A tank contains 300 gal of salt free water. A brine containing .5lb of salt per gallon of water runs into the tank at the rate of 2 gal/min and the well stirred mixtur runs out at the rate of 2 gal/min. What is the concentration of the salt in the tank at the end of the 10 min?

$\displaystyle A' = 2gal/min * .5lb/gal - 2gal/min * Alb/300gal$

$\displaystyle A' = 1 - (1/150)A$

$\displaystyle \frac{1}{1-1/150A} dA/dt = 1$

$\displaystyle \int t dt = \int \frac{150}{150-A} dA$

$\displaystyle t + c = 150 * ln |150-A|$

$\displaystyle \frac{t + c}{150} = ln |150-A|$

$\displaystyle +-e^{\frac{t + c}{150}} = 150-A$

$\displaystyle +-Ke^{\frac{t}{150}} = 150-A$

$\displaystyle +-150e^{\frac{t}{150}} = 150-A$

$\displaystyle A(t) = 150 +- 150 e^{t/150}$

Are my steps correct?