How to solve the equation $k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$ I've been struggling a little on this problem for a while. I'm not sure what I'm missing. It goes like this:
$$k = \frac{dv}{dt}(a + 2\int_0^t v(\tau) d\tau)^2$$
Here $k, a$ are constants and $v, t$ represent velocity and time respectively. I want to know $v$ as a function of $t$. Is it possible to solve this? Any help would be appreciated!
 A: Set $V(t)=\int_0^tv(s)ds$. Then your equation reads as
$$
k=V''(a+2V)^2.
$$
this can be separated and integrated once
$$
V''=\frac{k}{(a+2V)^2}\implies V'^2=c-\frac{k}{a+2V}.
$$
In principle you can now apply separation again, the resulting integral with square root might be difficult or impossible to solve.
A: If $v$ is velocity and is given by $v=\frac{ds}{dt}$ then
$$\int_0^tv(\tau)d\tau=\int_0^t\frac{ds}{d\tau}d\tau$$
By the fundamental theorem of calculus and Leibniz's rule:
$$\int_0^t\frac{ds}{dt}d\tau=\frac{d}{dt}\int_0^tsd\tau=\frac{d}{dt}st=s$$
Note: We are only allowed to do this because these specific limits of integration(zero at the left part, variable/parameter on the right part)
Substituting back in the original equation:
Which is 1st order in $v$
$$k=\frac{dv}{dt}(a+2s)^2$$
But because $v$ depends on $s$ we need to integrate it wrt. to $s$, then it's a 2nd order linear ODE:
$$k=\frac{d^2s}{dt^2}(a+2s)^2$$
$$\iint kdtdt=\iint(a+2s)^2dsds$$
$$k\frac{t^2}{2}=\int \frac{(a+2s)^3}{3a}ds+C=\frac{(a+2s)^4}{4a^2}+C$$
Now solve for $s$ and differentiate wrt. $t$ get back your solution to $v$.
Why doesn't distance depend on time?
To OP: try going through each step, i'll leave you to differentiate it on your own. Tell me if you got stuck.
