Differential Equations

Calculus Tutorial

Differential Equations

Intro

Handling of Materials

Suppose that we have a tank containing water. Suppose also that we
have 2 pumps — one pump can drain the tank in 6 hours, and the
other can drain the tank in 5 hours. How long will it take to drain
the tank with both pumps running?

Since one pump can drain the tank in 5 hours, this pump can drain
\frac{1}{5} of the tank in one hour (assuming constant time rate
of pumping). Using the same reasoning, the other pump can drain
\frac{1}{6} of the tank in one hour. Therefore, with both pumps
running we can drain

(1)   \begin{eqnarray*}     \frac{1}{6} + \frac{1}{5} &=& \frac{5 + 6}{30} \nonumber\\     \frac{1}{6} + \frac{1}{5} &=& \frac{11}{30} \nonumber   \end{eqnarray*}

\frac{11}{30} of the tank in one hour. Therefore, to
find the amount of time it will take to drain 1 whole tank, we
divide 1 whole tank by the number of units of the tank that can be
drained in one hour;

(2)   \begin{eqnarray*}     \frac{1}{\frac{11}{30}} &=& \frac{30}{11} \nonumber\\     \frac{1}{\frac{11}{30}} &=& 30 \times \frac{9}{99} \nonumber\\     \frac{1}{\frac{11}{30}} &=& 30 \times 0.\overline{09} \nonumber\\     \frac{1}{\frac{11}{30}} &=& 2.\overline{72} \nonumber   \end{eqnarray*}

Therefore, it will take approximately 2.73 hours to drain the tank
with both pumps running.

Chemical Solutions

In the branch of science known as chemistry, a solution is made
when we mix one substance, the solute, with another substance,
the solvent. Up to a certain concentration (mass of
solute per unit volume of solvent), the volume of the final solution
is that of the solvent.

Suppose we have at our disposal two salt water solutions — one
solution has a concentration of 0.5 kilograms salt per cubic meter
of water, and the other has a concentration of 0.2 kilograms salt
per cubic meter of water. Suppose also that we wish to make 0.7
cubic meters of salt water solution of concentration 0.3 kilograms
salt per cubic meter of water. How much of each salt water solution
should be mixed together to yield the desired amount and
concentration of the final solution?

Let the volume of 0.5 kilograms salt per cubic meter of water
solution be V_1 and the volume of 0.2 kilograms salt per cubic
meter of water solution be V_2. From the problem statement, we
know that

(3)   \begin{eqnarray*}     V_1 + V_2 &=& 0.7   \end{eqnarray*}

and

(4)   \begin{eqnarray*}     0.5 V_1 + 0.2 V_2 &=& 0.7 \times 0.3 \nonumber\\     0.5 V_1 + 0.2 V_2 &=& 0.21   \end{eqnarray*}

Multiplying both members of Equation 4 by 2, we
have

(5)   \begin{eqnarray*}     V_1 + 0.4 V_2 &=& 0.42   \end{eqnarray*}

Subtracting from the left and right members of Equation
3 the left and right members of Equation
5, respectively, we have

(6)   \begin{eqnarray*}     0.6 V_2 &=& 0.28 \nonumber\\     V_2 &=& \frac{0.28}{0.6} \nonumber\\     V_2 &=& \frac{2.8^40^40^40}{6} \nonumber\\     V_2 &=& 0.4\overline{6}   \end{eqnarray*}

From Equations 3 and 6, we have

(7)   \begin{eqnarray*}     V_1 + 0.4\overline{6} &=& 0.7 \nonumber\\     V_1 &=& 0.7 - 0.4\overline{6} \nonumber\\     V_1 &=& 0.6\overline{9} - 0.4\overline{6} \nonumber\\     V_1 &=& 0.2\overline{3}   \end{eqnarray*}

Therefore, from Equations 6 and 7,
the approximated desired amounts of the 0.5 and 0.2 kilograms salt
per cubic meter of water solutions are, respectively, 0.233 and
0.467 cubic meters.

Suppose we have a tank of volume V and containing a solution with
an initial mass concentration of \rho_0 kilograms per cubic meter.
Let us also suppose that one pump is pumping in, at a rate of \mu
cubic meters per second, another solution, of the same type, with a
mass concentration of \rho_1 kilograms per cubic meter. The tank
is also constantly stirred, to maintain uniform mass concentration,
and is pumped out at the same rate of \mu cubic meters per second.
The problem is to find how the mass concentration \rho, of the
contents of the tank, is related to the amount of time t that both
of the pumps are in operation.

To begin to solve this problem, let us consider the time rate of
mass of dissolved substance being pumped in, \frac{d m_0}{d t},
the time rate of mass of dissolved substance being pumped out,
\frac{d m_1}{d t}, and the rate of change of the total mass of
dissolved substance in the tank, \frac{d m}{d t}. Quite clearly,
these three quantities are related such that

(8)   \begin{eqnarray*}     \frac{d m}{d t} &=& \frac{d m_0}{d t} - \frac{d m_1}{d t} \nonumber\\     \frac{d m}{d t} &=& \mu \rho_1 - \mu \rho \nonumber\\     \frac{d m}{d t} &=& \mu (\rho_1 - \rho) \nonumber\\     \frac{d m}{d t} &=& \mu \left( \rho_1 - \frac{m}{V} \right) \nonumber\\     \frac{d m}{\left( \rho_1 - \frac{m}{V} \right)} &=& \mu d t \nonumber\\     \frac{\frac{-1}{V} d m}{\left( \rho_1 - \frac{m}{V} \right)} &=& \frac{-\mu}{V} d t \nonumber\\     \ln \left( \rho_1 - \frac{m}{V} \right) &=& \frac{-\mu}{V} t + C \nonumber\\     \ln \left( \rho_1 - \rho \right) &=& \frac{-\mu}{V} t + C \nonumber\\     \rho_1 - \rho &=& k e^{\frac{-\mu}{V} t} \nonumber\\     \rho &=& \rho_1 - k e^{\frac{-\mu}{V} t} \nonumber   \end{eqnarray*}

Since \rho = \rho_0 at t = 0;

(9)   \begin{eqnarray*}     \rho_0 &=& \rho_1 - k \nonumber\\     k &=& \rho_1 - \rho_0 \nonumber   \end{eqnarray*}

Using the above value for k;

(10)   \begin{eqnarray*}     \rho &=& \rho_1 - (\rho_1 - \rho_0) e^{\frac{-\mu}{V} t}   \end{eqnarray*}

Equation 10 is the
solution to our problem.

Sample Problem

See the intro section.

Solution

See the intro section.



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