It appears you have two pipes pumping brine into the tank totaling .09 kg
at 15L/min.
The rate out is 15L/min.
Let y(t) be the amount of salt at time t. We want the amount of salt at time t and t=60.
\(\displaystyle \L\\\frac{dy}{dt}\)=rate in-rate out. [1]
rate in=(.09 kg/L)(15 L/min)=\(\displaystyle \L\\\frac{27}{20}\)kg/min.
At time t, the mixture contains y(t) lbs of salt in 1000 L of water, so the
concentration of salt at time t is \(\displaystyle \L\\\frac{y(t)}{1000 kg/L}\)
rate out=\(\displaystyle \L\\(\frac{y(t)}{1000} kg/L)(15 L/min)=\frac{3y(t)}{200} kg/min\)
So we can write [1] as:
\(\displaystyle \L\\\frac{dy}{dt}=\frac{27}{20}-\frac{3y}{200}\)
\(\displaystyle \L\\\frac{dy}{dt}+\frac{3y}{200}=\frac{27}{20}\) [2]
Also, we have the initial condition y(0)=0.
Now, find the integrating factor:
\(\displaystyle \L\\e^{\int(\frac{3}{200})dt}=e^{\frac{3t}{200}}\)
Multiply both sides of [2] by the IF:
\(\displaystyle \L\\\frac{d}{dt}(e^{\frac{3t}{200}}y)=\frac{27}{20}e^{\frac{3t}{200}}\)
Integrate:
\(\displaystyle \L\\e^{\frac{3t}{200}}y=\int(\frac{27}{20})e^{\frac{3t}{200}}=90e^
{\frac{3t}{200}}+C\)
Solve for y:
\(\displaystyle \L\\y=90+Ce^{\frac{-3t}{200}}\)
From our initial condition:
\(\displaystyle \L\\90+C=0\)
so, C=-90
\(\displaystyle \H\\y(t)=90-90e^{\frac{-3t}{200}}\)
So at t=60 the amount of salt in the tank is:
\(\displaystyle \H\\y(60)=90-90e^{\frac{-3(60)}{200}}=53.41 kg\)
Please check my work.
