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Homework Help: Quick diff eq question

  1. Jan 22, 2005 #1
    I'm pretty sure I know the answer to this, just want to double check.

    For the problems that I'm currently working on, we are just solving the problems for an unknown constant C.

    I just finished one were I came up with
    [tex]\frac{x^2}{2}-y^2cos x-xy^3=C[/tex]

    The book shows the solution as
    [tex]y^2cos x+xy^3-\frac{x^2}{2}=C[/tex]

    Because C is an arbitrary, unknown constant, did they just multiply both sides by -1? Is there some reason for doing this other then getting rid of 2 negatives in the answer?
  2. jcsd
  3. Jan 22, 2005 #2


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    Homework Helper

    It's irrelevant which solution u chose,u can verify that both satisfy the same diff.eq.,the one which u were supposed to solve.In general,a smaller number of minuses is preferable.Par éxample:
    [tex] x+y=4 [/tex]
    would u like it more than
    [tex] -x-y=-4 [/tex]


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