A tank contains $100$ gal of water and $50$ oz of salt. Water containing a salt concentration of $\frac14[1+\frac12\sin t]$ oz/gal flows into the tank at a rate of $2$ gal/min, and the mixture in the tank flows out at the same rate.
(a) Find the amount of salt in the tank at any time.
(b) Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.
I already find the solution for (a) which is
[$25+\frac{25}{5002}\sin t-\frac{625}{2501}\cos t+\frac{63150}{2501}e^{-0.02t}$]
and I plotted the graph but not sure correct or not. 
Best Answer
For part $a)$
Let us take $Q(t)$ to be the amount of salt present in the tank at any time $t$. The salt enters the tank at a rate of $2\left[\dfrac14\left(1+\dfrac12\sin t\right)\right]=\dfrac12+\dfrac14\sin t\ \\ \dfrac{\mbox{ oz}}{\mbox{ min}}$
So, the differential equation governing the amount of salt in the tank at any given time is $$\dfrac{dQ}{dt}=\dfrac12+\dfrac14\sin t-\dfrac Q5$$
And the I.C $Q(0)=50$ oz. Now by solving the linear differential equation for $Q$ by using the I.F and I.C we get $$Q(t)=25+\dfrac{1}{2501}[12.5\sin t-625\cos t+63150e^{-t/50}]$$
So, the amount of salt approaches the steady state, with an oscillation of amplitude $\dfrac14$ with level $25$ oz
For part$b)$
Your graph looks fine.