TITLE: differential equation with reciprocal and a linear function in the derivative
QUESTION [5 upvotes]: My question is how to solve
$$ \frac{dx(at+b)}{dt} =\frac{c}{x(t)} $$
I'm aware that when $a=1, b = 0$, the solution to this problem is $x(t)=\sqrt{2(ct+k)}$ 
But in general, does this differential equation have an solution when $a \neq 1$ and $b \neq 0$?

REPLY [4 votes]: When $(a,b)\ne(1,0)$ this is not a differential equation. When $a=1$ and $b\ne0$ this is a delay differential equation; see the link provided by Michael Lee. When $a\ne1$ you can simplify the equation as follows: Introduce the new unknown function
$$y(\tau):=x\left(\tau-{b\over a-1}\right)\qquad(\tau\in{\mathbb R})$$
(this just amounts to a shift of the time scale). Then
$$\dot y(a\tau)=x'\left(a\tau-{b\over a-1}\right)=x'\left(a\left(\tau-{b\over a-1}\right)+b\right)={c\over x\left(\tau-{b\over a-1}\right)}={c\over y(\tau)}\ .$$
It follows that
$$y(\tau)\>\dot y(a\tau)=c\qquad(\tau\in{\mathbb R})\ .$$
Now you can try the Ansatz $y(\tau)= p\tau^\lambda$ with $p$,  $\lambda$ yet to be determined, and obtain something reasonable, albeit not analytic at $\tau=0$. Or you can try a power series
$$y(\tau)=\sum_{k=0}^\infty c_k\>\tau^k\ ,$$
whereby you choose $c=c_0=1$ for a start in order to simplify the resulting recursion formula.