Araceli Clay

Answered

2022-07-03

Absolute values in logarithms in a solution of differential equation

How have the moduli signs disappeared in the following step:

$\frac{1}{k}(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|)=-t$

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.

How have the moduli signs disappeared in the following step:

$\frac{1}{k}(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|)=-t$

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.

Answer & Explanation

gutinyalk

Expert

2022-07-04Added 11 answers

$g+kv$ 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 $\frac{1}{k}(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|)=-t$ 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 $\pm {e}^{C}$

Ideally, you would not arrive at $\frac{1}{k}(\mathrm{ln}|g+kv|-\mathrm{ln}|g+ku|)=-t$ 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 $\pm {e}^{C}$

rzfansubs87

Expert

2022-07-05Added 5 answers

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...

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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