Solving Newton's 2nd Law for Particle Falling in Medium

[TRB8985]

Homework Statement

Find the displacement and velocity of a particle undergoing vertical motion in a medium having a retarding force proportional to the velocity.

Relevant Equations

m*dv/dt = -mg - kmv

Good afternoon,

I have a question regarding this derivation that I'm covering in Thornton & Marion's "Classical Dynamics of Particles and Systems". In it, we're covering the most basic equation of motion for a particle falling in a medium.

I understand the process of starting with Newton's 2nd Law, using separation of variables, and ultimately solving for the velocity. The step right before my level of confusion was this one:

$$ -\int dt= \int \frac{dv}{kv+g}$$

This is what I wrote for the following step:

$$ -kt = ln(kv+g) + kc $$

But apparently, that's wrong. The constant c (with the k coefficient) should have been added on the left side of the time integral, like this:

$$ -kt + kc = ln(kv+g) $$

My question is: Why was the constant added on the time side? I'm used to solving differential equations where the constant is normally added on the x(t) or v(t) side of the equals sign, where the constant represents the initial amount of whatever is involved. There's another source I found online that does this same derivation that did things the same way, so there must be something fundamental involved there.Thank you for your time!

 

[etotheipi]

It doesn't matter, since it's just an as of yet undetermined constant of integration. You could just as easily transpose the $kc$ term to the other side of the equation, and define a new constant $d := -c$ so that now you have $kd$ on the RHS. You could also just change the $kc$ to another single arbitrary constant, say $A := kc$.

The point is that it's just an additive constant and it doesn't matter what you do with it, at least until you substitute in some boundary conditions that allow you to fix its value for certain.

 

[vanhees71]

Another trick in such cases of ODE's that can be solved by the "separation of variables" is to use definite integrals. The advantage is that you have the integration constants fixed such that the initial-value problem is automatically solved and that you get dimensionally sensible equations, because one should be very careful with a solution, where a dimensionful quantity occurs as the argument in a log function as in this example!

 

[etotheipi]

Another trick in such cases of ODE's that can be solved by the "separation of variables" is to use definite integrals. The advantage is that you have the integration constants fixed such that the initial-value problem is automatically solved and that you get dimensionally sensible equations, because one should be very careful with a solution, where a dimensionful quantity occurs as the argument in a log function as in this example!

This is a very good point. There is also some useful discussion here, specifically about how whenever we integrate to a logarithm a dimensional quantity like $\int \frac{dx}{x}$, we must explicitly include another arbitrary dimensional constant in the logarithm e.g. $\int \frac{dx}{x} = \ln \frac{x}{x_0}$, which is fine because $\frac{d}{dx} \ln \frac{x}{x_0} = \frac{x_0}{x} \frac{1}{x_0} = \frac{1}{x}$.

And once we tidy everything up, any lingering "dimensional" quantities inside the logarithm are really just the dimensionless coefficients of the unit, pretending to be a dimensional physical quantity

 

[haruspex]

A slightly different view maybe..
Adding it to the time side would be logical if there were some physical significance to the state in which the ln term vanishes, as in $k(t'-t)=\ln(...)$.
More likely, the state t=0 is interesting, so I would add it to the other side, but in the form of a ln: $kt=-\ln(\frac{kv+g}{kv_0+g})$.