Araceli Clay

2022-07-03

Absolute values in logarithms in a solution of differential equation
How have the moduli signs disappeared in the following step:
$\frac{1}{k}\left(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|\right)=-t$
Therefore
$\mathrm{ln}\left(\frac{g+kv}{g+ku}\right)=-kt$
$g$, $k$ and $u$ are positive constants. $t$ is time, $v$ is velocity.
Context: the above calculations are from solving the equation $dv/dt=-g-kv$ given that $v=u$ when $t=0$, and that $u$, $g$ and $k$ are positive constants.

gutinyalk

Expert

$g+kv$ need not be positive, but it will have the same sign as $g+ku$, 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 $\frac{1}{k}\left(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|\right)=-t$ at all; there is a cleaner way of solving this ODE. Namely, one goes from $\mathrm{ln}|g+kv|=-kt+C$ (with indefinite constant $C$) to $g+kv=A{e}^{-kt}$, where $A$ takes the place of $±{e}^{C}$

rzfansubs87

Expert

If $k$,$g$,$u$
are all positive, you need to have $v$ large enough (i.e., not too negative) to make $g+kv>0$. Maybe that results from some part that you haven't told us...

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