# Differential equation for air resistance

1. Sep 11, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Solve the differential equation $\displaystyle Cv^2 - mg = m\frac{d^2 y}{dt^2}$

2. Relevant equations

3. 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?

2. Sep 11, 2016

### Ray Vickson

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.

3. Sep 11, 2016

### Mr Davis 97

Actually, it does satisfy the original equation! So is my solution the correct one?

4. Sep 11, 2016

### Mr Davis 97

bump. I need a definitive answer

5. Sep 12, 2016

### Math_QED

If it satisfies the equation, of course it is.

6. Sep 12, 2016

### Ray Vickson

Cannot give you one until you say what YOU regard as a solution. I would personally regard a formula such as $v = f(t)$ or $t =h(v)$ or $y = F(t)$ or $t = H(y)$ as a solution, so that if I were given $t$ I could compute $v$ and/or $y$. So I myself would not say you were finished, but maybe your instructor has a different opinion.

However, your relationship between $y$ and $v$ MUST BE correct if it satisfies the DE.