4.4E: Exercises

In Exercises \((4.4E.1)\) to \((4.4E.15)\), find the general solution.

Exercise \(\PageIndex{1}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rr} 1&2\\2&1\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{2}\)

\(\displaystyle{{\bf y}'= {1\over4}\left[\begin{array}{rr}-5&3 \\3&-5\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{1\over4}{\left|\begin{array}{cc}-5-4\lambda&3\\
3&-5-4\lambda\end{array}\right|}
=(\lambda+1/2)(\lambda+2)\).
Eigenvectors  associated with \(\lambda_1=-1/2\)  satisfy
\(\displaystyle{\left[\begin{array}{rr}-3&3\\-4&4
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=x_2\).  Taking \(x_2=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{c}1 \\
1\end{array}\right]}e^{-t/2}\).
Eigenvectors  associated with \(\lambda_2=-2\)
satisfy
\(\displaystyle{\left[\begin{array}{rr}{3\over4}&{3\over4}\\1&1
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=-x_2\).  Taking \(x_2=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{r} -1\\
1\end{array}\right]}e^{-2t}\). Hence
 \({\bf y}=\displaystyle{ c_1\left[\begin{array}{r} 1\\1\end{array}
\right]e^{-t/2}+c_2\left[\begin{array}{r}-1\\1\end{array}
\right]e^{-2t}}\).

Exercise \(\PageIndex{3}\)

\(\displaystyle{{\bf y}'= {1\over5}\left[\begin{array}{rr}-4&3\\ -2&-11\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{4}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rr}-1&-4\\-1&-1\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{cc}-1-\lambda&-4\\-1&-1-\lambda\end{array}\right|}
=(\lambda-1)(\lambda+3)\).
Eigenvectors  associated with \(\lambda_1=-3\) satisfy
\(\displaystyle{\left[\begin{array}{rr}2&-4\\1&-2
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=2x_2\).  Taking \(x_2=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{c}
2\\1\end{array}\right]}e^{-3t}\).
Eigenvectors  associated with \(\lambda_2=1\)  satisfy
\(\displaystyle{\left[\begin{array}{rr}-2&-4\\-1&-2
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=-2x_2\).  Taking \(x_2=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{r} -2\\1\end{array}\right]}e^t\).
Hence  \({\bf y}=\displaystyle{ c_1\left[\begin{array}{r} 2\\1\end{array}
\right]e^{-3t}+c_2\left[\begin{array}{r}-2\\1\end{array}
\right]e^t}\).

Exercise \(\PageIndex{5}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rr} 2&-4\\-1&-1\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{6}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rr} 4&-3\\2&-1\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{cc}4-\lambda&-3\\2&-1-\lambda\end{array}\right|}
=(\lambda-2)(\lambda-1)\).
Eigenvectors  associated with \(\lambda_1=2\)  satisfy
\(\displaystyle{\left[\begin{array}{rr}2&-3\\2&-3
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=\displaystyle{3\over2}x_2\).  Taking \(x_2=2\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{r}3\\2\end{array}\right]}e^{2t}\).
Eigenvectors  associated with \(\lambda_2=1\) satisfy
\(\displaystyle{\left[\begin{array}{rr}3&-3\\2&-2
\end{array}\right]\left[\begin{array}{r}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=x_2\).  Taking \(x_2=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{r}1\\1\end{array}\right]}e^t\).
Hence
 \({\bf y}=\displaystyle{c_1\left[\begin{array}{r}3\\2\end{array}
\right]e^{2t}+c_2\left[\begin{array}{r}1\\1\end{array}\right]e^t}\).

Exercise \(\PageIndex{7}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rr}-6&-3\\1&-2\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{8}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr} 1&-1&-2\\1&-2&-3\\-4&1&-1\end{array}\right] {\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}1-\lambda&-1&-2\\1&-2-\lambda&-3 \\
-4&1&-1-\lambda\end{array}\right|}=-(\lambda+3)(\lambda+1)(\lambda-2)\).
The eigenvectors associated with
 with \(\lambda_1=-3\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}4&-1&-2&\vdots&0\\1&1
&-3&\vdots&0\\-4&1&2&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&-2&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_3\) and \(x_2=2x_3\).  Taking \(x_3=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr} 1\\2\\1
\end{array}\right]}e^{-3t}\).
The eigenvectors associated with
 with \(\lambda_2=-1\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}2&-1&-2&\vdots&0\\1&-1
&-3&\vdots&0\\-4&1&0&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&4&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-x_3\) and \(x_2=-4x_3\).  Taking \(x_3=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}-1\\-4\\ 1
\end{array}\right]}e^{-t}\).
The eigenvectors associated with
 with \(\lambda_3=2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}-1&-1&-2&\vdots&0\\1&
-4&-3&\vdots&0\\-4&1&-3&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-x_3\) and \(x_2=-x_3\).  Taking \(x_3=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}-1\\-1\\1
\end{array}\right]}e^{2t}\). Hence
 \({\bf y}=\displaystyle{c_1\left[\begin{array}{r} 1\\2\\1
\end{array}\right]e^{-3t}+c_2\left[\begin{array}{r}-1\\-4\\1
\end{array}
\right]e^{-t}+c_3\left[\begin{array}{r}-1\\-1\\1\end{array}
\right]e^{2t}}\).

