Let
denote the amount of salt (in lbs) in the tank at time
min up to the 10th minute. The tank starts with 100 gal of fresh water, so
.
Salt flows into the tank at a rate of

and flows out with rate

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

Solving with an integrating factor, we have



By the fundamental theorem of calculus, integrating both sides yields



After 10 min, the tank contains

of salt.
Now let
denote the amount of salt in the tank at time
min after the first 10 minutes have elapsed, with initial value
.
Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

so that
is given by the differential equation

We solve this equation in exactly the same way.





After another 10 min, the tank has

lb of salt.