Opening the book…
We saw in Section 10.3 that if an \(n\times n\) constant matrix \(A\) has \(n\) real eigenvalues \(\lambda_1\), \(\lambda_2\), …, \(\lambda_n\) (which need not be distinct) with associated linearly independent eigenvectors \({\bf x}_1\), \({\bf x}_2\), …, \({\bf x}_n\), then the general solution of \({\bf y}'=A{\bf y}\) is
\[{\bf y}=c_1{\bf x}_1e^{\lambda_1t}+c_2{\bf x}_2e^{\lambda_2 t} +\cdots+c_n{\bf x}_ne^{\lambda_n t}. \nonumber \]
In this section we consider the case where \(A\) has \(n\) real eigenvalues, but does not have \(n\) linearly independent eigenvectors. It is shown in linear algebra that this occurs if and only if \(A\) has at least one eigenvalue of multiplicity \(r>1\) such that the associated eigenspace has dimension less than \(r\). In this case \(A\) is said to be defective. Since it is beyond the scope of this book to give a complete analysis of systems with defective coefficient matrices, we will restrict our attention to some commonly occurring special cases.
Show that the system
\[\label{eq:10.5.1} {\bf y}'=\left[\begin{array}{cc}{11}&{-25}\\{4}&{-9}\end{array} \right] {\bf y}\]
does not have a fundamental set of solutions of the form \(\{{\bf x}_1e^{\lambda_1t},{\bf x}_2e^{\lambda_2t}\}\), where \(\lambda_1\) and \(\lambda_2\) are eigenvalues of the coefficient matrix \(A\) of Equation \ref{eq:10.5.1} and \({\bf x}_1\) and \({\bf x}_2\) are associated linearly independent eigenvectors.
Solution
The characteristic polynomial of \(A\) is
\[\begin{align*} \twochar{11}{-25}4{-9} &=(\lambda-11)(\lambda+9)+100\\ &=\lambda^2-2\lambda+1 \\ &=(\lambda-1)^2.\end{align*}\nonumber\]
Hence, \(\lambda=1\) is the only eigenvalue of \(A\). The augmented matrix of the system \((A-I){\bf x}={\bf 0}\) is
\[\left[\begin{array}{rrcr}10&-25&\vdots&0\\4& -10&\vdots&0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{cccc}{1}&{-\frac{5}{2}}&{\vdots }&{0}\\{0}&{0}&{\vdots }&{0} \end{array} \right] \nonumber\]
Hence, \(x_1=5x_2/2\) where \(x_2\) is arbitrary. Therefore all eigenvectors of \(A\) are scalar multiples of \(\bf {x}_1=\left[\begin{array}{c}{5}\\{2}\end{array} \right] \), so \(A\) does not have a set of two linearly independent eigenvectors.
From Example 10.4.1 , we know that all scalar multiples of \(\bf {y}_1= \twocol52 e^t\) are solutions of Equation \ref{eq:10.5.1}; however, to find the general solution we must find a second solution \(\bf {y}_2\) such that \(\{ \bf {y}_1, \bf {y}_2\}\) is linearly independent. Based on your recollection of the procedure for solving a constant coefficient scalar equation
\[ay''+by'+cy=0\nonumber\]
in the case where the characteristic polynomial has a repeated root, you might expect to obtain a second solution of Equation \ref{eq:10.5.1} by multiplying the first solution by \(t\). However, this yields \({\bf y}_2=\twocol52 te^t\), which does not quite work, since
\[{\bf y}_2'=\twocol52(te^t+e^t),\quad \text{while} \quad \left[\begin{array}{cc}{11}&{-25}\\{4}&{-9} \end{array} \right] {\bf y}_2=\twocol52te^t.\nonumber\]
However, the idea does incorporate this idea as the next theorem shows.
