TeachingCalculus – First-order differential equations – Mixing problems
Examples
1) A tank contains \(100\;\text{g}\) salt dissolved in \(250\;\ell\) water. This solution is kept thoroughly mixed and drains from the tank at a rate of \(5\;\ell/\text{min}\). Simultaneously, brine with a concentration of \(10\;\text{g}/\ell\) enters the tank at the same rate of \(5\;\ell/\text{min}\). After how many minutes is the amount of salt in the tank equal to \(1300\;\text{g}\)?
Solution:
Let \(y(t)\) denote the amount of salt (in \(\text{g}\)) in the tank at time \(t\) (in \(\text{min}\)). Then we have: \(y(0)=100\) (initial
condition) and \(\displaystyle\frac{dy}{dt}=50−\frac{y}{50}\). This differential equation is both separable and linear. Note that \(y(t)=2500\)
is a (constant) solution of this differential equation. For \(y(t)\neq2500\) we have:
Integration now leads to
\[\int\frac{dy}{2500-y}=\int\frac{dt}{50}\quad\Longleftrightarrow\quad-\ln|2500-y|=\frac{1}{50}t+C.\]This implies that
\[2500-y(t)=\pm e^{-\frac{1}{50}t-C}=\pm e^{-C}\cdot e^{-\frac{1}{50}t}.\]Note that \(\pm e^{-C}\) is an arbitrary positive or negative constant. If we replace this by an arbitrary constant \(K\) we retrieve the "lost" constant solution \(y(t)=2500\): \(y(t)=2500-Ke^{-\frac{1}{50}t}\). Finally, the initial condition \(y(0)=100\) leads to \(K=2400\).
So the solution of the initial-value problem is \(y(t)=2500-2400e^{-\frac{1}{50}t}\).
Now we are able to answer the question:
\[y(t)=1300\quad\Longleftrightarrow\quad e^{-\frac{1}{5}t}=\tfrac{1}{2}\quad\Longleftrightarrow\quad t=50\ln(2),\]which implies that after \(50\ln(2)\approx35\) minutes the amount of salt in the tank will be \(1300\;\text{g}\). It is easy to check that
\[y(50\ln(2))=2500-2400e^{-\ln(2)}=2500-1200=1300.\]2) A tank contains \(100\;\ell\) beer with \(5\%\) alcohol. Beer with \(7\%\) alcohol is pumped into the tank at a rate of \(1\;\ell/\text{min}\). The fluid in the tank is kept thoroughly mixed and drains from the tank at a rate of \(1\;\ell/\text{min}\). What is the alcohol percentage of the beer in the tank after \(1\) hour (\(60\) minutes)?
Solution:
Let \(y(t)\) denote the amount of alcohol (in \(\ell\)) in the tank at time \(t\) (in \(\text{min}\)). Then we have: \(y(0)=5\) (initial condition) and
\(\displaystyle\frac{dy}{dt}=\frac{7}{100}-\frac{y}{100}\). This differential is both separable and linear. Note that \(y(t)=7\) is a (constant)
solution of this differential equation. For \(y(t)\neq7\) we have:
Integration now leads to
\[\int\frac{dy}{7-y}=\int\frac{dt}{100}\quad\Longleftrightarrow\quad-\ln|7-y|=\frac{1}{100}t+C.\]This implies that
\[7-y(t)=\pm e^{-\frac{1}{100}t-C}=\pm e^{-C}\cdot e^{-\frac{1}{100}t}.\]Note that \(\pm e^{-C}\) is an arbitrary positive or negative constant. If we replace this by an arbitrary constant \(K\) we retrieve the "lost" constant solution \(y(t)=7\): \(y(t)=7-Ke^{-\frac{1}{100}t}\). Finally, the initial condition \(y(0)=5\) leads to \(K=2\).
So the solution of the initial-value problem is \(y(t)=7-2e^{-\frac{1}{100}t}\).
Now we obtain that \(y(60)=7-2e^{-\frac{3}{5}}\approx5.9\). Since the volume of the fluid in the tank is \(100\;\ell\), this is the alcohol percentage after \(1\) hour.
3) A tank contains \(100\;\ell\) brine with \(100\;\text{g}\) salt dissolved in water. This solution is kept thoroughly mixed and drains from the tank at a rate of \(3\;\ell/\text{min}\). Simultaneously, brine with a concentration of \(10\;\text{g}/\ell\) enters the tank at a rate of \(1\;\ell/\text{min}\). What is the amount of salt (in \(\text{g}\)) in the tank after \(25\) minutes?
Solution:
Let \(y(t)\) denote the amount of salt (in \(\text{g}\)) in the tank at time \(t\) (in \(\text{min}\)). Then we have: \(y(0)=100\) (initial
condition) and \(\displaystyle\frac{dy}{dt}=10−3\cdot\frac{y}{100-2t}\). This differential equation is linear (and not separable). So we look
for an integrating factor \(I(t)\):
Hence: \(I(t)=\exp\left(-\frac{3}{2}\ln(100-2t)\right)=(100-2t)^{-\frac{3}{2}}\) for instance. Then we have:
\[\frac{d}{dt}\left[(100-2t)^{-\frac{3}{2}}y(t)\right]=10(100-2t)^{-\frac{3}{2}}\quad\Longrightarrow\quad (100-2t)^{-\frac{3}{2}}y(t)=10(100-2t)^{-\frac{1}{2}}+C\]and therefore: \(y(t)=10(100-2t)+C(100-2t)^{\frac{3}{2}}\). The initial condition \(y(0)=100\) leads to: \(100=1000+1000C\) or equivalently \(C=-\frac{900}{1000}=-\frac{9}{10}\). Hence: \(y(t)=10(100-2t)-\frac{9}{10}(100-2t)^{\frac{3}{2}}\).
For \(t=25\) we obtain: \(y(25)=500-\frac{9}{10}\cdot50\sqrt{50}=500-225\sqrt{2}\).
Remark: Note that the differential equation only holds for \(0 < t < 50\). After \(50\) minutes the tank will be empty.
4) A tank with a volume of \(500\;\ell\) contains \(25\;\text{g}\) salt dissolved in \(100\;\ell\) water. This solution is kept thoroughly mixed and drains from the tank at a rate of \(1\;\ell/\text{min}\). Simultaneously, brine with a concentration of \(5\;\text{g}/\ell\) enters the tank at a rate of \(2\;\ell/\text{min}\). What is the amount of salt (in \(\text{g}\)) in the tank after \(25\) minutes?
Solution:
Let \(y(t)\) denote the amount of salt (in \(\text{g}\)) in the tank at time \(t\) (in \(\text{min}\)). Then we have: \(y(0)=25\) (initial
condition) and \(\displaystyle\frac{dy}{dt}=10−1\cdot\frac{y}{100+t}\). This differential equation is linear (and not separable). So we look
for an integrating factor \(I(t)\):
Hence: \(I(t)=\exp\left(\ln(100+t)\right)=100+t\) for instance. Then we have:
\[\frac{d}{dt}\left[(100+t)y(t)\right]=10(100+t)=1000+10t\quad\Longrightarrow\quad (100+t)y(t)=1000t+5t^2+C.\]The initial condition \(y(0)=25\) leads to \(C=2500\). Hence: \(y(t)=\displaystyle\frac{5t^2+1000t+2500}{100+t}\).
For \(t=25\) we obtain: \(y(25)=\displaystyle\frac{5\cdot625+25000+2500}{125}=25+200+20=245\).
Remark: Note that the differential equation only holds for \(0 < t < 400\). After \(400\) minutes the tank will overflow.
Last modified on March 8, 2021
