Differential equation for air resistance

[Mr Davis 97]

Homework Statement

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

Homework Equations

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 Statement

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

Homework Equations

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.

 

[Mr Davis 97]

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?

 

[Mr Davis 97]

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]

bump. I need a definitive answer

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.