Exercise \(\PageIndex{9}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr} -6&-4&-8\\-4&0&-4\\-8&-4&-6\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{10}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr}3&5&8\\1&-1& -2\\-1&-1&-1\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}3-\lambda&5&8\\1&-1-\lambda&-2\\
-1&-1&-1-\lambda\end{array}\right|}=-(\lambda-1)(\lambda+2)(\lambda-2)\).
The eigenvectors associated with
 with \(\lambda_1=1\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}2&5&8&\vdots&0\\1&-2
&-2&\vdots&0\\-1&-1&-2&\vdots&0
\end{array}\right]}\),
which is row equivalent to \(\displaystyle{\left[\begin{array}{rrrcr}1&0&\frac{2}{3}&\vdots&0\\0 &1& \frac{4}{3} &\vdots &0\\0 & 0 & 0&\vdots &0
\end{array}\right]}\)

Hence \(x_1=-{2\over3}x_3\) and \(x_2=-{4\over3}x_3\).  Taking \(x_3=3\)
yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}-2\\-4\\3
\end{array}\right]}e^t\).
The eigenvectors associated with
 with \(\lambda_2=-2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}5&5&8&\vdots&0\\1&1
&-2&\vdots&0\\-1&-1&1&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&1&0&\vdots&0\\0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-x_2\) and \(x_3=0\).  Taking \(x_2=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr} -1\\1\\0
\end{array}\right]}e^{-2t}\).
The eigenvectors associated with
 with \(\lambda_3=2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}1&5&8&\vdots&0\\1&-3
&-2&\vdots&0\\-1&-1&-3&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&{7\over4}&\vdots&0\\0&1&{5\over4}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-{7\over4}x_3\) and \(x_2=-{5\over4}x_3\).  Taking \(x_3=4\)
yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}-7\\-5\\4
\end{array}\right]}e^{2t}\). Hence
\({\bf y}=\displaystyle{ c_1\left[\begin{array}{r}-2\\-4\\
3\end{array}\right]e^t+c_2\left[\begin{array}{r}-1\\1\\0
\end{array}\right]e^{-2t}+c_3\left[\begin{array}{r}-7\\-5\\4
\end{array}\right]e^{2t}}\).

 

Exercise \(\PageIndex{11}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr} 1&-1&2\\12&-4 & 10\\-6&1&-7 \end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{12}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr} 4&-1&-4\\4&-3&-2\\1&-1&-1\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}4-\lambda&-1&-4\\4&-3-\lambda&-2\\
1&-1&-1-\lambda\end{array}\right|}=-(\lambda-3)(\lambda+2)(\lambda+1)\).
The eigenvectors associated
 with \(\lambda_1=3\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}1&-1&-4&\vdots&0\\4&-6
&-2&\vdots&0\\1&-1&-4&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-11&\vdots&0\\0&1&-7&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence  \(x_1=11x_3\) and \(x_2=7x_3\).  Taking \(x_3=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}11\\7\\1
\end{array}\right]}e^{3t}\).
The eigenvectors associated
 with \(\lambda_2=-2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}6&-1&-4&\vdots&0\\4&-1
&-2&\vdots&0\\1&-1&1&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&-2&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_3\) and \(x_2=2x_3\).  Taking \(x_3=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}1\\2\\1
\end{array}\right]}e^{-2t}\).
The eigenvectors associated
 with \(\lambda_3=-1\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}5&-1&-4&\vdots&0\\4&-2
&-2&\vdots&0\\1&-1&0&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&-1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence  \(x_1=x_3\) and \(x_2=x_3\).  Taking \(x_3=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}1\\1\\1
\end{array}\right]}e^{-t}\). Hence
 \({\bf y}=\displaystyle{c_1\left[\begin{array}{r}11\\\phantom{1}7
\\\phantom{1}1\end{array}\right]e^{3t}+c_2\left[\begin{array}{r}1\\2
\\1\end{array}\right]e^{-2t}+c_3\left[\begin{array}{r}1\\1\\1
\end{array}\right]e^{-t}}\).

