# Free fall with resistance: solution to the ODE

I'm having trouble solving this ODE:

$$\ddot x = \mu \dot x^2 - g, \space \space x(0)=x_0$$

This is the ODE that determines the equation of motion of an object with air resistance. $\mu$ is a positive constant. This is what I've done:

Let $\dot x = v$ we have $v(0) = 0$ by assumption and the equation becomes:

$$\dot v = \mu v^2 - g$$

This is a Riccati differential equation. Recall that a Riccati differential equation has the form

$$\dot x = h(t) + f(t) x + g(t)x^2$$

In our case is $h(t) = -g$, $f(t) = 0$ and $g(t) = \mu$

To solve this kind of differential equations one has to guess a solution, which in this case I found be:

$$v_p(t) = \sqrt{g \over \mu}$$

$\implies y = {1 \over {v- \sqrt{g \over n}}}$ $\implies y(0) = \sqrt{n\over g}$ to find the solution $v(t)$ one has to solve the following differential equation:

$$\dot y = -(f(t) + 2v_pg(t))y-g(t)$$

This is what I've done:

\begin{align} \dot y & = -(f(t) + 2v_pg(t))y-g(t) \\ & \dot y = -2 \sqrt{g\mu }\space y - \mu \space & (1) \end{align}

Now this is a linear inhomogeneous differential equation an the solution is given by the sum of the homogeneous solution $y_h(t)$ with the particular solution $y_p (t)$ The homogeneous solution comes from:

$$\dot y = -2 \sqrt{g\mu }\space y \implies y_h = y(0) e^{-2 \sqrt{g\mu }}$$

for the particular solution $y_p(t)$ is a little bit more complicated then to find it I substituted

$$y_p(t) = y(0)(t)e^{-2 \sqrt{g\mu }} = C_1 (t)e^{-2 \sqrt{g\mu }}$$

in the equation (1). This is what I get:

\begin{align} y_p(t)& =C_1(t)' e^{-2 \sqrt{g\mu }} -2 \sqrt{g\mu }C_1(t)e^{-2 \sqrt{g\mu }} = -2 \sqrt{g\mu }C_1(t)e^{-2 \sqrt{g\mu }} - \mu \end{align} \begin{align} \iff C_1(t)' e^{-2 \sqrt{g\mu }} & = -\mu \end{align}

$$\iff C_1(t) = {-\mu \over {2 \sqrt{g\mu }}}e^{2 \sqrt{g\mu }}$$

summarizing we have

$$y(t) = {1 \over v - \sqrt {g \over n}} = C_1(t)e^{-2 \sqrt{g\mu }} + y_h(t)= {-\mu \over {2 \sqrt{g\mu }}}e^{2 \sqrt{g\mu }}e^{-2 \sqrt{g\mu }} + y(0) e^{-2 \sqrt{g\mu }}$$

$$= {-\mu \over {2 \sqrt{g\mu }}} + y(0) e^{-2 \sqrt{g\mu }}$$

from here I tried to solve $y = {1 \over {v- \sqrt{g \over n}}}$ with respect to $v$ and then solve the last ODE $v = \dot x$ but I think that something has gone wrong.

Is there an alternative way to solve the original ODE? Is the method I've used the only one?

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Note that you can choose the initial velocity to be $0$, but it does not necessarily have to be. If your line $\dot{x}(0)=\dot{x}_0=0$ is meant as a justification, it is wrong like saying $f(0)=c\Rightarrow f'(0)=\frac{d}{dt}c=0.$ As your ODE is second order, you are free to choose both $x(0)=x_0$ and $\dot{x}(0)=v_0$. – flonk Mar 5 '14 at 13:49
Your are right, sorry, I'll edit it right away. However, the initial condition is v(0) = 0 – Bman72 Mar 5 '14 at 13:52

As you asked for methods I decided to post an alternative. Here we use the trick of eliminating the inhomogeneity.

First, consider only the equation for $v(t)=\dot{x}(t)$,

$$\dot{v}=\mu v^2-g.$$ The inhomogeneity $-g$ prevents us from simple integration of the ODE, so let's make it disappear by defining $w(t)=v(t)-\gamma$, which leads to

$$\dot{w}=\mu w^2+2\mu\gamma w+\mu\gamma^2-g.$$ We see that the choice $\gamma=\sqrt{g/\mu}$ kills the constant term, leadig to the homogenous equation for $w$: $$\frac{dw}{dt}=\mu w^2+2\sqrt{\mu g}w,$$ which now can directly be integrated:

$$\int_{w_0}^{w(t)} \frac{dw}{\mu w^2+2\sqrt{\mu g}w}=\int_0 ^t dt.$$ This can be solved using $\int dt/(at^2+bt)=-(2/b)\text{artanh} (2at/b+1)$ (for $b>0$) which gives

