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Solving a special type of a second order differential equation

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data

    y''[x] = y'[x] + x

    2. Relevant equations
    We were taught two special types of second order diff. equations:

    Type 1: Supposed to be when x is missing
    v = y'[x]
    v'[x] = y''[x]

    Type 2: Supposed to be when y is missing
    v = y'[x]
    v v'[x] = y''[x]

    3. The attempt at a solution
    The answer key reads:

    v'[x] - v = x

    d/dx{e^-x v} = x e^-x

    e^-x v = C1 * e^-x -x -1

    v = C1 * e^x -x -1

    y[x] = c2 - c1 * e^-x - (.5)x^2 - x

    I don't understand why we used type 1 to solve this problem since x is clearly stated in the problem. I was hoping someone could explain, thanks.
  2. jcsd
  3. Feb 18, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Are you sure you have listed the two types correctly? Usually, if y is missing you would do:

    v = y'
    v' = y''

    so your second order equation in y becomes a first order equation in v. That is the type of equation you have and also the method your manual gives.
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