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Just started a Differential equations class. Not sure if i'm understanding this right. Please verify if i'm correct, or please explain on how to go about this. Thanks!

Torricelli's Law.

Suppose a cylindrical container containing a fluid has a drain on the side. Torricelli's law states that the change in the height of the fluid above the middle of the drain is proportional to the square root of the height. Let $h(t)$ denote the height of the fluid above the middle of the drain. Determine a differential equation in $h(t)$ that models Torricelli's law.

What I came up with was...

$h(t) = \frac{2h^\frac{2}{3}}{3}$

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    It's asking for a *differential equation*, not a solution to a differential equation. So your result should involve $h'(t)$. – 2017-01-28
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    Okay, makes sense. – 2017-01-28
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    Can you further explain on how to approach the problem please? – 2017-01-28
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    Well, what does "$y$ is proportional to the square root of $x$" look like in symbols? Then replace $y$ with $h'$ and $x$ with $h$. – 2017-01-28
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    "$h' = \sqrt{h}"$ ? – 2017-01-28

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The question states that the change in height of the fluid is proportional to the square root of the height.

Therefore, we have: $$\frac{dh}{dt}\propto \sqrt{h} \tag{1}$$ Using a proportionality constant $k$, we obtain the following differential equation: $$\frac{dh}{dt}=-k\sqrt{h} \tag{2}$$ Note: I didn't use $\frac{dh}{dt}=k\sqrt{h}$, which is also correct, but I know that the height will always decrease. It is therefore convenient to use the form of $(2)$ instead since $k>0$.

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    Awesome. Thank you for explaining. – 2017-01-28
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    Is there difference between writing "$h(t)$" over "$h$" inside the square root? $\frac{dh}{dt}=-k\sqrt{h(t)} \tag{1}$ $\frac{dh}{dt}=-k\sqrt{h} \tag{2}$ – 2017-01-29
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    @cisco No, there is no difference. Writing $h(t)$ only emphasizes that $h$ is a function of the time $t$. – 2017-01-29