Exercise \(\PageIndex{13}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr}-2&2&-6\\2&6&2\\-2&-2& 2\end{array}\right]{\bf y}}\)

Exercise \(\PageIndex{14}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr}3&2&-2\\-2&7&-2\\ -10&10&-5\end{array}\right]{\bf y}}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}3-\lambda&2&-2\\-2&7-\lambda&-2\\-10
&10&-5-\lambda\end{array}\right|}=-(\lambda+5)(\lambda-5)^2\).
The eigenvectors associated
 with \(\lambda_1=-5\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}8&2&-2&\vdots&0\\-2&12
&-2&\vdots&0\\-10&10&0&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-{1\over5}&\vdots&0\\0&1&-{1\over5}&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence  \(x_1=\displaystyle{1\over5}x_3\) and \(x_2=\displaystyle{1\over5}x_3\).  Taking
\(x_3=5\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}1\\1\\5
\end{array}\right]}e^{-5t}\).
The eigenvectors associated
 with \(\lambda_2=5\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}-2&2&-2&\vdots&0\\-2&2
&-2&\vdots&0\\-10&10&-10&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&-1&1&\vdots&0\\0&0&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_2-x_3\).  Taking \(x_2=0\) and \(x_3=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}-1\\0\\1
\end{array}\right]}e^{5t}\).
 Taking \(x_2=1\) and \(x_3=0\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{5t}\). Hence
 \({\bf y}=\displaystyle{c_1\left[\begin{array}{r}1\\1\\5
\end{array}\right]e^{-5t}+c_2\left[\begin{array}{r}-1\\0\\
 1\end{array}
\right]e^{5t}+c_3\left[\begin{array}{r}1\\1\\0\end{array}
\right]e^{5t}}\).

Exercise \(\PageIndex{15}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr}3&1&-1\\3&5&1\\-6&2&4\end{array} \right]{\bf y}}\)

In Exercises \((4.4E.16)\) to \((4.4E.27)\), solve the initial value problem.

Exercise \(\PageIndex{16}\)

\({\bf y}'= \left[ \begin{array} \\ {-7} & 4 \\ {-6} & 7 \end{array} \right] {\bf y},\quad {\bf y}(0)= \left[ \begin{array} \\ 2 \\ {-4} \end{array} \right] \)

Answer

\(\displaystyle{\left|\begin{array}{cc}-7-\lambda&4\\-6&7-\lambda\end{array}\right|}
=(\lambda-5)(\lambda+5)\).
Eigenvectors  associated with \(\lambda_1=5\)  satisfy
\(\displaystyle{\left[\begin{array}{rr}-12&4\\-6&2
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=\displaystyle{x_2\over3}\).  Taking \(x_2=3\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{c} 1\\3\end{array}\right]}e^{5t}\).
Eigenvectors  associated with \(\lambda_2=5\)
satisfy
\(\displaystyle{\left[\begin{array}{rr}-2&4\\-6&12
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=2x_2\).  Taking \(x_2=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{c}2\\1\end{array}\right]}e^{-5t}\).
The general solution is
\({\bf y}=c_1\displaystyle{\left[\begin{array}{c}
1\\3\end{array}\right]}e^{5t}+
c_2\displaystyle{\left[\begin{array}{c}2\\1\end{array}\right]}e^{-5t}\).
Now \({\bf y}(0)=\displaystyle{\left[\begin{array}{c}2\\{-4}\end{array}\right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{c}
1\\3\end{array}\right]}+
c_2\displaystyle{\left[\begin{array}{c}2\\1\end{array}\right]}=\displaystyle{\left[\begin{array}{c}2\\{-4}\end{array}\right]}\),
so \(c_1=-2\) and \(c_2=2\). Therefore,
\({\bf y}=\displaystyle{-\left[\begin{array}{c}2\\6\end{array}\right] e^{5t}+\left[\begin{array}{c}4\\2 \end{array}\right] e^{-5t}}\).

Exercise \(\PageIndex{17}\)

\(\displaystyle{{\bf y}'={1\over6} \left[ \begin{array} \\ 7 & 2 \\ {-2} & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 0 \\ {-3} \end{array} \right] }\)

Exercise \(\PageIndex{18}\)

\(\displaystyle{{\bf y}'= \left[ \begin{array} \\ {21} & {-12} \\ {24} & {-15} \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 5 \\ 3 \end{array} \right]}\)

