Recently I missed a class in college regarding how to work out this question and now have no information on how to work this differential equation out and which methods I should use. Can anyone help me work out the method and answer as I'm really struggling. This is the question:

A tank full of water is being drained through an outlet. The height $H(m)$ of the water in the tank at any time $t(s)$ is given by the differential equation: $$\frac{\mathrm{d}H}{\mathrm{d}t} = −0.0054\sqrt{H}$$ Given the initial condition $H = 2$ when $t = 0$, find an expression for $H$.

Apparently I have to separate it I think and am I on the right tracks?

Separate Variables:

$$\frac{\mathrm{d}H}{\sqrt{H}} = -0.0054 \:\mathrm{d}t$$


$$2\sqrt{H} = -0.0054t + c$$

  • $\begingroup$ So, $\frac{dH}{dt}= a \sqrt{H}$, $\frac{dH}{\sqrt{H}} = a dt$, then, obtaining $H(t)=(2at+C)^{2}$ gives the result. Since $H(0)=2$, then $C = \pm \sqrt{2}$. $\endgroup$ – hyperkahler May 31 '15 at 21:07

Exactly as you said, separating variables and integrating we find that: \begin{equation} \begin{aligned} 2\sqrt{H}&=-0.0054t+c\\ \implies H&=\left(-0.0027t+d\right)^2 \end{aligned} \end{equation} where $d\equiv\frac{c}{2}$ is a constant that depends on the initial conditions. As we are given that $H=2$ when $t=0$ we find that: \begin{equation} \begin{aligned} 2&=\left(-0.0027\times 0+d\right)^2\\ \implies d&=\pm\sqrt{2} \end{aligned} \end{equation} A negative $d$ would imply a negative $c$ and therefore would a negative $\sqrt{H}$ at $t=0$ which is unphysical; we therefore take the positive $d$ and therefore find that: \begin{equation} H=\left(-0.0027t+\sqrt{2}\right)^2 \end{equation}

  • $\begingroup$ Thankyou so much for your help Eric. Best Regards! $\endgroup$ – Mister L May 31 '15 at 21:40
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    $\begingroup$ @MisterL, if you liked this answer and happy with it , then you can accept and upvote by clicking on the up arrow in the left hans corner. $\endgroup$ – abel Jun 1 '15 at 10:56

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