Ok, first order linear mixing problem. "A 400 gal. tank initially contains 200 gal. of water containing 2 ppb by weight of dioxin, an extremely potent carcinogen. Suppose water containing 5 ppb of dioxin flows into the top of the tank at a rate of 4 gal/min. The water in the tank is kept well mixed, and 2 gal/min are removed from the bottom of the tank. How much dioxin is in the tank when the tank is full? Ok. Vo = 200 gal., a = 2 ppb, b = 5 ppb, e = 4 gal/min, f = 2 gal/min (Vo = initial volume of water, a = initial amount of dioxin, b = amount of dioxin flowing in, e = rate of b, f = rate going out) So, if Q is the amount of dioxin in the mixture at any time t, dQ/dt + 2/(200 + 2t) = 20 (from dQ/dt + f/(Vo + (e-f)t) = be) So, my a(t) = 2/(200+2t), r(t) = 20. u(t) = e^((int)a(t)dt), u(t) = e^(int)(2/(200 + 2t))dt u(t) = e^(ln|200 + 2t|) = 200 + 2t //(int) -integral (too lazy for Latex) So, Q(t) = 1/u(t)[(int)u(t)r(t)dt] + c/u(t) Q(t) = 1/(200 + 2t)[(int)(200 + 2t)(20)dt] + c/(200 + 2t) Q(t) = [1/(200 + 2t)](4000t + 20t^2) + c/(200 + 2t) Q(t) = (20t^2 + 4000t + c)/(200 + 2t) Ok, since Q(0) = a = 2ppb then Q(0) = a = 2 = (c/200), so c = 400, then Q(t) = (20t^2 + 4000t + 400)/(200 + 2t) Now, when the tank is full (400 gal, I need the amount of dioxin, so 200 + 2t = 400, when t = 100min) Q(100) = (20(100^2) + 4000(100) + 400)/(400) Q(100) = 1501 ppb. Which is a little absurd(from 2ppb initially). Also, this is wrong because the answer is 4.25ppb. Can someone go through my math and see if I made a mistake? Because I can't seem to find what went wrong. Thanks.