# A tank contains 10 gallons of

A tank contains 10 gallons of water. Salt water containing aconcentration of 4t ounces per gallon flows into the tank at a rateof 4 gallons per minute and the mixture in the tank flows out atthe 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 thisas an initial condition to solve for the constant resulting fromintegration.

For our Tank

Initial water = 10 gallons

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

Note that dQ/dt should be in terms ofounce/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 ... PARTA

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(Qe0.4t​​​​​​​) = 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 = 100yeydy = 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!

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