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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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    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.2012-01-22

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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.

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    I was afraid that was the answer. No solution even in integral form.2012-02-20
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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 $y(t)$.2012-01-22