Originally Posted by

**AAKhan07** Hi, the two forces acting on an object are its weight and air resistance. In my differential equations coursework I begin with 'air resistance = -kv' where v is the velocity and k is a constant. This is easy enough to solve.

$\displaystyle

m\frac{d^2s}{dt^2} = mg - kv

$

$\displaystyle

m\frac{d^2s}{dt^2} + kv = mg

$

$\displaystyle

m\frac{d^2s}{dt^2} + k\frac{ds}{dt} = mg

$

Findgin auxiliary equation

$\displaystyle

m\lambda^2 +k\lambda = 0

$

$\displaystyle

\lambda(m\lambda + k) = 0

$

$\displaystyle

\lambda = 0 or \lambda = \frac{-k}{m}

$

$\displaystyle

\lambda = \frac{-k}{m}

$

$\displaystyle

s=A+Be^\frac{-kt}{m}

$

Finding particular integral

$\displaystyle

s=at

$

$\displaystyle

\frac{ds}{dt} =a

$

$\displaystyle

\frac{d^2s}{dt^2} =0

$

$\displaystyle

m(0) + k(a) = mg

$

$\displaystyle

a = \frac{mg}{k}

$

$\displaystyle

s= \frac{mg}{k}t

$

General Solution

$\displaystyle

s = A+Be^\frac{-kt}{m} + \frac{mg}{k}t

$

$\displaystyle

s=0

$

when

$\displaystyle

t=0

$

$\displaystyle

s = A+Be^\frac{-k(0)}{m} + \frac{mg}{k}(0)

$

$\displaystyle

0 = A+B

$

$\displaystyle

A = -B

$

Therefore

$\displaystyle

s = -B+Be^\frac{-kt}{m} + \frac{mg}{k}t

$

$\displaystyle

\frac{ds}{dt}=0

$

when

$\displaystyle

t=0

$

$\displaystyle

\frac{ds}{dt}=\frac{-kB}{m}e^\frac{-kt}{m} + \frac{mg}{k}

$

$\displaystyle

0=\frac{-kB}{m} + \frac{mg}{k}

$

$\displaystyle

B=\frac{m^2g}{k^2}

$

So the equation is

$\displaystyle

s = -\frac{m^2g}{k^2}+\frac{m^2g}{k^2}e^\frac{-kt}{m} + \frac{mg}{k}t

$

I would appreciate if you could first check the correctness of the above working.

My main question is: how do you find the differential equation when the air resistance = -kv^2

This time the air resistance is dependent on the square of the speed rather than the speed. Newton's second law becomes:

$\displaystyle

m\frac{d^2s}{dt^2} = mg - kv^2

$

I am looking to solve this for s as I did for the other equations.