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A differential equation

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve:
    y'''=(y'')2

    2. Relevant equations



    3. The attempt at a solution

    Basically easy, use the substitution y'' = p and solve through the equations.
    I end up having this:
    y=-(x+c1)ln|x+c1| + c2x + c3, which is what I listed in the answers as well.
    My question is about p≠0 which I get while solving the first integral.
    p=y''=0 is a solution to the equation.
    y''=0 means y'=c means y=cx+c
    Is this included in the solution? I don't think so, and then it should be listed as a solution as well, but it's not. Am I missing something, or it's already in the first solution?
     
  2. jcsd
  3. Dec 6, 2009 #2

    LCKurtz

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    I don't see a choice of constants in your solution yielding y = cx + d. (You don't mean cx + c). But, yes, it is a perfectly good solution as you can test by plugging it into the equation. It should be listed as an answer. If you divided by p in the process of solving the equation, that would explain why the solution was missed. Good work noticing that.
     
  4. Dec 6, 2009 #3
    yes, I meant y = cx + d of course.
    Well, in that case that's another mistake in the answer sheet - for the DE:
    2y y'' = (y')^2 + 1
    they list y = c as an answer, when it is obviously not.
     
  5. Dec 6, 2009 #4

    LCKurtz

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    Yes. You're right about that too.
     
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