# Differential equation for air resistance

## Homework Statement

Solve the differential equation ##\displaystyle Cv^2 - mg = m\frac{d^2 y}{dt^2}##

## The Attempt at a Solution

The problem is nonlinear, so we need to use unconventional methods. Specifically, if we can express the derivative of y with respect to v, then we might be able to integrate in order to find y.

So ##\displaystyle \frac{dy}{dv} = \frac{dy}{dt}\frac{dt}{dv} = v \frac{dt}{dv} = \frac{v}{\frac{dv}{dt}}##

But ##\displaystyle \frac{dv}{dt}## is given by ##\displaystyle \frac{C}{m}v^2 - g##, so
##\displaystyle \frac{dy}{dv} = \frac{mv}{Cv^2 - mg}##

If we solve this, we get ##\displaystyle y = \frac{m}{2c} \ln{|1 - \frac{Cv^2}{mg}|}## where ##V_0 = 0##.

Is this the correct solution?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Solve the differential equation ##\displaystyle Cv^2 - mg = m\frac{d^2 y}{dt^2}##

## The Attempt at a Solution

The problem is nonlinear, so we need to use unconventional methods. Specifically, if we can express the derivative of y with respect to v, then we might be able to integrate in order to find y.

So ##\displaystyle \frac{dy}{dv} = \frac{dy}{dt}\frac{dt}{dv} = v \frac{dt}{dv} = \frac{v}{\frac{dv}{dt}}##

But ##\displaystyle \frac{dv}{dt}## is given by ##\displaystyle \frac{C}{m}v^2 - g##, so
##\displaystyle \frac{dy}{dv} = \frac{mv}{Cv^2 - mg}##

If we solve this, we get ##\displaystyle y = \frac{m}{2c} \ln{|1 - \frac{Cv^2}{mg}|}## where ##V_0 = 0##.

Is this the correct solution?

Check this solution by taking the derivative and seeing if the differential equation is satisfied. That is something you should always do, whenever it is possible.

Check this solution by taking the derivative and seeing if the differential equation is satisfied. That is something you should always do, whenever it is possible.
Actually, it does satisfy the original equation! So is my solution the correct one?

bump. I need a definitive answer

member 587159
bump. I need a definitive answer

If it satisfies the equation, of course it is.

Ray Vickson