$$\frac{-1}{\sqrt{\mu g}}\text{artanh}\left( \mu w+1 \right)-C=t$$ with some integration constant $C$. We can solve this for $w$ and obtain $$w(t)=\sqrt{\frac{g}{\mu}}\left( \tanh[-\sqrt{\mu g}(t+C)]-1 \right).$$ Transforming back to our original function $v$ gives $$v(t)=w(t)+\gamma=\sqrt{\frac{g}{\mu}} \tanh[-\sqrt{\mu g}(t+C)],$$ and from the initial condition $v(0)=0$ it follows that $C=0$.

Now you can integrate $v=\dot{x}$ to get the position solution using that $\int dt \tanh(t)=\ln (\cosh x)$ arriving at

$$x(t)=x_0-\frac{1}{\mu}\ln\left[ \cosh(-\sqrt{\mu g}t) \right],$$

where the integration constant has been chosen such that $x(0)=x_0$.

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+1 I like this approach, as it is more succinct. I will check my method to see how the two answers can be resolved. – Chinny84 Mar 5 '14 at 16:06
In particular, one does not have to guess a solution to the differential equation, as the original poster suggested. It can be integrated as shown here. – Jonathan Mar 5 '14 at 16:12
@Ale Your last assumtion is right, I have put all constants into one $C$ to make life easier. You just need to remember that this $C$ determines the initial condition $w_0$ (or $v_0$). – flonk Mar 6 '14 at 8:57
@Ale Just to be precise, note that if $F$ is the antiderivative function, your equation after integrating is $F(w(t))-F(w_0)=t-t_0$, so you have two free constants, the starting time $t_0$ and your initial condition $w_0$. According to your question, $t_0=0$ is already chosen, so in the equation of your last comment, you should not have two constants, your $C$ corresponds to the already chosen $t_0$. – flonk Mar 6 '14 at 9:01
@Ale I can't tell how often the method will work, but it's always a good idea to remember new tricks and keep them in your treasure chest for the rest of your life :) – flonk Mar 6 '14 at 10:26

Use the fact $$\ddot{x} = \dot{x}\frac{d}{dx}\dot{x} = \frac{d}{dx}\frac{\dot{x}^{2}}{2}$$ then the equation becomes $$\frac{d}{dx}\frac{v^{2}}{2} = \mu v^{2} - g$$ Then you can solve directly to get $$v^{2}\mathrm{e}^{-2\mu x} = \int 2g \mathrm{e}^{-2\mu x} dx + C_{1} = -\frac{g}{\mu}\mathrm{e}^{-2\mu x} + C_{1}$$ if $\dot{x}(0)=x(0)=0$ then $$C_{1} = \frac{g}{\mu}$$ Hence $$v =\sqrt{\frac{g}{\mu}}\sqrt{\mathrm{e}^{2\mu x} - 1}$$ then $$\frac{dx}{dt} = \sqrt{\frac{g}{\mu}}\sqrt{\mathrm{e}^{2\mu x} - 1}$$ this leads $$\int \frac{1}{\sqrt{\mathrm{e}^{2\mu x} - 1} }dx = \sqrt{\frac{g}{\mu}}t + C_{1}$$ solving the right hand side yields $$\frac{1}{\mu}\arctan{\sqrt{\mathrm{e}^{2\mu x} - 1} } = \sqrt{\frac{g}{\mu}}t + C_{1},\\ \sqrt{\mathrm{e}^{2\mu x} - 1} = \tan{\left(\sqrt{\mu g}t + C_{2}\right)}$$ and then finally $$x(t) = \frac{1}{\mu}\ln{\left[\sec{\left(\sqrt{\mu g}t + C_{2}\right)}\right]}$$

Check it out if it works out correctly. I did this without pen and paper..but the method holds (I think).

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How is $\ddot{x} = \dot{x}\frac{d}{dx}\dot{x}$? First I guess you mean $\ddot{x} = \dot{x}\frac{d}{dt}\dot{x}$, but still I think it's wrong. – flonk Mar 5 '14 at 13:57
@flonk $\frac{dx}{dt}\frac{du}{dx}=\frac{du}{dt}$ – Chinny84 Mar 5 '14 at 14:04
Ah I see, thanks! – flonk Mar 5 '14 at 14:08
I didn't find the mistake yet, but your solution $x(t)$ is periodic in time, whereas the particle should fall down monotonously $\to-\infty$. – flonk Mar 5 '14 at 16:01
@Jonathan Oh thanks. However, aren't monotonic functions quite boring? – flonk Mar 5 '14 at 16:18