Answer:
the required differential equation is dA/dt - [tex]\frac{2A}{t+500}[/tex] = 6
A(0) = 50lb
Step-by-step explanation:
Given the data in the question;
A(t) is the mass of the salt in in the tank at time t
the new volume of the solution in the tank increases by one gallon each minute
so
dA/dt = ( Rate in) - ( Rate out)
dA/dt = ( 3 gal/min × 2 lb/gal ) - ( 2 gal/min × [tex]\frac{A}{t+500}[/tex] lb/gal )
dA/dt = 6 - [tex]\frac{2A}{t+500}[/tex]
dA/dt - [tex]\frac{2A}{t+500}[/tex] = 6
A(0) = 50lb
Therefore, the required differential equation is dA/dt - [tex]\frac{2A}{t+500}[/tex] = 6
A(0) = 50lb