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The ratio of two total differentials equals the dervative? Why?

  1. Jan 30, 2010 #1
    I was stuck on a problem in my Diff Eq book and I asked my teacher for some guidance. This was the problem:

    Let M dx + N dy = 0 be a homogeneous diff eq. Using the substitution y = r sin(t), x = r cos(t), show that this equation is a separable equation of r and t.

    Now it was obvious that we would have dy/dx = F(y/x) = F(tan(t)). But I couldn't see how to get dy/dx into a separable equation of r and t. I asked my teacher about this and he didn't bother actually telling me what was going on but just very vaguely mentioned the total differential was used. So tried it out and got

    dy = sin(t) dr + r cos(t) dt
    dx = cos(t) dr - r sin(t) dt

    I then thought, well maybe I can take the ratio of these two equations to get dy/dx, which got me

    dy/dx = (sin(t) dr + r cos(t) dt) / (cos(t) dr - r sin(t) dt) = F(tan(t))

    And from there I saw that it was separable. I wasn't 100% sure of my answer so I checked the student solutions and sure enough that was how it was done.

    So my question is, why is this allowed? I can see how the limit {h -> 0} f(x + h)/h = df(x)/dx could somehow imply to someone that dx is a piece of notation to denote this quantity h -> 0 in the same way df(x) is represents f(x + h) for h -> 0. But is there a formal proof that shows this? Are there hypothesis that must be satisfied or is it true for all such continuous functions? I suppose my curiosity in this means I'm ready for some analysis text book that shows why its true, but a quick informal proof would at least calm me down for now :)
     
  2. jcsd
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