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First order ODE problem

  1. May 4, 2008 #1
    Let be the first order ODE's

    [tex] y'(x)g(x)=0 [/tex] and [tex] y'(x)g(x)=\delta (x-a) [/tex]

    except when x=a the two equations are equal , however the solutions are very different

    [tex] y(x)=C [/tex] and [tex] y(x)= C+ \int dx \frac{\delta (x-a)}{g(x)} [/tex]

    or using the properties of Dirac delta [tex] y(x)=C+\frac{1}{g(a)} [/tex]

    the second equation depends on the form of g(x) whereas the first does not, however except at the point x=a the 2 ODE's are completely equal.
     
  2. jcsd
  3. May 4, 2008 #2

    HallsofIvy

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    Why are you saying the two solutions are different? You should be writing C for one and, say, C' for the other- the two constants are not necessarily the same. In fact, all you are saying is that C= C'+ 1/g(a). Which is perfectly reasonable since 1/g(a) is itself a constant.
     
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