Question

\({\bf{11}}\). Consider the large mixing tanks shown in Figure \({\bf{8}}.{\bf{3}}.{\bf{1}}.\) suppose that both tanks \(A\) and \(B\) initially contain \({\bf{100}}\) gallons of brine.

Liquid is pumped in and out of the tanks as indicated in the figure; the mixture pumped between and out of the tanks is assumed to be well-stirred.

(a) Construct a mathematical model in the form of a linear system of tirst-order differential equations for the number of pounds \({x_1}(t)\)and \({x_2}(t)\) of salt in tanks \(A\) and\(B\), respectively, at time\(t\). Write the system in matrix form. [Hint: Review Section 3.3.]

(b) Use the method of undetermined coefficients to solve the linear system in part (a) subject to\({x_1}(0) = 60,{x_2}(0) = 10\).

(c) What are \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t)\) and\(\mathop {\lim }\limits_{t \to \infty } {x_2}(t)\) ? Interpret this result.

(d) Use a graphing utility to plot the graphs of \({x_1}(t)\)and \({x_2}(t)\) on the same coordinate axes.

Short answer

(a) The system in matrix form is \({X^\prime } = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}}&{\frac{1}{{100}}}\\{\frac{1}{{50}}}&{ - \frac{1}{{25}}}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

(b) The linear system is \(X(t) = \frac{{80}}{3}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{ - 0.02t}} - \frac{{70}}{3}\left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right){e^{ - 0.05t}} + \left( {\begin{array}{*{20}{l}}{10}\\{30}\end{array}} \right)\)

(c) The \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t)\) and \(\mathop {\lim }\limits_{t \to \infty } {x_2}(t)\)is \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t) = 10\)and \(\mathop {\lim }\limits_{t \to \infty } {x_2}(t) = 30\)

(d) The coordinate axes of the graph is

Step-by-step solution

Determine the complex Eigen values

The polynomial of degree \(n\) of the variable \(\lambda \) is the characteristic polynomial of the \(n \times n\) matrix\(A\).

\(p(\lambda ) = \det (\lambda I - A)\)

Its root is the eigenvalue of\(A\).

Determine the matrix form of the system

(a) For tank \(A,\)

\(\frac{{d{x_1}}}{{dt}} = \)Input Rate of Salt - Output Rate of Salt

\( = (2{\rm{gal}}/{\rm{min}})(0{\rm{lb}}/{\rm{gal}}) + (1{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_2}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right)\)

\( - (2{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_1}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right) - (1{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_1}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right)\)\( = \frac{{{x_2}}}{{100}}{\rm{lb}}/\min - \frac{{3{x_1}}}{{100}}{\rm{lb}}/\min \)

For tank\(B\),

\(\frac{{d{x_2}}}{{dt}} = {\rm{ Input Rate of Salt - Output Rate of Salt}}\)\({\rm{ }} = (2{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_1}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right) + (2{\rm{gal}}/{\rm{min}})\left( {\frac{1}{2}{\rm{lb}}/{\rm{gal}}} \right)\)\( - (1{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_2}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right) - (3{\rm{gal}}/{\rm{min}})\left( {\frac{{{x_2}}}{{100}}{\rm{lb}}/{\rm{gal}}} \right)\)

\( = \frac{{{x_1}}}{{50}}{\rm{lb}}/{\rm{min}} - \frac{{{x_2}}}{{25}}{\rm{lb}}/{\rm{min}} + 1{\rm{lb}}/{\rm{min}}\)

So now we have

\(\left\{ {\begin{array}{*{20}{l}}{\frac{{d{x_1}}}{{dt}} = - \frac{3}{{100}}{x_1} + \frac{1}{{100}}{x_2}}\\{\frac{{d{x_2}}}{{dt}} = \frac{1}{{50}}{x_1} - \frac{1}{{25}}{x_2} + 1}\end{array}} \right.\)

Which can be written in the matrix form

\({X^\prime } = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}}&{\frac{1}{{100}}}\\{\frac{1}{{50}}}&{ - \frac{1}{{25}}}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

Where

\(A = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}}&{\frac{1}{{100}}}\\{\frac{1}{{50}}}&{ - \frac{1}{{25}}}\end{array}} \right)\)

And

\(F(t) = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

Therefore, the system in matrix form is \({X^\prime } = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}}&{\frac{1}{{100}}}\\{\frac{1}{{50}}}&{ - \frac{1}{{25}}}\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

Determine the Eigen values of the matrix

Obtaining the eigenvalues of the matrix coefficient,

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{ - 0.03 - \lambda }&{0.01}\\{0.02}&{ - 0.04 - \lambda }\end{array}} \right| = 0\)

\(( - 0.03 - \lambda )( - 0.04 - \lambda ) - (0.01)(0.02) = 0\)

\(0.0012 + 0.03\lambda + 0.04\lambda + {\lambda ^2} - 0.0002 = 0\)

\({\lambda ^2} + 0.07\lambda + 0.001 = 0\)

So our eigenvalues are

\({\lambda _1} = - \frac{1}{{50}}\)

\( = - 0.02\)

And

\({\lambda _2} = - \frac{1}{{20}}\)

\( = - 0.05\)

Therefore, the Eigen values are \({\lambda _1} = - 0.02,{\lambda _2} = - 0.05\)

Determine the Eigen vector and a corresponding solution vector

For\({\lambda _1} = - 0.02\):

\((A + 0.02I\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 0.03 - ( - 0.02)}&{0.01}&0\\{0.02}&{ - 0.04 - ( - 0.02)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 0.01}&{0.01}&0\\{0.02}&{ - 0.02}&0\end{array}} \right)\)

Apply row operation \({R_2} + 2{R_1} \to {R_2}\)

\( = \left( {\begin{array}{*{20}{c}}{ - 0.01}&{0.01}&0\\0&0&0\end{array}} \right)\)

So from the first row we have

\( - 0.01{k_1} + 0.01{k_2} = 0\;\;\; \to \;\;\;{k_1} = {k_2}\)

Choosing \({k_2} = 1\)yields\({k_1} = 1\).

