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This equation describes leaking water form a conical tank. We are interested in finding $t$ when $h(t) = 0$ (time it takes to empty the tank).

$$ \frac{dh}{dt} = - \frac{5}{6h^{3/2}}, h(0) = 20 $$

Since this is separable, I separate the equation and solve:

$$ 6h^{3/2}dh = −5dt $$

$$ \frac{12}{5}h^{5/2} = −5t + c $$

Using $h(0) = 20$ I find that $c = 1920\sqrt{5}$. Then, solving $h(t) = 0$, I find that $t= 384\sqrt{5}$.

Why is that the way I distribute the 6/5 have an impact on the solution? All of thesee equations results in different values for $t$:

$$ -6h^{3/2}dh = 5dt $$ $$ \frac{6}{5}h^{3/2}dh = -dt $$ $$ -\frac{6}{5}h^{3/2}dh = dt $$ $$ etc. $$

Why? I understand distributing 6/5 affects $c$, but since it's an arbitrary constant, why does it matter?

Physically, it makes no sense (there can't be an infinite number of times it takes to empty a tank of water).

Thoughts? I must be getting this wrong.

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up vote 3 down vote accepted

Are you "solving for $h(t)$" before trying to figure out $h(t)=0$? You can't just plug in $h(t)=0$ and then solve for $t$ in the resulting equation, because $h$ is not independent of $t$. So you must first rewrite $\frac{12}{5}h^{5/2} = -5t + 1920\sqrt{5}$ as $$h(t) = \left(-\frac{25}{12}t + 800\sqrt{5}\right)^{2/5},$$ and then solve $h(t)=0$ for $t$.

Likewise with the other ways of solving the differential equation: before you can solve $h(t)=0$ for $t$, you need to have an explicit expression for $h$ in terms of $t$, rather than an implicit one. Well, you don't have to go all the way to the expression above, it is enough to get to $$(h(t))^{5/2} = -\frac{25}{12}t + 800\sqrt{5}$$

I think you will find that if you clear that constant factor from $h$ first, you will always get the same $t$ for $h(t)=0$.

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Thanks! That clears it up a bit. – David Chouinard Oct 10 '11 at 11:12

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