A tank contains 30 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 5 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal. A(t) = Incorrect: Your answer is incorrect. lb Without actually graphing, conjecture what the solution curve of the IVP should look like. Then use a graphing utility to plot the graph of the solution on the interval [0, 300].

If [tex]A(t)[/tex] is the amount of salt in the tank at any time [tex]t[/tex], then the solution has a concentration of [tex]\dfrac{A(t)}{500}\dfrac{\rm lb}{\rm gal}[/tex].

The net rate of change of the amount of salt in the solution, [tex]A'(t)[/tex], is the difference between the amount flowing in and the amount getting pumped out:

[tex]A'(t)=\left(5\dfrac{\rm gal}{\rm min}\right)\left(\left(2+\sin\dfrac t4\right)\dfrac{\rm lb}{\rm gal}\right)-\left(5\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{A(t)}{50}\dfrac{\rm lb}{\rm gal}\right)[/tex]

Dropping the units and simplifying, we get the linear ODE

[tex]A'=10+5\sin\dfrac t4-\dfrac A{10}[/tex]

[tex]10A'+A=100+50\sin\dfrac t4[/tex]

Multiplying both sides by [tex]e^{10t}[/tex] allows us to identify the left side as a derivative of a product:

[tex]10e^{10t}A'+e^{10t}A=\left(100+50\sin\dfrac t4\right)e^{10t}[/tex]

[tex]\left(e^{10}tA\right)'=\left(100+50\sin\dfrac t4\right)e^{10t}[/tex]

[tex]e^{10t}A=\displaystyle\int\left(100+50\sin\dfrac t4\right)e^{10t}\,\mathrm dt[/tex]

Integrate and divide both sides by [tex]e^{10t}[/tex] to get

[tex]A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+Ce^{-10t}[/tex]

The tanks starts off with 30 lb of salt, so [tex]A(0)=30[/tex] and we can solve for [tex]C[/tex] to get a particular solution of

[tex]A(t)=10-\dfrac{200}{1601}\cos\dfrac t4+\dfrac{8000}{1601}\sin\dfrac t4+\dfrac{32,220}{1601}e^{-10t}[/tex]