undetermined constant, A as a specific solution to the entire equation: If y= A, y'= 0 and y"= 0 so the equation becomes A= 1. Just add 1 to your general solution to the homogeneous equation.
  6. Oct 8, 2009 #5
    Thank you so very much for the help.
    I first found the complementary solution then add it to the particular solution and I am now solving for the constants.
    It's all starting to come back to me. It's been a handful of years since I took diff eq.
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