Suppose the \(n\times n\) matrix \(A\) has an eigenvalue \(\lambda_1\) of multiplicity \(\ge2\) and the associated eigenspace has dimension \(1;\) that is\(,\) all \(\lambda_1\)-eigenvectors of \(A\) are scalar multiples of a single eigenvector \({\bf x}.\) Then there are infinitely many vectors \({\bf u}\) such that
\[\label{eq:10.5.2} (A-\lambda_1I){\bf u}={\bf x}.\]
Moreover\(,\) if \({\bf u}\) is any such vector then
\[\label{eq:10.5.3} {\bf y}_1={\bf x}e^{\lambda_1t}\quad\mbox{and }\quad {\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}\]
are linearly independent solutions of \({\bf y}'=A{\bf y}.\)
A complete proof of this theorem is beyond the scope of this book. The difficulty is in proving that there’s a vector \({\bf u}\) satisfying Equation \ref{eq:10.5.2}, since \(\det(A-\lambda_1I)=0\). We’ll take this without proof and verify the other assertions of the theorem. We already know that \({\bf y}_1\) in Equation \ref{eq:10.5.3} is a solution of \({\bf y}'=A{\bf y}\). To see that \({\bf y}_2\) is also a solution, we compute
\[\begin{align*} {\bf y}_2'-A{\bf y}_2&=\lambda_1{\bf u}e^{\lambda_1t}+{\bf x} e^{\lambda_1t} +\lambda_1{\bf x} te^{\lambda_1t}-A{\bf u}e^{\lambda_1t}-A{\bf x} te^{\lambda_1t}\\ &=(\lambda_1{\bf u}+{\bf x} -A{\bf u})e^{\lambda_1t}+(\lambda_1{\bf x} -A{\bf x} )te^{\lambda_1t}.\end{align*}\nonumber\]
Since \(A{\bf x}=\lambda_1{\bf x}\), this can be written as
\[{\bf y}_2'-A{\bf y}_2=- \left((A-\lambda_1I){\bf u}-{\bf x}\right)e^{\lambda_1t},\nonumber\]
and now Equation \ref{eq:10.5.2} implies that \({\bf y}_2'=A{\bf y}_2\). To see that \({\bf y}_1\) and \({\bf y}_2\) are linearly independent, suppose \(c_1\) and \(c_2\) are constants such that
\[\label{eq:10.5.4} c_1{\bf y}_1+c_2{\bf y}_2=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t} +{\bf x}te^{\lambda_1t})={\bf 0}.\]
We must show that \(c_1=c_2=0\). Multiplying Equation \ref{eq:10.5.4} by \(e^{-\lambda_1t}\) shows that
\[\label{eq:10.5.5} c_1{\bf x}+c_2({\bf u} +{\bf x}t)={\bf 0}.\]
By differentiating this with respect to \(t\), we see that \(c_2{\bf x}={\bf 0}\), which implies \(c_2=0\), because \({\bf x}\ne{\bf 0}\). Substituting \(c_2=0\) into Equation \ref{eq:10.5.5} yields \(c_1{\bf x}={\bf 0}\), which implies that \(c_1=0\), again because \({\bf x}\ne{\bf 0}\)
Use Theorem 10.4.1 to find the general solution of the system
\[\label{eq:10.5.6} {\bf y}'=\left[\begin{array}{cc}{11}&{-25}\\{4}&{-9}\end{array} \right]{\bf y}\]
considered in Example 10.4.1 .
Solution
In Example 10.4.1 we saw that \(\lambda_1=1\) is an eigenvalue of multiplicity \(2\) of the coefficient matrix \(A\) in Equation \ref{eq:10.5.6}, and that all of the eigenvectors of \(A\) are multiples of
\[{\bf x}=\twocol52.\nonumber\]
Therefore
\[{\bf y}_1=\twocol52e^t\nonumber\]
is a solution of Equation \ref{eq:10.5.6}. From Theorem 10.4.1 , a second solution is given by \({\bf y}_2={\bf u}e^t+{\bf x}te^t\), where \((A-I){\bf u}={\bf x}\). The augmented matrix of this system is
\[\left[\begin{array}{rrcr}10&-25&\vdots&5\\4&-10&\vdots&2\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrcr}1&-\frac{5}{2}&\vdots&\frac{1}{2}\\0&0&\vdots&0\end{array}\right],\nonumber\]
Therefore the components of \({\bf u}\) must satisfy
\[u_1-{5\over2}u_2={1\over2},\nonumber\]
where \(u_2\) is arbitrary. We choose \(u_2=0\), so that \(u_1=1/2\) and
\[{\bf u}=\twocol{1\over2}0.\nonumber\]
Thus,
\[{\bf y}_2=\twocol10{e^t\over2}+\twocol52te^t.\nonumber\]
Since \({\bf y}_1\) and \({\bf y}_2\) are linearly independent by Theorem 10.4.1 , they form a fundamental set of solutions of Equation \ref{eq:10.5.6}. Therefore the general solution of Equation \ref{eq:10.5.6} is
\[{\bf y}=c_1\twocol52e^t+c_2\left(\twocol10{e^t\over2}+\twocol52te^t\right).\nonumber\]
Find the general solution of
