Question

WATER. ASSUME TMAT WATER ConTAnMG RATE OF 드 GAL/min nD TXE WELL-SRRED

Answer 1

The rate of flow into the tank = concentration of salt times flow rate = 2 times 5 = 10 The rate of flow out of tank = concentrated of salt in tank times flow rate = y/100 times 5 = 5y/100 let y (t) the amount of salt present in tank at time t therefore The rate change of salt solution in the tank is given dy/dt = Rate in flow - Rate out flow dy/dt = 10 - 5y/100 As these use 25 lb of salt present in solution initially, ie y (0) = 25 therefore dy/dt = 10 - 0.05y, y (0) = 25 find the integrating factor, I. F = e^integral 0.05 dt = e^0.05 t The general solution is given by, y e^0.05 t = integral 10 e^0.05 t dt y e^0.05 t = 10/0.05 e^0.05t + c \r\n Now dividing by e^+ 0.05 t, y = 200 + c/e^0.05 t therefore y =