Answer

\(\displaystyle{\left|\begin{array}{cc}21-\lambda&-12\\24&-15-\lambda\end{array}\right|}
=(\lambda-9)(\lambda+3)\).
Eigenvectors  associated with \(\lambda_1=9\)  satisfy
\(\displaystyle{\left[\begin{array}{rr}12&-12\\24&-24
\end{array}\right]\left[\begin{array}{c}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=x_2\).  Taking \(x_2=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{r}1\\1\end{array}\right]}e^{9t}\).
Eigenvectors  associated with \(\lambda_2=-3\)
\(\displaystyle{\left[\begin{array}{rr}24&-12\\24&-12
\end{array}\right]\left[\begin{array}{r}
x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}
\right]}\),
so \(x_1=\displaystyle{1\over2}x_2\).  Taking \(x_2=2\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{r}1\\2\end{array}\right]}e^{-3t}\).
The general solution is
\({\bf y}=c_1\displaystyle{\left[\begin{array}{c}1\\1 \end{array}\right]}e^{9t}+c_2\displaystyle{\left[\begin{array}{c}1\\2\end{array}\right]}e^{-3t}\).
Now  \({\bf y}(0)=\displaystyle{\left[\begin{array}{c}5\\3\end{array}\right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{c}1\\1\end{array}\right]}+c_2\displaystyle{\left[\begin{array}{c}1\\2\end{array}\right]}=\displaystyle{\left[\begin{array}{c}5\\3\end{array}\right]}\), so \(c_1=7\)
and \(c_2=-2\). Therefore,
\({\bf y}=\displaystyle{\left[\begin{array}{c}7\\7 \end{array}\right]e^{9t}-\left[\begin{array}{c}2\\4 \end{array}\right]e^{-3t}}\).

Exercise \(\PageIndex{19}\)

\(\displaystyle{{\bf y}'= \left[ \begin{array} \\ {-7} & 4 \\ {-6} & 7 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ {-1} \\ 7 \end{array} \right] }\)

Exercise \(\PageIndex{20}\)

\(\displaystyle{{\bf y}'={1\over6} \left[ \begin{array} \\ 1 & 2 & 0 \\ 4 & {-1} & 0 \\ 0 & 0 & 3 \end{array} \right] {\bf y}, \quad {\bf
y}(0)= \left[ \begin{array} \\ 4 \\ 7 \\ 1 \end{array} \right]}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}{1\over6}-\lambda&{1\over3}&0\\
{2\over3}&-{1\over6}-\lambda&0\\
0&0&{1\over2}-\lambda\end{array}\right|}=-(\lambda+1/2)(\lambda-1/2)^2\).
The eigenvectors associated with
 with \(\lambda_1=-1/2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}{2\over3}&{1\over3}&0&\vdots&0
\\ {2\over3}&{1\over3}&0&\vdots&0\\0&0&1&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&{1\over2}&0&\vdots&0\\ 0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-\displaystyle{x_2\over2}\) and \(x_3=0\).  Taking \(x_2=2\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}-1\\2\\0
\end{array}\right]}e^{-t/2}\).
The eigenvectors associated with
 with \(\lambda_2=\lambda_3=1/2\) satisfy the system with  augmented
matrix
\(\displaystyle{\left[\begin{array}{rrrcr}-{1\over3}&{1\over3}&0&\vdots&0\\
{2\over3}&-{2\over3}&0&\vdots&0\\0&0&0&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&-1&0&\vdots&0\\0&0&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_2\) and \(x_3\) is arbitrary.  Taking \(x_2=1\) and \(x_3=0\)
yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{t/2}\).
Taking \(x_2=0\) and \(x_3=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}0\\0\\1
\end{array}\right]}e^{t/2}\).
 The general solution is
\({\bf y}=\displaystyle{c_1\left[\begin{array}{c} {-1}\\2\\0 \end{array}\right]e^{-t/2}+
c_2\left[\begin{array}{c} 1\\1\\0 \end{array}\right]e^{t/2}+c_3\left[\begin{array}{c} 0\\0\\1\end{array}\right]e^{t/2}}\).
Now \({\bf y}(0)=\displaystyle{\left[\begin{array}{c} 4\\7\\1 \end{array}\right]}\Rightarrow
\displaystyle{c_1\left[\begin{array}{c} {-1}\\2\\0\end{array}\right]+
c_2\left[\begin{array}{c} 1\\1\\0\end{array}\right]+c_3\left[\begin{array}{c} 0\\0\\1\end{array}\right]e^{t/2}}=\displaystyle{\left[\begin{array}{c} 4\\7\\1\end{array}\right]}\),
so \(c_1=1\), \(c_2=5\), and \(c_3=1\). Hence
\({\bf y}=\displaystyle{\left[\begin{array}{c} {-1}\\2\\0 \end{array}\right]e^{-t/2}+
\left[\begin{array}{c} 5\\5\\0 \end{array}\right]e^{t/2}+\left[\begin{array}{c} 0\\0\\1\end{array}\right]e^{t/2}}\).

