How Does a Nonlinear Retarding Force Affect Particle Movement in a Medium?

[Mindscrape]

Homework Statement

A particle moves in a medium under the influence of a retarding force equal to $$mk(v^3 + a^2 v)$$, where k and a are constants. Show that for any value of the initial speed the particle will never move a distance greater than pi/2ka and that the particle comes to rest only for $$t \rightarrow \infty$$

Homework Equations

Legrangian seems overkill, so I used Newton's.

The Attempt at a Solution

$$\frac{dv}{dt} = k (v^3 + a^2 v)$$

then separate the ODE and integrate

$$\frac{lnv}{a^2} - \frac{ln(a^2 + v^2)}{2a^2} = kt + C$$

multiply by 2a^2

$$2lnv - ln(a^2 + v^2) = 2kta^2 + C$$

use log properties and combine the natural logs

$$ln ( \frac{v^2}{a^2 + v^2} ) = 2kta^2 + C$$

exponentiate and carry constant down

$$\frac{v^2}{a^2 + v^2} = Ce^{2kta^2}$$

add and subtract a^2 in numerator to simplify, and then subtract the one

$$\frac{-a^2}{v^2+a^2} = Ce^{2kta^2}$$

use algebra to isolate v

$$v^2 = a^2 (1 - \frac{1}{1-Ce^{2kta^2}})$$

Now I am at the point where I can solve for the integration constant, but I don't see it giving me anything close to revealing a max distance of pi/2ka. Also, in my solution as t approaches infinity v = a, and not zero. Maybe I made an algebraic mistake?

 

[dextercioby]

The error is clearly in the integration of the ODE.

 

[HallsofIvy]

Presumably, though you didn't show it, you used partial fractions to write
$$\frac{dx}{k(v^3+ a^2v)}= \frac{A}{v}+ \frac{Bv+C}{v^2+ a^2}$$

How did you integrate $$\frac{Bv+C}{v^2+ a^2}$$?

 

[Mindscrape]

I used mathematica, though partial fractions is the obviously the way to do it by hand.

 

[AlephZero]

The error is clearly in the integration of the ODE.

The first error is integrating the wrong ODE.

$$\frac{dv}{dt} = - k (v^3 + a^2 v)$$ would be better, since the form of the answer implies k is positive.

 

[Mindscrape]

The first error is integrating the wrong ODE.

$$\frac{dv}{dt} = - k (v^3 + a^2 v)$$ would be better, since the form of the answer implies k is positive.

Yep, that was actually the problem. I went through it again and figured out what I did wrong. You end up with v = +/- a/sqrt(-1-Ce^(2kta^2).