Changing Units to Solve Differential Equation

Question

If this should be on SE Physics instead, let me know, and I'll post there instead.

I have a diff eq. of the form $\ddot x + \beta \dot x^2 = 0$

Maybe there is a simple analytic solution, but I haven't been able to find it (can't use the method for Bernoulli diff eqs because $P(x) = 0 $). If it is the case that there is no analytic solution, this makes this problem even more interesting.

$\beta > 0$ here is a factor of units $[m^{-1}]$, and it looks something like $\beta = c\frac{A}{V}$, where $c$ is a dimensionless physical constant, $A$ is area, $V$ is volume.

Now, there is a simple solution to $\ddot x + \dot x ^2 = 0$, namely $$x = ln(t + c_1)+c_2$$ (I did not solve this myself and I'm not sure how I would, but it's easy to check that it works. I got this from Wolfram Alpha, which did not give me an analytic solution for the original problem*).

Now, depending on the object's actual size and shape, it should always (or at least in most cases) be possible to come up with a unit of distance such that measuring in this system gives $\frac{A}{V} = c^{-1}$.

If we let $x(0) = 0,\ \dot x(0) = v_0$, we get: $$c_1 = v_0^{-1}$$ $$c_2= ln(v_0)$$

Now if we convert from, say meters, to our new units, write that in to our $v_0$ and then convert our general result back to meters, it seems we've solved this problem.

So I have 2 questions:

1) Is this approach valid? The results I get don't make physical sense, but I don't see any flaws in the argument (there may be problems with my physical assumptions).

2) If this is a valid approach, does it reduce to any other method?Is it somehow equivalent to other approaches? It feels like something essentially different in some way. If it does have unique power, where does that power come from? It feels sort of like black magic, getting something out of nothing.

One other thing: if I'm totally wrong, and there is an easy way to solve this, I'd appreciate any approach. Also, if that's the case, are there any situations where this trick allows us to solve things we couldn't otherwise?

Answer 1

I'm gonna make an attempt but it was long time ago I solved these on paper.

You can solve by rewriting $y''=-by'^2$ and then taking logarithms: $\log(y'')=\log(-b)+2\log(y')$ and then substitute $y'=\exp(g)$ giving:

$$y''=/\text{ chain rule }/=g'\exp(g)$$ inserting $$\log(g')+g=\log(-b)+2g$$ simplifying $$g'=-b\exp(g)$$ which has solution $$g=-\log(bt+c_1)$$

now what remains is to substitute back from $g$ and find $y'(t)$ and from there $y(t)$.

You can use wolfram alpha to verify the solution: our solution, Wolfram's solution

The only thing differing is the name of the constants

1. Wolfram has $c_1,c_2$
2. We have $c,k_1$

Answer 2

Using @user108128's hint $$x''+\beta (x')^2=0 \implies y'+\beta y^2=0\implies y=\frac{1}{\beta t+c_1}$$ Then $$x'=\frac{1}{\beta t+c_1}\implies x=\frac 1 \beta \log(\beta t+c_1)+c_2$$

So, using the conditions at $t=0$ $$0=\frac 1 \beta \log(c_1)+c_2$$ $$v_0=\frac 1 {c_1}$$ then $$c_1=\frac 1 {v_0}\qquad \text{and} \qquad c_2=\frac 1 \beta \log(v_0)$$