Exercise \(\PageIndex{21}\)

\(\displaystyle{{\bf y}'={1\over3} \left[ \begin{array} \\ 2 & {-2} & 3 \\ {-4} & 4 & 3 \\ 2 & 1 & 0 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 1 \\ 1 \\ 5 \end{array} \right]}\)

Exercise \(\PageIndex{22}\)

\(\displaystyle{{\bf y}'= \left[ \begin{array} \\ 6 & {-3} & {-8} \\ 2 & 1 & {-2} \\ 3 & {-3} & {-5} \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 0 \\ {-1} \\ {-1} \end{array} \right]}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}6-\lambda&-3&-8\\2&1-\lambda&-2\\
3&-3&-5-\lambda\end{array}\right|}=-(\lambda-1)(\lambda+2)(\lambda-3)\).
The eigenvectors associated
 with \(\lambda_1=1\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}5&-3&-8&\vdots&0\\2&0
&-2&\vdots&0\\3&-3&-6&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence  \(x_1=x_3\) and \(x_2=-x_3\).  Taking \(x_3=\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}1\\-1\\1
\end{array}\right]}e^t\).
The eigenvectors associated
 with \(\lambda_2=-2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}8&-3&-8&\vdots&0\\2&3
&-2&\vdots&0\\3&-3&-3&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&0&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_3\) and \(x_2=0\).  Taking \(x_3=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}1\\0\\1
\end{array}\right]}e^{-2t}\).
The eigenvectors associated
 with \(\lambda_3=3\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}3&-3&-8&\vdots&0\\2&-2
&-2&\vdots&0\\3&-3&-8&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&-1&0&\vdots&0\\0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence  \(x_1=x_2\) and \(x_3=0\).  Taking \(x_2=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{3t}\).
The general solution is
\({\bf y}=c_1\displaystyle{\left[\begin{array}{rrr}1\\-1\\1
\end{array}\right]}e^t+
c_2\displaystyle{\left[\begin{array}{rrr}1\\0\\1
\end{array}\right]}e^{-2t}
+c_3\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{3t}\).
Now \({\bf y}(0)=\displaystyle{\left[\begin{array}{c} 0{-1}{-1} \end{array}\right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{rrr}1\\-1\\1
\end{array}\right]}+
c_2\displaystyle{\left[\begin{array}{rrr}1\\0\\1
\end{array}\right]}
+c_3\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}=\displaystyle{\left[\begin{array}{c} 0{-1}{-1}\end{array}\right]}\), so
\(c_1=2\), \(c_2=-3\), and \(c_3=1\). Therefore,
\({\bf
y}=\displaystyle{\left[\begin{array}{c} 2\\{-2}\\2\end{array}\right] e^t-\left[\begin{array}{c} 3\\0\\3 \end{array}\right]e^{-2t}+\left[\begin{array}{c} 1\\1\\0\end{array}\right]e^{3t}}\).

Exercise \(\PageIndex{23}\)

\(\displaystyle{{\bf y}'={1\over3} \left[ \begin{array} \\ 2 & 4 & {-7} \\ 1 & 5 & {-5} \\ {-4} & 4 & {-1} \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 4 \\ 1 \\ 3 \end{array} \right]}\)

Exercise \(\PageIndex{24}\)

\(\displaystyle{ {\bf y}'= \left[ \begin{array} \\ 3 & 0 & 1 \\ {11} & {-2} & 7 \\ 1 & 0 & 3 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 2 \\ 7 \\ 6 \end{array} \right]}\)

Answer

\(\displaystyle{\left|\begin{array}{ccc}3-\lambda&0&1\\11&-2-\lambda&7\\
1&0&3-\lambda\end{array}\right|}=-(\lambda-2)(\lambda+2)(\lambda-4)\).
The eigenvectors associated with
 with \(\lambda_1=2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\11&-4
&7&\vdots&0\\1&0&1&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&1&\vdots&0\\0&1&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=-x_3\) and \(x_2=-x_3\).  Taking \(x_3=1\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}-1\\-1\\1
\end{array}\right]}e^{2t}\).
The eigenvectors associated with
 with \(\lambda_2=-2\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}5&0&1&\vdots&0\\11&0
&7&\vdots&0\\1&0&5&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&0&\vdots&0\\0&0&1&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_3=0\) and \(x_2\) is arbitrary.  Taking \(x_3=1\) yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}0\\1\\0
\end{array}\right]}e^{-2t}\).
The eigenvectors associated with
 with \(\lambda_3=4\) satisfy the system with  augmented matrix
\(\displaystyle{\left[\begin{array}{rrrcr}-1&0&1&\vdots&0\\11&-6
&7&\vdots&0\\1&0&-1&\vdots&0
\end{array}\right]}\),
which is row equivalent to
\(\displaystyle{\left[\begin{array}{rrrcr}1&0&-1&\vdots&0\\0&1&-3&
\vdots&0\\0&0&0&\vdots&0\end{array}\right]}\).
Hence \(x_1=x_3\) and \(x_2=3x_3\).  Taking \(x_3=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}1\\3\\1
\end{array}\right]}e^{4t}\).
 The general solution is
\({\bf y}=c_1\displaystyle{\left[\begin{array}{rrr}-1\\-1\\1
\end{array}\right]}e^{2t}
+c_2\displaystyle{\left[\begin{array}{rrr}0\\1\\0
\end{array}\right]}e^{-2t}
+c_3\displaystyle{\left[\begin{array}{rrr}1\\3\\1
\end{array}\right]}e^{4t}\).
Now \({\bf y}(0)=\displaystyle{\left[\begin{array}{c} 2\\7\\6 \end{array}\right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{rrr}-1\\-1\\1
\end{array}\right]}
+c_2\displaystyle{\left[\begin{array}{rrr}0\\1\\0
\end{array}\right]}
+c_3\displaystyle{\left[\begin{array}{rrr}1\\3\\1
\end{array}\right]}=\displaystyle{\left[\begin{array}{c} 2\\7\\6\end{array}\right]}\), so \(c_1=2\), \(c_2=-3\), and
\(c_3=4\). Hence
\({\bf y}=\displaystyle{\left[\begin{array}{c} {-2}\\{-2}\\2 \end{array}\right]e^{2t}-\left[\begin{array}{c} 0\\3\\0\end{array}\right]e^{-2t}+
\left[\begin{array}{c} 4\\{12}\\4\end{array}\right]e^{4t}}\)

