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Math 0250 Exam 2Solutions:

November 6, 1998

F. Beatrous

1.
Find eigenpairs for the matrix

\begin{displaymath}\left[\begin{array}{cc}a&a^2\\ 1&a\end{array}\right]
\end{displaymath}

where a is a non-zero number.

Solution:
The characteristic polynomial is

\begin{displaymath}\det(A-\lambda I) = \left\vert\begin{array}{cc}a-\lambda&a^2\\ 1&a-\lambda
\end{array}\right\vert
= \lambda(\lambda-2a)
\end{displaymath}

so the eigenvalues are 0 and 2a.

For the eigenvalue $\lambda=0$,

\begin{displaymath}A-\lambda I = \left[\begin{array}{cc}a&a^2\\ 1&a\end{array}\r...
...\rightarrow \left[\begin{array}{rr}1&a\\ 0&0\end{array}\right]
\end{displaymath}

which gives the eigenvector

\begin{displaymath}\left[\begin{array}{c}a\\ -1\end{array}\right]
\end{displaymath}

and the eigenpair

\begin{displaymath}\left(0,\left[\begin{array}{c}a\\ -1\end{array}\right]\right)
\end{displaymath}

For the eigenvalue $\lambda=2a$,

\begin{displaymath}A-\lambda I = \left[\begin{array}{cc}-a&a^2\\ 1&-a\end{array}...
...rightarrow \left[\begin{array}{cc}1&-a\\ 0&0\end{array}\right]
\end{displaymath}

which gives the eigenvector

\begin{displaymath}\left[\begin{array}{c}a\\ 1\end{array}\right]
\end{displaymath}

and the eigenpair

\begin{displaymath}\left(2a,\left[\begin{array}{c}a\\ 1\end{array}\right]\right)
\end{displaymath}

2.
 
(a)
 The matrix

\begin{displaymath}A= \left[\begin{array}{rr}1&1\\ 4&1\end{array}\right]
\end{displaymath}

has eigenvalues 3 and -1 with corresponding eigenvectors

\begin{displaymath}\left[\begin{array}{r}1\\ 2\end{array}\right] \quad\hbox{and}\quad
\left[\begin{array}{r}1\\ -2\end{array}\right]
\end{displaymath}

Find the general solution to the system X'=AX.

Solution:

\begin{displaymath}X = c_1 \left[\begin{array}{r}1\\ 2\end{array}\right] e^{3t}
+ c_2 \left[\begin{array}{r}1\\ -2\end{array}\right] e^{-t}.
\end{displaymath}

(b)
Solve the initial value problem

\begin{displaymath}X' = A X
\qquad X(0) = \left[\begin{array}{r}2\\ 3\end{array}\right]
\end{displaymath}

Solution:
Substituting the initial condition into the general solution from 2a gives

\begin{displaymath}\left[\begin{array}{rr}1&1\\ 2&-2\end{array}\right]
\left[\b...
...rray}\right] =
\left[\begin{array}{r}2\\ 3\end{array}\right]
\end{displaymath}

Solving for c1 and c2 gives

\begin{displaymath}\left[\begin{array}{c}c_1\\ c_2\end{array}\right] =
\left[\begin{array}{c}1/4\\ 1/4\end{array}\right]
\end{displaymath}

so the required solution is

\begin{displaymath}X = \frac74 \left[\begin{array}{r}1\\ 2\end{array}\right] e^{...
...\frac14 \left[\begin{array}{r}1\\ -2\end{array}\right] e^{-t}
\end{displaymath}

3.
(a)
 With A as in problem 2, find all solutions to the system

\begin{displaymath}X' = A X + \left[\begin{array}{r}2\\ 1\end{array}\right] e^{-2t}.
\end{displaymath}

Solution: Look for a particular solution of the form X=Ce-2t. Substituting into the above system gives

\begin{displaymath}-2 C e^{-2t} = A C e^{-2t} +
\left[\begin{array}{r}2\\ 1\end{array}\right] e^{-2t}
\end{displaymath}

and canceling the exponential factors gives

\begin{displaymath}-2 C = A C +
\left[\begin{array}{r}2\\ 1\end{array}\right]
\end{displaymath}

and rearranging gives

\begin{displaymath}(A+2I)C = \left[\begin{array}{r}-2\\ -1\end{array}\right]
\end{displaymath}

