'm struggling with a). Any help would be greatly appreciated!!! I tried setting up a equation and using separable ODE and then using integration by parts,however i'm getting a weird answer.
Whisky consists mainly of water and alcohol. Before bottling, the alcohol is diluted with water to the standard bottle concentration of (44−46)%. Suppose a 2500 gallon tank initially contains 600 gallons of pure alcohol. An alcohol-water mix of $\frac{1+\cos t}{6}$ % alcohol is fed in at 6 gallons per minute and the mixture stirred; simultaneously the mixture is withdrawn at 3 gallons per minute.
(a) Find the amount of alcohol (in %) in the tank at time t. (b) Approximately how many hours will it take for the mixture in the tank to reach 44%? (c) Approximately how many hours will it take for the mixture in the tank to overflow?
Let $V(t)$ be the volume of solution in your tank at time $t$. We have $V(0) = 600$. Let $Q(t)$ be the volume of alcohol in your tank at time $t$, so we have $Q(0) = 600$. The concentration at time $t$ is: $$C(t) = \frac{Q(t)}{V(t)}.$$ We measure the volume and the quantity of alcohol in gallons. The concentration has no unit. We measure time in minutes.
The volume is increased by $R_{in} = 6$ gallons per minute and decreased by $R_{out} = 3$ gallons per minute. Hence we have $$\frac{dV}{dt} = R_{in} - R_{out} = 3.$$ So, this gives us equations for the volume: $$\begin{align} V(0) &= 600 \\ \frac{dV}{dt} &= 3 \end{align}$$ which has solution $V(t) = 600 + 3t$.
At time $t$, the quantity of alcohol entering the tank is $R_{in} \frac{1 + \cos t}{6}$ (flow rate in times alcohol concentration in the inflow) and the amount of alcohol exiting the tank is $R_{out} C(t) = R_{out}\frac{Q(t)}{600 + 3t}$ (flow rate out times alcohol concentration in the tank). So we can now setup the equations for $Q(t)$: $$\begin{align} Q(0) &= 600 \\ \frac{dQ}{dt} &= R_{in} \frac{1 + \cos t}{6} - R_{out} C(t) = 1 + \cos t - \frac{Q(t)}{200 + t} \end{align}$$ The solution to this is: $$ Q(t) = \frac{239998+400 t+t^2+2 \cos t}{400+2 t}+\sin t $$ and we get that the concentration is: $$ C(t) = \frac{Q(t)}{V(t)} = \frac{\frac{239998+400 t+t^2+2 \cos t}{400+2 t}+\sin t}{600+3t} = \frac{239998+400t+t^2+2 \cos t+(400+2t) \sin t}{6 (200+t)^2}. $$