Question

Solve the differential equation.

Answer 1

A large tank is initially filled to capacity with 10 gallons of saltwater solution that has a salt concentration of 2 lbs/gallon. A saltwater solution containing 3 lbs of salt per gallon is pumped into the tank at a rate of 2 gal/min. The well-mixed solution is pumped out of the tank at the rate of 3 gal/min. Set up a differential equation for y(t), the amount of salt water in the tank after t minutes have elapsed. Also specify the initial condition. Solve the differential equation. Your answer should not have any unspecifed constants in it.

Answer 2

Heres what i have:
V(0) = 10 gal
V(t) = 10-t
y(0) = 20
y(t) = ???
in(t) = 2 gal/min
out(t) = 3 gal/min
k(t) = 3 lbs/gal

Answer 3

Ive solved the entire equation, im just looking to check my answer really so i dont necessarily need a full explanation

Answer 4

whats your differential equation ?

Answer 5

you want to solve the amount of salt in the tank
not the amount of "salt water" in the tank

Answer 6

say \(y(t)\) is your function for salt in the tank at any given time \(t\)
To start with we have `10 gallons of saltwater solution that has a salt concentration of 2 lbs/gallon.` so \(y(0) = 20\)

Answer 7

`A saltwater solution containing 3 lbs of salt per gallon is pumped into the tank at a rate of 2 gal/min`

2 gals/min and each gallon has 3lb of salt so incoming salt rate = \(6 ~\mathrm{lb / min}\)

Answer 8

` The well-mixed solution is pumped out of the tank at the rate of 3 gal/min.`

outgoing salt depends on amount of solution in the tank,
the outoing salt rate works out to : \(\frac{y(t)}{10-t}\times 3 ~\mathrm{lb / min}\)

Answer 9

putting the pieces together, the differential eqn/IVP should be
\[y' = 6 - \dfrac{3y}{10-t}~~~; y(0) = 20\]