Integrating the separable, first-order ordinary differential equation m d v d t = m g − a vI can't solve the very first problem from Slater &amp; Frank's book, and have no one to help me (I'm self-studying it in these vacations):1. A particle moves in a vertical line under the action of gravity and a viscous force (−av) where v is its velocity. Show that the velocity at any time is given by v = ( v 0 + m g a ) e − a m t − m g a .Show that this solution reduces [...].What I did is:The resultant force is m g − a v, and so by Newton's Second Law, m d v d t = m g − a v .Since m, a, and g are constants in this case, this is a first-order ODE, which also happens to be a separable one.We rewrite it as a relation between differentials: m d v m g − a v = d t .But I have no idea how to integrate the left side relative to v. Despite the fact that the exercise already gave the solution, v(t).Am I doing it right? Or maybe there is a simpler way using methods specials from mathematical mechanics?

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Integrating the separable, first-order ordinary differential equation   m            d      v              d      t        =  m  g  −  a  vI can&#039;t solve the very first problem from Slater &amp;amp; Frank&#039;s book, and have no one to help me (I&#039;m self-studying it in these vacations):1. A particle moves in a vertical line under the action of gravity and a viscous force (−av) where v is its velocity. Show that the velocity at any time is given by   v  =      (          v      0        +                  m        g            a        )        e          −              a        m            t        −            m      g        a    .Show that this solution reduces [...].What I did is:The resultant force is   m  g  −  a  v, and so by Newton&#039;s Second Law,  m            d      v              d      t        =  m  g  −  a  v  .Since m, a, and g are constants in this case, this is a first-order ODE, which also happens to be a separable one.We rewrite it as a relation between differentials:  m            d      v              m      g      −      a      v        =  d  t  .But I have no idea how to integrate the left side relative to v. Despite the fact that the exercise already gave the solution, v(t).Am I doing it right? Or maybe there is a simpler way using methods specials from mathematical mechanics?

Tania Alvarado · Accepted Answer

Hint The denominator of l.h.s. of the differential equation            m            d      v              m      g      −      a      v        =  d  tis linear in v, so this can be readily integrated. To see things a little more clearly, make the (linear) change of variables   u  =  m  g  −  a  v,   d  u  =  −  a    d  vThis gives            m            d      v              m      g      −      a      v        =  −      m    a              d      u        u    .As you probably recall,   ∫            d      u        u    =  log  ⁡      |    u      |    +  C

Integrating the separable, first-order ordinary differential equation m (dv)/(dt) = mg - av

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