Question

(2 pts) A 150 L tank contains 100 L of pure water. Brine that contains 0.1 kg of salt/L enters the tank at 5 L/min. The solution is kept thoroughly mixed and drains from the tank at the rate of 4 L/min. Find the concentration of the salt in the tank at the moment it is full. (2 pts) Separate variables and use partial fractions to solve the following initial value problem. da T = x (- 1), x(0) = 3

Answer 1

(a)

Let be amount of salt (in kg) in solution in the tank at time

Then let us consider $\frac{dS}{dt}$ at time

The inflow rate of salt at time is equal to the concentration of salt that enters per min at time times the inflow rate kg/min

Concentration of salt in the tank at time = (Amount of salt at time ) / (Volume of water in the tank at time )

Concentration of salt in the tank at time

S(t) 100+ (5-4) s(t) 100+t

The outflow rate of salt at time which is equal to the concentration of salt in the tank at time times the outflow rate.

The outflow rate of salt at time

45(t) 100 +t

Then

ds di = 0.5 - 4.5(t) 100++

ds dt 4s(t) 100 15

Multiplying by integrating factor $(100+t)^4$ on both sides we get

$\frac{d}{dt}\left( S(t)(100+t)^4\right)=0.5(100+t)^4$

Thus

s(t)(100+t) = 0.1(100+ t) 5 + 0

$S(t)=0.1(100+t)+C(100+t)^{-4}$

Since $S(0)=0$ ,

0 = S(0) = 0.1(100) + C(100) - 4

$C=-10^9$

$S(t)=0.1(100+t)-10^9(100+t)^{-4}$

The tank is full at $t=50$ min since the volume of water in tank is L.

Amount of salt in the tank when the tank is full

$=S(50)=0.1(100+50)-10^9(100+50)^{-4}=13.025$

Concentration of salt in the tank when the tank is full $=13.025/150=0.087$

(b)

$\frac{dx}{x(x-1)}=dt$

$\frac{dx}{x-1}-\frac{dx}{x}=dt$

$\ln(x-1)-\ln(x)=t+c$

$\ln(\frac{x-1}{x})=t+c$

$\frac{x-1}{x}=\lambda e^t$

$x(1-\lambda e^t)=1$

$x=(1-\lambda e^t)^{-1}$

Since $x(0)=3$ , we get $\lambda=2/3$ .