Question

ank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentrationof 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute.The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute.The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallonsper minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons perminute. Find the differential equations with initial conditions for the amounts,x(t)andy(t), of saltin tanks A and B, respectively, at time t.

Answer 1

Answer:

$\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}$

$\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}$

x(0)=20,y(0)=5

Step-by-step explanation:

We are given that

Tank A contains water=80 gallons

x(0)=20,y(0)=5

Tank B contains water=30 gallons

Rate=4 gallon/min

Concentration of salt  is pumped into tank=0.5 pound /gallon of water

Solution pumped from tank A to tank B at the rate=6 gallons/min

Solution pumped from tank B to tank A at the rate=2gallon/min

Solution from tank B is pumped out of the system at the rate=4 gallon/min

We have to find the DE at time t

For x

Rate in=$0.5\times 4+y(t)/30\times 2$

Rate in=$2+y/15$

Rate out=$x/80\times 6$

Rate out=3x/40[/tex]

$\frac{dx}{dt}=$Rate in-Rate out

$\frac{dx}{dt}=2+y/15-3x/40$

$\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}$

For y

Rate in=$x/80\times 6$

Rate in=3/40x

Rate out=$y/30\times (4+2)$

Rate out=y/5

$\frac{dy}{dt}=$Rate in-Rate out

$\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}$