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I have the differential equation $${{dh} \over {dt}} = - K\sqrt h $$

describing water flowing out of the bottom of a tank of uniform cross-sectional area under the action of gravity. $h(t)$ is the water depth at time t with ${h_0}$ being the initial depth. K is a positive constant.

I have no idea how to solve this. I only know how to separate variables to solve DEs. How would I solve this one? This is one of the starred questions in the book. I can't find another example like this. Thanks!

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You can still solve this with separation of variables. divide both sides by $\sqrt{h}$ and you have a separated equation.

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  • $\begingroup$ I did that but I got that $t = - {{2\sqrt h } \over K}$ so when I plug in h = 0, that means the tank empties instantaneously... What am I doing wrong? $\endgroup$ – SumMathGuy Feb 24 '16 at 18:23
  • $\begingroup$ After dividing by $\sqrt{h}$ and integrating both sides we have the equation $2\sqrt{h} = -Kt + C$ where C is a constant. solving for h here gives $h = ((-Kt+C)/2)^2$ $\endgroup$ – CrazyIvan Feb 24 '16 at 18:27
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    $\begingroup$ @SumMathGuy You have to integrate using proper limits. You will get $t=-\frac{2\sqrt h}{K} + c$, where $c$ has to be found using the initial condition given. $\endgroup$ – GoodDeeds Feb 24 '16 at 18:27
  • $\begingroup$ I did get that but I don't have an initial condition. As I said above, it just says ${h_0}$ so I can't solve for c. Ivan, I am solving for t not for h. $\endgroup$ – SumMathGuy Feb 24 '16 at 18:33
  • $\begingroup$ h(0) = $h_0$ so when we plug in 0 for t we can solve for c. in particular $c = 2\sqrt{h_0}/K$. $\endgroup$ – CrazyIvan Feb 24 '16 at 18:35
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I present this way for solution your answer. Following the below way dh/dt=-k√(h,) You can solve this equation very easy. Divide both side of equation to √(h,). In this case, equation is changed to a separated of variable. By integrating both of them relative to separation variable and using initial condition, at which h(t=0)=h_0 , you can find integration constant, c=(h_0 )^1/2 and finally its answer: h=(-1/2*kt+h_0^(1/2))^1/2 This equation shows the h versus t variation. By the way, you can draw it to linearity with h^2 versus t.

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  • $\begingroup$ here is a mathjax reference. $\endgroup$ – Siong Thye Goh Apr 29 '19 at 7:17

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