Beginning at time t = 0, fresh water is pumped at the rate of 3 gal/min into a 6

Question

Beginning at time t = 0, fresh water is pumped at the rate of 3 gal/min into a 60-gal tank initially filled with brine. The resulting less-and-less salty mixture overflows at the same rate into a second 60-gal tank that initially contained only pure water, and from there it eventually spills onto the ground. Assuming perfect mixing in both tanks, A) when will the water in the second tank taste saltiest? B) And exactly how salty will it then be, compared with the original brine?

Explanation / Answer

PLEASE RATE ME FIRST The integrating factor is u(t) = exp(? dt/20) = e^(t/20). Multiplying the equation through by this integrating factor gives e^(t/20) (dy/dt + y/20) = X_0/20 ==> d/dt (e^(t/20) y) = X_0/20. When you integrate both sides you get e^(t/20) y = (X_0/20) t + C. Then multiplying by the reciprocal of the exponential gives the general solution of the differential equation y = (X_0/20) t e^(-t/20) + C e^(-t/20). Only now is it possible to apply the initial condition. Perhaps it's more clear if you write the original equation as dy/dt + (1/20) y = .... (A) after 20 minutes the water in the second tank taste saltiest (B)And exactly it wil be salty e^-1 then be, compared with the original brine

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