A pond contains 2760 L of pure water and an uknown amount of an undesirable chemical. Water contaninig 0.01 kg of this chemical per liter flows into the pond at a rate of 4 L/h. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. Let Q(t) be the amount of chemical (in kg) in the pond at time t hours. (a) Write a differential equation for the amount of chemical in the pond? at any time time (enter Q for Q(t)) (b) How much chemical will be in the pond after a long time?

Let [tex]q[/tex] be the unknown amount of the chemical originally in the pond, so [tex]Q(0)=q[/tex].a. The incoming water introduces the chemical at a rate of[tex]Q'_{\rm in}=\left(0.1\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac25\dfrac{\rm kg}{\rm hr}[/tex]and the mixture flows out at a rate of[tex]Q'_{\rm out}=\left(\dfrac Q{2760}\dfrac{\rm kg}{\rm L}\right)\left(4\dfrac{\rm L}{\rm hr}\right)=\dfrac Q{690}\dfrac{\rm kg}{\rm hr}[/tex]so that the net rate of change (in kg/hr) of the chemical in the pond is given by the differential equation,[tex]\boxed{Q'=\dfrac25-\dfrac Q{690}}[/tex]b. The ODE is linear; multiplying both sides by [tex]e^{t/690}[/tex] gives[tex]e^{t/690}Q'+\dfrac{e^{t/690}}{690}Q=\dfrac{2e^{t/690}}5[/tex]Condense the left side into the derivative of a product:[tex]\left(e^{t/690}Q\right)'=\dfrac{2e^{t/690}}5[/tex]Integrate both sides to get[tex]e^{t/690}Q=276e^{t/690}+C[/tex]and solve for [tex]Q[/tex] to get[tex]Q=276+Ce^{-t/690}[/tex]The pond starts with [tex]q[/tex] kg of the chemical, so when [tex]t=0[/tex] we have[tex]q=276+C\implies C=q-276[/tex]so that the amount of chemical in the water at time [tex]t[/tex] is[tex]Q(t)=276+(q-276)e^{-t/690}[/tex]As [tex]t\to\infty[/tex], the exponential term will converge to 0, leaving a fixed amount of 276 kg of the chemical in the pond.