Solve the problem.
-A tank initially contains 120 gal of brine in which 40 lb of salt are dissolved. A brine containing 2 lb/gal of salt runs into the tank at the rate of 9 gal/min. The mixture is kept uniform by stirring and flows out of the tank at
The rate of 6 gal/min. Write, in standard form, the differential equation that models the mixing process.
A) $\frac { d y } { d t } = 18 - \frac { 6 } { 120 + 3 t } y$
B) $\frac { d y } { d t } + \frac { 9 } { 120 + 3 t } y = 12$
C) $\frac { d y } { d t } + \frac { 6 } { 120 + 3 t } y = 18$
D) $\frac { d y } { d t } + \frac { 6 } { 120 - 3 t } y = 18$

Question

Solve the problem.
-A tank initially contains 120 gal of brine in which 40 lb of salt are dissolved. A brine containing 2 lb/gal of salt runs into the tank at the rate of 9 gal/min. The mixture is kept uniform by stirring and flows out of the tank at
The rate of 6 gal/min. Write, in standard form, the differential equation that models the mixing process. 
A) $\frac { d y } { d t } = 18 - \frac { 6 } { 120 + 3 t } y$
B) $\frac { d y } { d t } + \frac { 9 } { 120 + 3 t } y = 12$
C) $\frac { d y } { d t } + \frac { 6 } { 120 + 3 t } y = 18$
D) $\frac { d y } { d t } + \frac { 6 } { 120 - 3 t } y = 18$

Quizplus · Accepted Answer

The rate of salt entering the tank is $9 \text{ gal/min} \times 2 \text{ lb/gal} = 18 \text{ lb/min}$. The rate of salt leaving the tank is proportional to the concentration of salt in the tank, which is $y/(120 + 3t)$ lb/gal (since the volume increases by 3 gal/min due to the difference between the inflow and outflow rates), times the outflow rate of 6 gal/min, giving a rate of $(6y)/(120 + 3t)$ lb/min. Thus, the differential equation modeling the salt amount $y$ over time $t$ is $\frac{dy}{dt} + \frac{6}{120 + 3t}y = 18$.

Solve the Problem dydt=18−6120+3ty\frac { d y } { d t } = 18 - \frac { 6 } { 120 + 3 t } ydtdy​=18−120+3t6​y

Solve the Problem $\frac { d y } { d t } = 18 - \frac { 6 } { 120 + 3 t } y$