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Unformatted text preview: Solution: A(t) = kg of salt in the tank at time t minutes A'(t) = The rate at which the amount of salt in the tank is changing = (rate of salt going in) - (rate of salt going out) rate of salt going in = 0.242 x 5-= 1 kg min min rate of salt going out = 4(2) kg of salt 500 x 5 A(t) min 100 A' (t) = 1 - A(t) 100 A'(t) + (t) 100 = 1 For equation of y' + by = g(t) , Solution isy = e-bt febt g(t) dt + De-bt A(t) = e 100 fe10o dt + De 100 A(t) = e 100 (100 e100 ) + De 100 A(t) = (100 ) + De 100 Using the initial condition A (0) = 5 5 = (100 ) + D , D = -95 A(t) = (100 ) - 95 e 100 10 A(10) = (100 ) - 95 e 100 = 14.04kg Let C(t) be the concentration of salt and V(t) be the volume of brine solution in the tank. Then the amount of salt is given by A(t) = C(t) . V(t) Where V(t) = 500 C(t) = 500 (100 ) - 95 e 100 C(10) = 1 10 500 (100 ) - = 0.02808 kg/l...
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