Exercise \(\PageIndex{25}\)

\(\displaystyle{ {\bf y}'= \left[ \begin{array} \\ {-2} & {-5} & {-1} \\ {-4} & {-1} & 1 \\ 4 & 5 & {3} \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 8 \\ {-10} \\ {-4} \end{array} \right]}\)

Exercise \(\PageIndex{26}\)

\(\displaystyle{ {\bf y}'= \left[ \begin{array} \\ 3 & {-1} & 0 \\ 4 & {-2} & 0 \\ 4 & {-4} & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 7 \\ {10} \\ 2 \end{array} \right]}\)

Answer

\(\left|\begin{array}{cccc}3-\lambda&-1&0\\4&-2-\lambda&0
\\4&-4&2-\lambda\end{array}\right|
=-(\lambda+1)(\lambda-2)^2\).
The eigenvectors associated
 with \(\lambda_1=-1\) satisfy the system with  augmented matrix
\(\left[\begin{array}{ccccc}4&-1&0&\vdots&0\\4&-1&0&\vdots&0\\4&-4&3&\vdots&0\end{array}\right]\),
which is row equivalent to
\(\left[\begin{array}{ccrcc}1&0&-{1\over4}&\vdots&0\\0&1&-1&\vdots&0\\0&0&0&\vdots&0\end{array}\right]\)
Hence \(x_1=x_2/4\) and \(x_2=x_3\).  Taking \(x_3=4\) yields
\({\bf y}_1=\displaystyle{\left[\begin{array}{rrr}1\\4\\4
\end{array}\right]}e^{-t}\).
The eigenvectors associated with
 with \(\lambda_2=\lambda_3=2\) satisfy the system with  augmented
matrix
\(\left[\begin{array}{ccccc}1&-1&0&\vdots&0\\4&-4&0&\vdots&0\\4&-4&0&\vdots&0\end{array}\right]\),
which is row equivalent to
\(\left[\begin{array}{crccc}1&-1&0&\vdots&0\\0&0&0&\vdots&0\\0&0&0&\vdots&0\end{array}\right]\).
Hence \(x_1=x_2\) and \(x_3\) is arbitrary.  Taking \(x_2=1\) and \(x_3=0\)
yields
\({\bf y}_2=\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{2t}\).
Taking \(x_2=0\) and \(x_3=1\) yields
\({\bf y}_3=\displaystyle{\left[\begin{array}{rrr}0\\0\\1
\end{array}\right]}e^{2t}\).
 The general solution is
\({\bf y}=c_1\displaystyle{\left[\begin{array}{rrr}1\\4\\4
\end{array}\right]}e^{-t}+
c_2\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}e^{2t}+
c_3\displaystyle{\left[\begin{array}{rrr}0\\0\\1
\end{array}\right]}e^{2t}\).
Now \({\bf y}(0)=\displaystyle{\left[\begin{array}{c} 7\\{10}\\{2}\end{array}\right]}\Rightarrow
c_1\displaystyle{\left[\begin{array}{rrr}1\\4\\4
\end{array}\right]}+
c_2\displaystyle{\left[\begin{array}{rrr}1\\1\\0
\end{array}\right]}+
c_3\displaystyle{\left[\begin{array}{rrr}0\\0\\1
\end{array}\right]}=\displaystyle{\left[\begin{array}{c} 7\\{10}\\{2} \end{array}\right] }\),
 so \(c_1=1\), \(c_2=6\), and
\(c_3=-2\). Hence
 \({\bf y}=\displaystyle
\left[\begin{array}{c} 1\\4\\4\end{array}\right]e^{-t}+\left[\begin{array}{c} 6\\6\\{-2}\end{array}\right]e^{2t}\).

