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?

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It is easy to set up an integral which provides $t$ given $y$... – Fabian Jan 21 '12 at 23:14
Yes, then what? – jnm2 Jan 21 '12 at 23:35
You'll never solve this equation in common. – Evgeny Savinov Jan 22 '12 at 2:09
I fear you are in trouble unless you content yourself with some iterative techniques. These will grant a solution in t even if an approximate one. – Jon Jan 22 '12 at 10:36

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

$$\frac{{dx}}{{dy}} = \sqrt {\frac{{a - y}}{y}}$$

$$\frac{{dy}}{{dt}} = \sqrt {\frac{{ay - b}}{y}}$$

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.

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I don't think the cycloid has a closed form for $y$ in terms of $x$, just $x$ in terms of $y$ like I was trying to get away from with this question. – jnm2 Feb 20 '12 at 14:29
@jnm2 Then at least now you know there is no solution. Yes, I mean $x$ in terms of $y$. However, you get a cycloid running through the $y$ axis if you switch $y$ and $x$. However, see here a plot of your function (for $\phi \geq 0$) . I seems quite 1 to 1. – Pedro Tamaroff Feb 20 '12 at 14:36
I was afraid that was the answer. No solution even in integral form. – jnm2 Feb 20 '12 at 15:19

You'd better do the next: $\frac{dy}{\sqrt{a - \frac{b}{y(t)}}} = dt$ then after integrating you have solution $t(y)$

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I already have $t(y)$. I'm asking how to set up an integral for $y(t)$. – jnm2 Jan 22 '12 at 0:07