1. ## Differential Equation problems

A tank initially holds 100gal of a brine solution containing 20 lb of salt. At t=0 another brine solution containing 1lb of salt per gallon is poured into the tank at the rate of 3gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t.

Any help on this problem will be appreciated,

Cheers.

2. Originally Posted by shinn
A tank initially holds 100gal of a brine solution containing 20 lb of salt. At t=0 another brine solution containing 1lb of salt per gallon is poured into the tank at the rate of 3gal/min, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time t.

Any help on this problem will be appreciated,

Cheers.
Let x be the amount of salt (in lb) in the tank at time t.

Then $\frac{dx}{dt} = \,$ (rate of salt in) - (rate of salt out).

Rate in = (concentration of salt in inflow)(rate of inflow) = (1 lb/gal)(3 gal/min) = 3 lb/min.

Rate out = (concentration of salt in tank at time t)(rate of outflow)

= $\left (\frac{\text{amount of salt in tank at time } t}{\text{volume of tank at time }t} \right)$ (3 gal/min) = $\frac{x}{100} \times 3 = \frac{3x}{100}$.

Therefore the differential equation you need to solve is

$\frac{dx}{dt} = 3 - \frac{3x}{100} = \frac{3(100 - x)}{100}$

subject to the boundary condition x(0) = 20.

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### tank initially holds 80 gal of a brine solution containing 1/8 lb of salt per gallon. at t=0, another brine solution containing 1 lb of salt per gallon of solution is poured into the tank at the rate of 4 gal/min, while well-stirred mixture leaves the tan

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