\[\label{eq:10.5.7} {\bf y}'=\left[\begin{array}{ccc}{3}&{4}&{-10}\\{2}&{1}&{-2}\\{2}&{2}&{-5}\end{array} \right] {\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.5.7} is
\[\left|\begin{array}{ccc} 3-\lambda & 4 & -10\\ 2 & 1-\lambda & -2\\ 2 & 2 &-5-\lambda\end{array}\right| =- (\lambda-1)(\lambda+1)^2.\nonumber\]
Hence, the eigenvalues are \(\lambda_1=1\) with multiplicity \(1\) and \(\lambda_2=-1\) with multiplicity \(2\). Eigenvectors associated with \(\lambda_1=1\) must satisfy \((A-I){\bf x}={\bf 0}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} 2 & 4 & -10 &\vdots & 0\\ 2& 0 & -2 &\vdots & 0\\ 2 & 2 & -6 & \vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 0 & -1 &\vdots& 0\\ 0 & 1 & -2 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Hence, \(x_1 =x_3\) and \(x_2 =2 x_3\), where \(x_3\) is arbitrary. Choosing \(x_3=1\) yields the eigenvector
\[{\bf x}_1=\threecol121.\nonumber\]
Therefore
\[{\bf y}_1 =\threecol121e^t\nonumber\]
is a solution of Equation \ref{eq:10.5.7}. Eigenvectors associated with \(\lambda_2 =-1\) satisfy \((A+I){\bf x}={\bf 0}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & 0\\ 2 & 2 & -2 & \vdots & 0\\2 & 2 & -4 &\vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 1 & 0 &\vdots& 0\\ 0 & 0 & 1 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Hence, \(x_3=0\) and \(x_1 =-x_2\), where \(x_2\) is arbitrary. Choosing \(x_2=1\) yields the eigenvector
\[{\bf x}_2=\threecol{-1}10,\nonumber\]
so
\[{\bf y}_2 =\threecol{-1}10e^{-t}\nonumber\]
is a solution of Equation \ref{eq:10.5.7}. Since all the eigenvectors of \(A\) associated with \(\lambda_2=-1\) are multiples of \({\bf x}_2\), we must now use Theorem 10.4.1 to find a third solution of Equation \ref{eq:10.5.7} in the form
\[\label{eq:10.5.8} {\bf y}_3={\bf u}e^{-t}+\threecol{-1}10te^{-t},\]
where \({\bf u}\) is a solution of \((A+I){\bf u=x}_2\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & -1\\ 2 & 2 & -2 & \vdots & 1\\ 2 & 2 & -4 &\vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 1 & 0 &\vdots& 1\\ 0 & 0 & 1 &\vdots& {1\over2} \\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Hence, \(u_3=1/2\) and \(u_1 =1-u_2\), where \(u_2\) is arbitrary. Choosing \(u_2=0\) yields
\[{\bf u} =\threecol10{1\over2},\nonumber\]
and substituting this into Equation \ref{eq:10.5.8} yields the solution
\[{\bf y}_3=\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}\nonumber\]
of Equation \ref{eq:10.5.7}. Since the Wronskian of \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) at \(t=0\) is
\[\left|\begin{array}{rrr} 1&-1&1\\2&1&0\\1&0&1\over2\end{array}\right|={1\over2},\nonumber\]
\(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is a fundamental set of solutions of Equation \ref{eq:10.5.7}. Therefore the general solution of Equation \ref{eq:10.5.7} is
\[{\bf y}=c_1\threecol121e^t+c_2\threecol{-1}10e^{-t}+c_3\left (\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}\right).\nonumber\]
Suppose the \(n\times n\) matrix \(A\) has an eigenvalue \(\lambda_1\) of multiplicity \(\ge 3\) and the associated eigenspace is one–dimensional\(;\) that is\(,\) all eigenvectors associated with \(\lambda_1\) are scalar multiples of the eigenvector \({\bf x}.\) Then there are infinitely many vectors \({\bf u}\) such that
\[\label{eq:10.5.9} (A-\lambda_1I){\bf u}={\bf x},\]
and, if \({\bf u}\) is any such vector\(,\) there are infinitely many vectors \({\bf v}\) such that
\[\label{eq:10.5.10} (A-\lambda_1I){\bf v}={\bf u}.\]
If \({\bf u}\) satisfies Equation \ref{eq:10.5.9} and \({\bf v}\) satisfies Equation \ref{eq:10.5.10}, then
\[\begin{aligned} {\bf y}_1 &={\bf x} e^{\lambda_1t},\\ {\bf y}_2&={\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\ {\bf y}_3&={\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}\end{aligned}\nonumber\]
are linearly independent solutions of \({\bf y}'=A{\bf y}\).
