1 Solutions27and simplifying givesy= (5+t)+c(5+t)1. The initial conditiony(0) = 0impliesc=25soy= (5+t)25(5+t)1;for0t45. Att=T= 45we gety(45) = 502550= 49:5lbs salt. Once the tank is full, the inflowand outflow rates will be equal and the brine in the tank will (in thelimit ast! 1) stabilize to the concentration of the incoming brine, i.e.,0:5lb/gal. Since the tank holds 100 gal, the total amount present willapproach0:5100 = 50lbs. Thuslimt!1y(t) = 50.31. input rate:input rate=rcoutput rate:output rate=rP(t)VLetP0denote the amount of pollutant at timet= 0. SinceP0=input rateoutput rate it follows thatP(t)is a solution of the initialvalue problemP0=rcrP(t)V;P(0) =P0:Rewriting this equation in standard form gives the differential equationP0+rVP=rc. The coefficient function isp(t) =r=Vand the integratingfactor is(t) =ert=V. Thus(ertVP)0=rcertV. Integrating and simplifyinggivesP(t) =cV+kertV;wherekis the constant of integration. The initialconditionP(0) =P0impliesc=P0cVsoP(t) =cV+ (P0cV)ertV:(a)limt!1P(t) =cV:(b) When the river is cleaned up att= 0we assume the input con-centration isc=0. The amount of pollutant is therefore given byP(t) =P
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Constant of integration, Lake Erie, Lake Ontario, output rate, 100 gal