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Absolute values resulting in diff-eqs

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider a simple first-order linear differential equation, such as

    [tex]y' + \tan x y = 0[/tex]

    With initial condition y(0)=C for some constant C. Find all solutions y which satisfy this differential equation on the entire real line.

    2. Relevant equations

    General method for solving such an equation:
    Given
    [tex]y' + P(x)y = 0[/tex]
    and an initial condition y(a)=b for some constant b, the solution will be given by
    [tex]y=b \exp{\left(-\int_a^x P(t) dt\right)}[/tex]

    3. The attempt at a solution
    [tex]\int_0^x \tan x dx = -ln|\cos x|[/tex]
    so
    [tex] y=C \exp(ln|\cos x|) = C | \cos x | [/tex]

    However, the absolute value actually presents a problem. Leaving it in means that this is, in fact, not a solution for the differential equation, however nothing in the original question indicates that it should be removed.


    To quote from http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx" [Broken],
    A similar problem is present with something like this:

    [tex]y' - \frac {1}{x} y = 0[/tex]

    Wolfram Alpha has removed the absolute value signs on both of these, but they often remove absolute value signs when it is convenient in the answer without justification.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 24, 2011 #2

    LCKurtz

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    In your first example, you can't get a solution beyond (-π/2,π/2) because tan(x) isn't defined at the end points. And on that interval you can indeed drop the || signs because cos(x) is positive there.

    Similarly with 1/x in the equation, you aren't going to get a solution crossing the y axis. I would say you have to look at your particular equation and be careful about it.
     
  4. Sep 24, 2011 #3
    OK, sounds good. Thanks for the clarification.
     
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