Exercise \(\PageIndex{27}\)

\(\displaystyle{{\bf y}'= \left[\begin{array}{rrr}-2&2&6\\2&6&2\\-2&-2& 2\end{array}\right]{\bf y}},\quad{\bf y}(0)= \left[ \begin{array} \\ 6 \\ {-10} \\ 7 \end{array} \right]\)

Exercise \(\PageIndex{28}\)

Let \(A\) be an \(n\times n\) constant matrix. Then Theorem \((4.2.1)\) implies that the solutions of

\begin{equation} \label{eq:4.4E.1}
{\bf y}' = A{\bf y}
\end{equation}

are all defined on \((-\infty,\infty)\).

(a) Use Theorem \((4.2.1)\) to show that the only solution of \eqref{eq:4.4E.1} that can ever equal the zero vector is \({\bf y}\equiv{\bf0}\).

(b) Suppose \({\bf y}_1\) is a solution of \eqref{eq:4.4E.1} and \({\bf y}_2\) is defined by \({\bf y}_2(t)={\bf y}_1(t-\tau)\), where \(\tau\) is an arbitrary real number. Show that \({\bf y}_2\) is also a solution of \eqref{eq:4.4E.1}.

(c) Suppose \({\bf y}_1\) and \({\bf y}_2\) are solutions of \eqref{eq:4.4E.1} and there are real numbers \(t_1\) and \(t_2\) such that \({\bf y}_1(t_1)={\bf y}_2(t_2)\). Show that \({\bf y}_2(t)={\bf y}_1(t-\tau)\) for all \(t\), where \(\tau=t_2-t_1\).
Hint: Show that \({\bf y}_1(t-\tau)\) and \({\bf y}_2(t)\) are solutions of the same initial value problem for \eqref{eq:4.4E.1}, and apply the uniqueness assertion of Theorem \((4.2.1)\).

Answer

{\bf(a)}  If \({\bf y}(t_0)=0\), then \({\bf y}\) is the solution
of the initial value problem \({\bf y}'=A{\bf y},\ {\bf y}(t_0)={\bf
0}\). Since \({\bf y}\equiv0\) is a solution of this problem,
Theorem~4.2.1 implies the conclusion.

{\bf(b)} It is given that \({\bf y}_1'(t)=A{\bf y}_1(t)\) for all \(t\).
Replacing \(t\) by \)t-\tau\) shows that \({\bf y}_1'(t-\tau)=A{\bf
y}_1(t-\tau)=A{\bf y}_2(t)\) for all \(t\). Since \({\bf y}_2'(t)={\bf
y}_1'(t-\tau)\) by the chain rule, this implies that \({\bf
y}_2'(t)=A{\bf y}_2(t)\) for all \(t\).

{\bf(c)} If \({\bf z}(t)={\bf y}_1(t-\tau)\), then \({\bf z}(t_2)={\bf
y}_1(t_1)={\bf y}_2(t_2)\); therefore \({\bf z}\) and \({\bf y}_2\) are
both solutions of the initial value problem \({\bf y}'=A{\bf y},\ {\bf
y}(t_2)={\bf k}\), where \({\bf k}={\bf y}_2(t_2)\).

In Exercises \((4.4E.29)\) to \((4.4E.34)\), describe and graph trajectories of the given system.

Exercise \(\PageIndex{29}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ 1 & 1 \\ 1 & {-1} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{30}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-4} & 3 \\ {-2} & {-11} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{31}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ 9 & {-3} \\ {-1} & {11} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{32}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-1} & {-10} \\ {-5} & 4 \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{33}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ 5 & {-4} \\ 1 & {10} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{34}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-7} & 1 \\ 3 & {-5} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{35}\)

Suppose the eigenvalues of the \(2\times 2\) matrix \(A\) are \(\lambda=0\) and \(\mu\ne0\), with corresponding eigenvectors \({\bf x}_1\) and \({\bf x}_2\). Let \(L_1\) be the line through the origin parallel to \({\bf x}_1\).

(a) Show that every point on \(L_1\) is the trajectory of a constant solution of \({\bf y}'=A{\bf y}\).

(b) Show that the trajectories of nonconstant solutions of \({\bf y}'=A{\bf y}\) are half-lines parallel to \({\bf x}_2\) and on either side of \(L_1\), and that the direction of motion along these trajectories is away from \(L_1\) if \(\mu>0\), or toward \(L_1\) if \(\mu<0\).

The matrices of the systems in Exercises \((4.4E.36)\) to \((4.4E.41)\) are singular. Describe and graph the trajectories of nonconstant solutions of the given systems.

