Absolute values in logarithms in a solution of differential equationHow have the moduli signs disappeared in the following step: 1 k ( ln ⁡ | g + k v | − ln ⁡ | g + k u | ) = − tTherefore ln ⁡ ( g + k v g + k u ) = − k t g, k and u are positive constants. t is time, v is velocity.Context: the above calculations are from solving the equation d v / d t = − g − k v given that v = u when t = 0, and that u, g and k are positive constants.

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Absolute values in logarithms in a solution of differential equationHow have the moduli signs disappeared in the following step:      1    k        (    ln    ⁡          |        g    +    k    v          |        −    ln    ⁡          |        g    +    k    u          |        )    =  −  tTherefore  ln  ⁡      (                  g        +        k        v                    g        +        k        u              )    =  −  k  t  g,   k and   u are positive constants.   t is time,   v is velocity.Context: the above calculations are from solving the equation   d  v      /    d  t  =  −  g  −  k  v given that   v  =  u when   t  =  0, and that   u,   g and   k are positive constants.

gutinyalk · Accepted Answer

g  +  k  v need not be positive, but it will have the same sign as   g  +  k  u, because the solutions cannot cross the equilibrium at   v  =  −  g      /    k. Hence, the quotient inside the logarithm is positive.Ideally, you would not arrive at       1    k        (    ln    ⁡          |        g    +    k    v          |        −    ln    ⁡          |        g    +    k    u          |        )    =  −  t at all; there is a cleaner way of solving this ODE. Namely, one goes from   ln  ⁡      |    g  +  k  v      |    =  −  k  t  +  C (with indefinite constant   C) to   g  +  k  v  =  A      e          −      k      t      , where   A takes the place of   ±      e    C

rzfansubs87 · Accepted Answer

If   k,  g,  u are all positive, you need to have   v large enough (i.e., not too negative) to make   g  +  k  v  &amp;gt;  0. Maybe that results from some part that you haven&#039;t told us...

Absolute values in logarithms in a solution of differential equation How have the moduli signs disa

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