Question

Consider a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute. The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

(a) Find a differential equation for the amount of salt x(t) in the tank at time t.
(b) Solve the DE in (a).

Answer 1

Answer:

Step-by-step explanation:

given that  a fish tank that initially contains 300 liters of water with 100 grams of salt dissolved in it. Water containing 4 grams of salt per liter is allowed to run into the tank at a rate of 2 liters per minute.

The thoroughly stirred mixture in the tank is also allowed to drain out at a rate of 3 liters per minute.

Let X(t) be the amount of salt in the tank at time t

X(0) = 100 gms.

Volume of tank at time t = initial volume+t(incoming rate-outgoing rate)

= 300+(2-3)t= 300 -t

Rate of change of X would be=incoming salt rate-outgoing salt rate

$= 4(2) -\frac{3X(t)}{300-t}$,X(0) = 100 gm

i.e. $\frac{dx}{dt} = 8-\frac{3X(t)}{300-t}$

$\frac{dx}{dt} +\frac{3X(t)}{300-t}= 8$

this is linear DE

$e^{\int\limits {\frac{3}{300-t} } \, dt } =e^{-3ln (300-t)} \\=\frac{1}{(300-t)^3}$

Solution woul dbe

$X*\frac{1}{(300-t)^3} =\int\limits{\frac{8}{(300-t)^3} } \, dt\\=\frac{4}{(300-t)^2} +C\\X(t) = 4(300-t)+C(300-t)^3$

Use initial condition to find t

X(0) = 100 = 1200+300^3$X(t) = 4(300-t)-4.074 (300-t)^3$C

C = -4.074(10^-5)