Exercise \(\PageIndex{36}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\{ -1} & 1 \\ 1 & {-1} \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{37}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-1} & {-3} \\ 2 & 6 \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{38}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ 1 & {-3} \\ {-1} & 3 \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{39}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ 1 & {-2} \\ {-1} & 2 \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{40}\)

\({\bf y}'=\displaystyle{ \left[ \begin{array} \\ {-4} & {-4} \\ 1 & 1 \end{array} \right]}{\bf y}\)

Exercise \(\PageIndex{41}\)

\( {\bf y}'=\displaystyle{ \left[ \begin{array} \\ 3 & {-1} \\ {-3} & 1 \end{array} \right] }{\bf y}\)

Exercise \(\PageIndex{42}\)

Let \(P=P(t)\) and \(Q=Q(t)\) be the populations of two species at time \(t\), and assume that each population would grow exponentially if the other didn't exist; that is, in the absence of competition,

\begin{equation} \label{eq:4.4E.2}
P' \ aP \quad \mbox{and} \quad Q' = bQ,
\end{equation}

where \(a\) and \(b\) are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so \eqref{eq:4.4E.2} is replaced by

\begin{eqnarray*}
P'&=&\phantom{-}aP-\alpha Q\\
Q'&=&-\beta P+bQ,
\end{eqnarray*}

where \(\alpha\) and \(\beta\) are positive constants. (Since negative population doesn't make sense, this system holds only while \(P\) and \(Q\) are both positive.) Now suppose \(P(0)=P_0>0\) and \(Q(0)=Q_0>0\).

(a) For several choices of \(a\), \(b\), \(\alpha\), and \(\beta\), verify experimentally (by graphing trajectories of \eqref{eq:4.4E.2} in the \(P\)-\(Q\) plane) that there's a constant \(\rho>0\) (depending upon \(a\), \(b\), \(\alpha\), and \(\beta\)) with the following properties:
(i) If \(Q_0>\rho P_0\), then \(P\) decreases monotonically to zero in finite time, during which \(Q\) remains positive.
(ii) If \(Q_0<\rho P_0\), then \(Q\) decreases monotonically to zero in finite time, during which \(P\) remains positive.

(b) Conclude from part (a) that exactly one of the species becomes extinct in finite time if \(Q_0\ne\rho P_0\). Determine experimentally what happens if \(Q_0=\rho P_0\).

(c) Confirm your experimental results and determine \(\gamma\) by expressing the eigenvalues and associated eigenvectors of

\begin{eqnarray*}
A = \left[ \begin{array} \\ a & {-\alpha} \\ {-\beta} & b \end{array} \right]
\end{eqnarray*}

in terms of \(a\), \(b\), \(\alpha\), and \(\beta\), and applying the geometric arguments developed at the end of this section.

Answer

The characteristic polynomial of \)A\) is
\)p(\lambda)=\lambda^2-(a+b)+ab-\alpha\beta\), so the eigenvalues of \)A\)
are \(\lambda_1=\displaystyle{a+b-\gamma\over2}\) and
\(\lambda_1=\displaystyle{a+b+\gamma\over2}\), where
\(\gamma=\sqrt{(a-b)^2+4\alpha\beta}\); \({\bf
x}_1=\ctwocol{b-a+\gamma}{2\beta}\) and \({\bf
x}_2=\ctwocol{b-a-\gamma}{2\beta}\) are associated eigenvectors. Since
\(\gamma>|b-a|\), if \)L_1\) and \)L_2\) are lines through the origin
parallel to \({\bf x}_1\) and \({\bf x}_2\), then \)L_1\) is in the first and
third quadrants and \)L_2\) is in the second and fourth quadrants. The
slope of \)L_1\) is \(\rho=\displaystyle{2\beta\over b-a+\gamma}>0\). If \)Q_0=\rho
P_0\) there are three possibilities: (i) if \(\alpha\beta=ab\), then
\(\lambda_1=0\) and \)P(t)=P_0\), \)Q(t)=Q_0\) for all \)t>0\); (ii) if
\(\alpha\beta<ab\), then \(\lambda_1>0\) and
\(\lim_{t\to\infty}P(t)=\lim_{t\to\infty}Q(t)=\infty\) (monotonically);
(iii) if \(\alpha\beta>ab\), then \(\lambda_1<0\) and
\(\lim_{t\to\infty}P(t)=\lim_{t\to\infty}Q(t)=0\) (monotonically). Now
suppose \)Q_0\ne\rho P_0\), so that the trajectory cannot intersect
\)L_1\), and assume for the moment that (A) makes sense for all \)t>0\);
that is, even if one or the other of \)P\) and \)Q\) is negative. Since
\(\lambda_2>0\) it follows that either \(\lim_{t\to\infty}P(t)=\infty\) or
\(\lim_{t\to\infty}Q(t)=\infty\) (or both), and the trajectory is
asymptotically parallel to \)L_2\). Therefore,the trajectory must cross
into the third quadrant (so \)P(T)=0\) and \)Q(T)>0\) for some finite \(t\))
if \)Q_0>\rho P_0\), or into the fourth quadrant (so \)Q(T)=0\) and
\)P(T)>0\) for some finite \(t\)) if \)Q_0<\rho P_0\).