Question

A tank contains 10 gallons of water. Salt water containing a concentration of 4t ounces per gallon flows into the tank at a rate of 4 gallons per minute and the mixture in the tank flows out at the same rate.
(a)Construct the mathematical model for this flow process

(b)Use integrating factors to solve for Q(t).
(c)If the tank contains Q0 amount of salt at time t = 0, use this as an initial condition to solve for the constant resulting from integration.

Answer 1

For our Tank

Initial water = 10 gallons

dQ/dt = rate in - rate out = (salt concentration in) x (flow rate in) - (tank salt concentration) x (flow rate out)

Note that dQ/dt should be in terms of ounce/min

Q(t) : the amount of salt at time t (ounces)

rate in = rate out = 4 gal/min

=> the total volume remains same.

salt concentration in = 4t  oz/gal

tank salt concentration = Q/current amount of water solution = Q/10 oz/gal

so

dQ/dt = ( 4t ) x 4 - ( Q/10 ) x 4

=> dQ/dt = 16t - 0.4Q ... PART A

Solving,

dQ/dt + 0.4Q = 16t

I.F. = e0.4dt = e0.4t

=>

e0.4t dQ/dt + 0.4Q e0.4t​​​​​​​ = 16te0.4t​​​​​​​

e0.4t​​​​​​​ dQ + 0.4Q e0.4t​dt​​​​​​ = 16te0.4t​​​​​​​dt

d(Q e0.4t​​​​​​​) = 16te0.4t​​​​​​​dt

d(Q e0.4t​​​​​​​) = 16te0.4t​​​​​​​dt

Q e0.4t​​​​​​​ = 16te0.4t​​​​​​​dt

For 16te0.4t​​​​​​​dt

put 0.4t = y

16te0.4t​​​​​​​dt = 16x y/0.4 x ey​​ x dy/0.4 = 100y eydy = 100 ey( y -1 )

100 ey( y -1 ) = 100 e0.4t ( 0.4t - 1 )

so =>

Q e0.4t​​​​​​​ = 100 e0.4t ( 0.4t - 1 ) + C

=> Q(t) = 100( 0.4t - 1 ) + Ce-0.4t

we have @t = 0 and Q(0) = 0

Q(0) = 0 = -100 + C

C = 100

=> Q(t) = 100( 0.4t - 1 ) + 100e-0.4t

=-=-=-=-

Hope it helps!

Do give feedback!

​​​​​​​Have a nice day!