Use Theorem 10.4.2 to find the general solution of
\[\label{eq:10.5.11} {\bf y}'=\left[\begin{array}{ccc}1&1&1 \\ 1&3&-1 \\ 0&2&2 \end{array} \right] {\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.5.11} is
\[\left|\begin{array}{ccc} 1-\lambda & 1 & \phantom{-}1\\ 1 & 3-\lambda & -1\\ 0 & 2 & 2-\lambda\end{array}\right| =-(\lambda-2)^3.\nonumber\]
Hence, \(\lambda_1=2\) is an eigenvalue of multiplicity \(3\). The associated eigenvectors satisfy \((A-2I){\bf x=0}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots & 0\\ 1& 1 & -1 &\vdots & 0\\ 0 & 2 & 0 & \vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& 0\\ 0 & 1 & 0 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Hence, \(x_1 =x_3\) and \(x_2 = 0\), so the eigenvectors are all scalar multiples of
\[{\bf x}_1=\threecol101.\nonumber\]
Therefore
\[{\bf y}_1=\threecol101e^{2t}\nonumber\]
is a solution of Equation \ref{eq:10.5.11}. We now find a second solution of Equation \ref{eq:10.5.11} in the form
\[{\bf y}_2={\bf u}e^{2t}+\threecol101te^{2t},\nonumber\]
where \({\bf u}\) satisfies \((A-2I){\bf u=x}_1\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots & 1\\ 1& 1 & -1 &\vdots & 0\\ 0 & 2 & 0 & \vdots & 1\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -{1\over2}\\ 0 & 1 & 0 &\vdots& {1\over2}\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Letting \(u_3=0\) yields \(u_1=-1/2\) and \(u_2=1/2\); hence,
\[{\bf u}={1\over2}\threecol{-1}10\nonumber\]
and
\[{\bf y}_2=\threecol{-1}10{e^{2t}\over2}+\threecol101te^{2t}\nonumber\]
is a solution of Equation \ref{eq:10.5.11}. We now find a third solution of Equation \ref{eq:10.5.11} in the form
\[{\bf y}_3={\bf v}e^{2t}+\threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}\nonumber\]
where \({\bf v}\) satisfies \((A-2I){\bf v}={\bf u}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots &-{1\over2}\\ 1& 1 & -1 &\vdots & {1\over2}\\ 0 & 2 & 0 & \vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& {1\over2}\\ 0 & 1 & 0 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Letting \(v_3=0\) yields \(v_1=1/2\) and \(v_2=0\); hence,
\[{\bf v}={1\over2}\threecol100.\nonumber\]
Therefore
\[{\bf y}_3=\threecol100{e^{2t}\over2}+ \threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}\nonumber\]
is a solution of Equation \ref{eq:10.5.11}. Since \({\bf y}_1\), \({\bf y}_2\), and \({\bf y}_3\) are linearly independent by Theorem 10.4.2 , they form a fundamental set of solutions of Equation \ref{eq:10.5.11}. Therefore the general solution of Equation \ref{eq:10.5.11} is
\[\begin{aligned} {\bf y} = c_{1}\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right]e^{2t}+c_{2}\left(\left[ \begin{array}{c}{-1}\\{1}\\{0}\end{array} \right]\frac{e^{2t}}{2}+\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right] te^{2t} \right) + c_{3}\left(\left[\begin{array}{c}{1}\\{0}\\{0}\end{array} \right]\frac{e^{2t}}{2}+\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array} \right]\frac{te^{2t}}{2}+\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right]\frac{t^{2}e^{2t}}{2} \right) \end{aligned}\nonumber \]
Suppose the \(n\times n\) matrix \(A\) has an eigenvalue \(\lambda_1\) of multiplicity \(\ge 3\) and the associated eigenspace is two–dimensional; that is, all eigenvectors of \(A\) associated with \(\lambda_1\) are linear combinations of two linearly independent eigenvectors \({\bf x}_1\) and \({\bf x}_2\)\(.\) Then there are constants \(\alpha\) and \(\beta\) \((\)not both zero\()\) such that if
\[\label{eq:10.5.12} {\bf x}_3=\alpha{\bf x}_1+\beta{\bf x}_2,\]
then there are infinitely many vectors \({\bf u}\) such that
\[\label{eq:10.5.13} (A-\lambda_1I){\bf u}={\bf x}_3.\]
If \({\bf u}\) satisfies Equation \ref{eq:10.5.13}, then
\[\label{eq:10.5.14}\begin{aligned} {\bf y}_{1}&={\bf x}_{1}e^{\lambda _{1}t},\\{\bf y}_{2}&={\bf x}_{2}e^{\lambda _{1}t},\text{and}\\{\bf y}_{3}&={\bf u}e^{\lambda _{1}t}+{\bf x}_{3}te^{\lambda _{1}t}\end{aligned}\]
are linearly independent solutions of \({\bf y}'=A{\bf y}.\)
We omit the proof of this theorem.
