free falling with resistance

[waytogo]

I reviewed the classical problem about an object falling through some medium. Resistance is propotional to the square of speed.
Corresponding differential equation is as follows

\(\displaystyle \frac{dv}{dt}=-mg+k v^{2} \), with k=2, m=10, g=9.8.

In some step of solution
\(\displaystyle \frac{|v+7|}{|v-7|}=ce^{\frac{14}{5}t}\)
is obtained and it is written that left side of equation cannot change sign because of some existence and uniqueness theorem, so modulus can be given up.

Can somebody explain this step?

 

[HallsofIvy]

I reviewed the classical problem about an object falling through some medium. Resistance is propotional to the square of speed.
Corresponding differential equation is as follows

\(\displaystyle \frac{dv}{dt}=-mg+k v^{2} \), with k=2, m=10, g=9.8.

In some step of solution
\(\displaystyle \frac{|v+7|}{|v-7|}=ce^{\frac{14}{5}t}\)
is obtained and it is written that left side of equation cannot change sign because of some existence and uniqueness theorem, so modulus can be given up.

Can somebody explain this step?

With k= 2, m= 10, g= 9.8, the equation is dv/dt= 2v^2- 98= 2(v^2- 49)= 2(v- 7)(v+ 7). Now, the initial value of v is either less than 7, equal to 7, or greater than 7.

If the initial value of v is equal to 7, dv/dt= 0 and v is always equal to 7. (Note that \(\displaystyle \frac{|v+ 7|}{|v- 7|}= ce^{\frac{14}{5}t}\) is NOT the general solution because it does not include this solution.)

If the initial value of v is greater than 7, dv/dt is positive so v is increasing and stays greater than 7.

If the initial value of v is less than 7, dv/dt is negative so v is decreasing and stays less than 7.