Integral for an inverse with no closed-form solution I have a function that gives $t$ in terms of $y$ that has no closed-form solution for $y$ (W|A).  
I have that $$\frac{dy}{dt} = \sqrt{a - \frac{b}{y(t)}}.$$  Is there some way to set up an integral that I can evaluate that would output $y$ given $t$?
Integrating $dt$:
$$y = \int {\sqrt{a-\frac{b}{y(t)}}} dt$$
Is there anywhere I can go from here?
 A: Have you heard of the cycloid? It is parametrized by
$x = a\dfrac{\theta + \sin \theta}{2}$
$y = a\dfrac{1-\cos \theta}{2}$
and is the solution to 
$$\frac{{dx}}{{dy}} = \sqrt {\frac{{a - y}}{y}} $$
Since your equation is
$$\frac{{dy}}{{dt}} = \sqrt {\frac{{ay - b}}{y}} $$ 
we can go like this:
Put 
$$\frac{{dt}}{{dy}} = \sqrt {\frac{y}{{ay - b}}} $$
Now let
$$y = \frac{b}{a}{\cosh ^2}\theta $$
We get
$$dt = \frac{{2b}}{{{a^{3/2}}}}{\cosh ^2}\theta d\theta $$
So
$$dt = \frac{{2b}}{{{a^{3/2}}}}\left( {\frac{1}{2} + \frac{{\cosh 2\theta }}{2}} \right)d\theta $$
and integrating gives
$$t = \frac{{2b}}{{{a^{3/2}}}}\left( {\frac{{2\theta }}{4} + \frac{{\sinh 2\theta }}{4}} \right)+C$$
So your solution is parametrized by ($\phi = 2\theta$, suppose initial conditions make $C=0$) 
$$\eqalign{
  & y = \frac{b}{{2a}}\left( {1 + \cosh \phi } \right)  \cr 
  & t = \frac{b}{{2{a^{3/2}}}}\left( {\phi  + \sinh \phi } \right) \cr} $$
In the same way the cycloid is a "deformed" circumeference your solution is a "deformed" hiperbola. The cycloid has a closed form for $(x,y)$ coordinates so you might be able to find one for the above curve.
A: You'd better do the next:
$\frac{dy}{\sqrt{a - \frac{b}{y(t)}}} = dt $ then after integrating you have solution $t(y)$
