A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank (a) after minutes and (b) after one hour?

Answer:(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153Step-by-step explanation:To start with,Let s(t) = amount of salt in kg of salt at time t.Then we have:dt/ds =  (rate of salt into tank) − (rate of salt out of tank)= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/minSo we get the differential equationdt/ds = 0.65 − 15s / 1000= 65 / 100 − 15s / 1000 dt/ds = 130 - 3s / 200We separate s and t to get1 / 130 - 3s ds = 1 / 200 dtThen we Integrate,Thus we have∫1 / 130 - 3s ds = ∫1 / 200 dt- 1/3  · ln |130 − 3s| =1 / 200 t + C1ln |130 − 3s| = - 3 /200 t + C2|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2|130 − 3s| = C3e − ∧3t/200130 − 3s = C4e  ∧−3t/200−3s = −130 + C4e  ∧−3t/200s = 130 - C4e ∧−3t/200 / 3Since we begin with pure water, we have s(0) = 0. Substituting, 0 = 130 − C4e  ∧−3·0/200 / 30 = 130 − C4C4 = 130So our function iss(t) = 130 − 130e ∧−3t/200 / 3After one hour (60 min), we haves(60) = 130 − 130e ∧−3·60/200 / 3s(60) ≈ 25.7153Thus, after one hour there is about 25.72 kg of salt in the tank.