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.