Balance constant inflow against volume-dependent outflow

$dV/dt=4-kV$. At $V=6$ and $V^{\prime}=1$, $1=4-6k$, giving $k=0.5\,\mathrm{s}^{-1}$. Hence $V^{\prime}=4-0.5V$, with $t$ measured in seconds.

$\int dV/(4-0.5V)=\int dt$ gives $-2\ln|4-0.5V|=t+C$, so $4-0.5V=Ae^{-0.5t}$. With $V(0)=2$, $A=3$, and therefore $V=8-6e^{-0.5t}$.

As $t\to\infty$, $V\to8\,\mathrm{m}^3$. The volume increases towards this limit; at the limit, the outflow $0.5V$ balances the constant inflow $4$.

For $t\ge0$, the graph begins at $(0,2)$, increases and is concave down, approaching the horizontal asymptote $V=8$ from below. Here $V^{\prime}=3e^{-0.5t}>0$ and $V^{\prime\prime}=-1.5e^{-0.5t}<0$.