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Differential equation help

The square horizontal cross section of a container has side 2m. Water is poured in at the constant rate of 0.08 m^3/s and, at the same time, leaks out of a whole in the base of the container at the rate of 0.12x m^3/s, where X metres is the depth of the water in the container at time t seconds. So the volume Vm^3, of the water in the container at time t is given by V= 4x and rate of change of volume is given by: dV/dt= 0.08 - 0.12x.
Use these results to find an equation for dx/dt in terms of X, and solve this to find X in terms of t if the container is initially empty.

I have no idea where to even start, any help would be greatly appreciated.
you need to set up a chain... for instance dQ/dt = dQ/dW*dW/dt
do you have the answers?
Reply 3
Original post by Squiidsquid
do you have the answers?


I don't unfortunately
Reply 4
Original post by the bear
you need to set up a chain... for instance dQ/dt = dQ/dW*dW/dt



Ah, that never occurred to me.

So is it dx/dt= 1/4 x (0.08-0.12x)=

(0.02-0.03x)

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