Question

Two tanks of brine are connected via two tubes. Tank 1 contains x(t) pounds of salt in 500 gallons of brine, and tank 2 contains y(t) pounds of salt in 500 gallons of brine. Brine is pumped from tank 1 to tank 2 at a rate of 50 gallons/min and brine is pumped from tank 2 to tank 1 at the same rate of 50 gallons/min, ensuring that the total volume of brine in the tanks remains constant over time.

a) Show that x(t) and y(t) satisfy the differential equations x'= − (1/10)x + (1/10)y ; y' = (1/10)x − (1/10)y.


b) If initially tank 1 contains no salt, and tank 2 contains 10 pounds of salt, use the method of elimination to find x(t) and y(t).

Answer 1

a. Salt flows into tank A at a rate of

(y(t)/500 lb/gal) * (50 gal/min) = y(t)/10 lb/gal

and out at a rate of

(x(t)/500 lb/gal) * (50 gal/min) = x(t)/10 lb/gal

so the net rate of change of the amount of salt in tank A is

`x'(t)=(y(t))/(10)-(x(t))/(10)`

Similarly, you would find

`y'(t)=(x(t))/(10)-(y(t))/(10)`

b. We have

`x'=-\frac x{10}+\frac y{10}\implies x''=-(x')/(10)+(y')/(10)`

Notice that x' = -y', so

`x''=-{2x'}{10}\implies 5x''+x'=0`

Solve for x: the characteristic equation

`5r^2+r=r(5r+1)=0`

has roots
`r=0` and
`r=-\frac15`, so

`x(t)=C_1+C_2e^(-t/5)`

Again using the fact that y' = -x', we then find

`x'(t)=-\frac{C_2}5e^(-t/5)\implies y'(t)=\frac{C_2}5e^(-t/5)\implies y(t)=C_1-C_2e^(-t/5)`

Given that x(0) = 0 and y(0) = 10, we find

`\begin{cases}0=C_1+C_2\\10=C_1-C_2\end{cases}\implies C_1=5,C_2=-5`