Use Theorem 10.4.3 to find the general solution of
\[\label{eq:10.5.15} {\bf y}'=\left[\begin{array}{ccc}{0}&{0}&{1}\\{-1}&{1}&{1}\\{-1}&{0}&{2}\end{array} \right]{\bf y}.\]
Solution
The characteristic polynomial of the coefficient matrix \(A\) in Equation \ref{eq:10.5.15} is
\[\left|\begin{array}{ccc} -\lambda & 0 & 1\\ -1 & 1-\lambda & 1\\ -1 & 0 & 2-\lambda\end{array}\right| =-(\lambda-1)^3.\nonumber\]
Hence, \(\lambda_1=1\) is an eigenvalue of multiplicity \(3\). The associated eigenvectors satisfy \((A-I){\bf x=0}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} -1 & 0 & 1 &\vdots & 0\\ -1& 0 & 1 &\vdots & 0\\ -1 & 0 & 1 & \vdots & 0\end{array}\right],\nonumber\]
which is row equivalent to
\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& 0\\ 0 & 0 & 0 &\vdots& 0 \\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]
Hence, \(x_1 =x_3\) and \(x_2\) is arbitrary, so the eigenvectors are of the form
\[{\bf x}_1=\threecol{x_3}{x_2}{x_3}=x_3\threecol101+x_2\threecol010.\nonumber\]
Therefore the vectors
\[\label{eq:10.5.16} {\bf x}_1 =\threecol101\quad\mbox{and }\quad {\bf x}_2=\threecol010\]
form a basis for the eigenspace, and
\[{\bf y}_1 =\threecol101e^t \quad \text{and} \quad {\bf y}_2=\threecol010e^t\nonumber\]
are linearly independent solutions of Equation \ref{eq:10.5.15}. To find a third linearly independent solution of Equation \ref{eq:10.5.15}, we must find constants \(\alpha\) and \(\beta\) (not both zero) such that the system
\[\label{eq:10.5.17} (A-I){\bf u}=\alpha{\bf x}_1+\beta{\bf x}_2\]
has a solution \({\bf u}\). The augmented matrix of this system is
\[\left[\begin{array}{rrrcr} -1 & 0 & 1 &\vdots &\alpha\\ -1& 0 & 1 &\vdots &\beta\\ -1 & 0 & 1 & \vdots &\alpha\end{array}\right],\nonumber\]
which is row equivalent to
\[\label{eq:10.5.18} \left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -\alpha\\ 0 & 0 & 0 &\vdots&\beta-\alpha\\ 0 & 0 & 0 &\vdots&0\end{array} \right].\]
Therefore Equation \ref{eq:10.5.17} has a solution if and only if \(\beta=\alpha\), where \(\alpha\) is arbitrary. If \(\alpha=\beta=1\) then Equation \ref{eq:10.5.12} and Equation \ref{eq:10.5.16} yield
\[{\bf x}_3={\bf x}_1+{\bf x}_2= \threecol101+\threecol010=\threecol111,\nonumber\]
and the augmented matrix Equation \ref{eq:10.5.18} becomes
\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -1\\ 0 & 0 & 0 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array} \right].\nonumber\]
This implies that \(u_1=-1+u_3\), while \(u_2\) and \(u_3\) are arbitrary. Choosing \(u_2=u_3=0\) yields
\[{\bf u}=\threecol{-1}00.\nonumber\]
Therefore Equation \ref{eq:10.5.14} implies that
\[{\bf y}_3={\bf u}e^t+{\bf x}_3te^t=\threecol{-1}00e^t+\threecol111te^t\nonumber\]
is a solution of Equation \ref{eq:10.5.15}. Since \({\bf y}_1\), \({\bf y}_2\), and \({\bf y}_3\) are linearly independent by Theorem 10.4.3 , they form a fundamental set of solutions for Equation \ref{eq:10.5.15}. Therefore the general solution of Equation \ref{eq:10.5.15} is
\[{\bf y}=c_1\threecol101e^t+c_2\threecol010e^t +c_3\left(\threecol{-1}00e^t+\threecol111te^t\right).\bbox\nonumber\]
This page titled 10.4: Constant Coefficient Homogeneous Systems II was last modified on Tue, 30 Apr 2024 03:30:40 GMT and is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.
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