A tank contains \(2\) m \(^{3}\)of water initially. Water flows into the tank at a constant rate of \(4\) m\(^{3}\)s\(^{-1}\)and flows out at a rate which is proportional to the amount of water \(V\) m\(^{3}\) in the tank. The rate of change of volume of water in the tank is equal to \(1\) m\(^{3}\)sx at the instant when \(V=6\). By setting up and solving a differential equation, show that \(V=8-6{{\mathrm{e}}^{-0.5t}}\).[7]
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