Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

I've been musing on a problem I came across whilst decanting some sloe gin (that time of the year here in the UK and all that). Essentially, I want to know given a vessel of known dimensions, open to the atmosphere at the top, with a hole in the bottom, and a pipe at the top filling with liquid at a steady rate, what height the liquid in the vessel will be at a given time. Anyway, the meat of this boils down to an equation of the form:

[tex]\frac{dz}{dt} = a - b\sqrt{z(t)},~~~z(0)=0[/tex]

I've tried a couple of methods to try and quell this beast, for example a simple substitution

[tex]u(t) = \sqrt{z(t)},~~~\frac{dz}{dt} = 2u(t)\frac{du}{dt}[/tex]

results in the equally horrible

[tex]u(t)\left(2\frac{du}{dt} + b\right) = a[/tex]

I've tried throwing it in Maple too, but no joy (ends up with a Lambert W function, which I don't count as an exact solution), am I missing some insight into this? A quick numerical simulation shows the equation behaves the way I would expect it to.

(P.S. This isn't homework, I'm just amusing myself with a problem...)

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# Diff. eq. with square root

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