Solving gives

\begin{displaymath}C = \left[\begin{array}{r}-1\\ 1\end{array}\right]
\end{displaymath}

so the required particular solution is

\begin{displaymath}X_p = \left[\begin{array}{r}-1\\ 1\end{array}\right] e^{-2t}
\end{displaymath}

The general solution is obtained by adding the complementary solution from problem 2a:

\begin{displaymath}X = X_p + X_h = \left[\begin{array}{r}-1\\ 1\end{array}\right...
...
+ c_2 \left[\begin{array}{r}1\\ -2\end{array}\right] e^{-t}.
\end{displaymath}

(b)
Find the solution to the system of part 3a satisfying the initial condition

\begin{displaymath}X(0) = \left[\begin{array}{r}0\\ 0\end{array}\right].
\end{displaymath}

Solution:
Substituting the initial condition into the general solution form 3a gives

\begin{displaymath}\left[\begin{array}{r}-1\\ 1\end{array}\right] +
\left[\beg...
...array}\right]
= \left[\begin{array}{r}0\\ 0\end{array}\right]
\end{displaymath}

and solving gives

\begin{displaymath}\left[\begin{array}{c}c_1\\ c_2\end{array}\right]
= \left[\begin{array}{c}1/4\\ 3/4\end{array}\right]
\end{displaymath}

so

\begin{displaymath}X = \left[\begin{array}{r}-1\\ 1\end{array}\right] e^{-2t}
+...
...\frac34 \left[\begin{array}{r}1\\ -2\end{array}\right] e^{-t}.
\end{displaymath}

4.
Two 100 gallon tanks are connected by pipes as indicated. Both tanks are initially filled with pure water. Starting at time t=0 minutes, a brine solution containing 10 pounds of salt per gallon of solution is pumped into the left tank at a rate of 10 gallons per minute. Fluid is pumped from the left tank to the right tank at a rate of 15 gallons per minute, and from the right to the left at a rate of 5 gallons per minute. Fluid is allowed to drain from the right tank at a rate of 10 gallons per minute. Set up a system of differential equations and initial conditions for the amount of salt in each tank. What is the coefficient matrix for the system? What is the forcing function? Do not solve the system.

\includegraphics{tanks.eps}

Solution: Let x1 and x2 denote the amount of salt (in pounds) in the left and right tank respectively. Then the salt concentrations in the two tanks are x1/100 and x2/100. The rate of change of the amount of salt in the left tank is

\begin{eqnarray*}\frac{dx_1}{dt}
&=& -(15 \hbox{ gal/min})\left(\frac{x_1}{100}...
...eft(10 \hbox{ lb/gal}\right)\\
&=& -0.15 x_1 + 0.05 x_2 + 100.
\end{eqnarray*}


Similarly, the rate of change in the right tank is

\begin{eqnarray*}\frac{dx_2}{dt} &=&(15 \hbox{ gal/min})\left(\frac{x_1}{100} \h...
...frac{x_2}{100} \hbox{ lb/gal}\right)\\
&=& 0.15 x_1 - 0.15 x_2
\end{eqnarray*}


In matrix notation,

\begin{displaymath}X' = \left[\begin{array}{cc}-0.15&0.05\\ 0.15&-0.15\end{array}\right] X
+ \left[\begin{array}{c}100\\ 0\end{array}\right]
\end{displaymath}

Since the initial amount of salt in both tanks is zero, the initial condition is

\begin{displaymath}X(0) = \left[\begin{array}{r}0\\ 0\end{array}\right]
\end{displaymath}

The coefficient matrix is

\begin{displaymath}\left[\begin{array}{cc}-0.15&0.05\\ 0.15&-0.15\end{array}\right]
\end{displaymath}

and the forcing function is

\begin{displaymath}\left[\begin{array}{c}100\\ 0\end{array}\right]
\end{displaymath}



 
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Frank Beatrous
1998-11-10