Application of Differential Equation: mixture problem

Volume at any time
$V = 600 + (1 - 2)t$

$V = 600 - t$
 

Let Q = amount of salt (in lbs) in the tank at any time.
$\dfrac{Q}{600 - t}$ = concentration of salt (in lbs/gal) at any time.
 

$\dfrac{dQ}{dt} = 0 - \dfrac{Q}{600 - t}(2)$

$\displaystyle \int \dfrac{dQ}{Q} = 2 \int \dfrac{-dt}{600 - t}$

$\ln Q = 2 \ln (600 - t) + \ln C$

$\ln Q = \ln (600 - t)^2 + \ln C$

$\ln Q = \ln C(600 - t)^2$

$Q = C(600 - t)^2$
 

When t = 0, Q = 1500 lbs
$1500 = C(600 - 0)^2$

$C = \frac{1}{240}$
 

Hence,
$Q = \frac{1}{240}(600 - t)^2$
 

When t = 1 hr = 60 min
$Q = \frac{1}{240}(600 - 60)^2 = 1215 ~ \text{lbs}$
 

When t = 9 hrs and 59 min = 599 min
$Q = \frac{1}{240}(600 - 599)^2 = \frac{1}{240} ~ \text{lbs}$