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A doubt in the DE

  1. Sep 18, 2013 #1
    I am still in the first few pages of my differential course. I just passed through the term equilibrium solution. Apparently, it meant for me as when the solution of the DE is zero. For example, dv/dt= 9.8-(v/5) ... has an equilibrium solution of 49 where dv.dt is actually zero. Is this the case or not??
  2. jcsd
  3. Sep 18, 2013 #2


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    Homework Helper

    An equilibrium solution is a solution which is a constant. Since the derivative of a constant is zero, a constant solution [itex]v(t) = v_0[/itex] of [itex]\dot v(t) = f(v(t))[/itex] must satisfy [itex]f(v_0) = 0[/itex].

    Thus any equilibrium solution of [itex]\dot v = 9.8 - \frac15 v[/itex] must satisfy [itex]9.8 - \frac15 v = 0[/itex], ie. [itex]v(t) = 49[/itex].
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