How to show dx/dt for this?

Question

A rectangular tank has a horizontal base. Water is flowing into the tank at a constant rate, and flows out at a rate which is proportional to the depth of water in the tank. At time t seconds the depth of water in the tank is x metres. If the depth is 0.5m, it remains at this constant value. Show that

dx/dt = - K(2x - 1)

Answer 1

dx/dt has two components:
1) a positive component due to the inflow, which will be a constant K
2) a negative component due to the outflow, which is proportional to the height. Therefore this term has the form -Cx, where C is a positive constant.

So, we have
dx/dt = K - Cx
but when x = 1/2, dx/dt = 0, so we must have K - C/2 = 0. Therefore C = 2K and we have

dx/dt = K - 2Kx = -K(2x-1).

Answer 2

wait. what is K?