This gives an eigenvector and a corresponding solution vector

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{ - 0.02t}}\)

For \({\lambda _2} = - 0.05:\)

\((A + 0.05I\mid 0){\rm{ }} = \left( {\begin{array}{*{20}{c}}{ - 0.03 - ( - 0.05)}&{0.01}&0\\{0.02}&{ - 0.04 - ( - 0.05)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{0.02}&{0.01}&0\\{0.02}&{0.01}&0\end{array}} \right)\)

Apply row operation \({R_2} - {R_1} \to {R_2}\)

\( = \left( {\begin{array}{*{20}{c}}{0.02}&{0.01}&0\\0&0&0\end{array}} \right)\)

So from the first row we have

\(0.02{k_1} + 0.01{k_2} = 0\;\;\; \to \;\;\;{k_1} = - \frac{1}{2}{k_2}\)

Choosing \({k_2} = 2\)yields\({k_1} = - 1\).

This gives an eigenvector and a corresponding solution vector

\({K_2} = \left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right){e^{ - 0.05t}}\)

Hence, the complementary solution is \({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{ - 0.02t}} + {c_2}\left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right){e^{ - 0.05t}}\)

Determine the general solution of the system and the linear system of equation

Since\(F(t) = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\), we shall try to find a particular solution of the system that possesses the same form:

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\end{array}} \right)\)

Differentiating,

\(X_{\rm{p}}^\prime = \left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right)\)

Substituting into the system we obtained above,

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}}&{\frac{1}{{100}}}\\{\frac{1}{{50}}}&{ - \frac{1}{{25}}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - \frac{3}{{100}}{a_1} + \frac{1}{{100}}{a_2}}\\{\frac{1}{{50}}{a_1} - \frac{1}{{25}}{a_2}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

Here we have two equations,

\( - \frac{3}{{100}}{a_1} + \frac{1}{{100}}{a_2} = 0......(1)\)

\(\frac{1}{{50}}{a_1} - \frac{1}{{25}}{a_2} = - 1........(2)\)

Multiplying equation \(\left( 1 \right)\) by\(\frac{2}{3}\),

\( - \frac{1}{{50}}{a_1} + \frac{1}{{150}}{a_2} = 0\)

Adding the equation we obtained to equation\(\left( 2 \right)\),

\( - \frac{1}{{30}}{a_2} = - 1\;\;\; \to \;\;\;{a_2} = 30\)

To solve for \({a_1}\)we substitute this value we obtained for \({a_2}\) into one of the equations we have above,

\( - \frac{3}{{100}}{a_1} + \frac{1}{{100}}(30) = 0\;\;\; \to - \frac{3}{{100}}{a_1}{\rm{ }} = - \frac{3}{{10}}{a_1}{\rm{ }} = 10\)

Our particular solution is therefore,

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{10}\\{30}\end{array}} \right)\)

Finally, the general solution of the system is

\(X(t) = {X_{\bf{c}}} + {X_{\bf{p}}} = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{ - 0.02t}} + {c_2}\left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right){e^{ - 0.05t}} + \left( {\begin{array}{*{20}{l}}{10}\\{30}\end{array}} \right)\)

Now, applying the initial value condition

\(X(0) = \left( {\begin{array}{*{20}{l}}{60}\\{10}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}{60}\\{10}\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right) + \left( {\begin{array}{*{20}{l}}{10}\\{30}\end{array}} \right)\)

Here we have

\({c_1} - {c_2} = 50\)

\({\rm{ }}{c_1} + 2{c_2} = - 20\)

Multiplying \(\left( 1 \right)\) by \(2\) , we get

\(2{c_1} - 2{c_2} = 100\)

Adding equation \((2)\) to the equation we obtained,

\(3{c_1} = 80\;\;\; \to \;\;\;{c_1} = \frac{{80}}{3}\)

To solve for\({c_2}\), we substitute the value of \({c_1}\) that we obtained into one of the equations,

\(\frac{{80}}{3} + 2{c_2} = - 20\;\;\; \to \;\;\;{c_2} = - \frac{{70}}{3}\)

Therefore, the linear system of the equation is \(X(t) = \frac{{80}}{3}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{ - 0.02t}} - \frac{{70}}{3}\left( {\begin{array}{*{20}{r}}{ - 1}\\2\end{array}} \right){e^{ - 0.05t}} + \left( {\begin{array}{*{20}{l}}{10}\\{30}\end{array}} \right)\)

Determine  \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t)\) and  \(\mathop {\lim }\limits_{t \to \infty } {x_2}(t)\)

\(\mathop {\lim }\limits_{t \to \infty } {x_1}(t) = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{80}}{3}{e^{ - 0.02t}} + \frac{{70}}{3}{e^{ - 0.05t}} + 10} \right]\)

\( = 0 + 10\)

\( = 10\)

\(\mathop {\lim }\limits_{t \to \infty } {x_2}(t) = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{{80}}{3}{e^{ - 0.02t}} - \frac{{140}}{3}{e^{ - 0.05t}} + 30} \right]\)

\( = 0 + 30\)

\( = 30\)

Therefore, the \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t)\) and \(\mathop {\lim }\limits_{t \to \infty } {x_2}(t)\)is \(\mathop {\lim }\limits_{t \to \infty } {x_1}(t) = 10\)and \(\mathop {\lim }\limits_{t \to \infty } {x_2}(t) = 30\)

Determine the graph

The Coordinate of axes of the graph is