
# 10.4: Constant Coefficient Homogeneous Systems I



where $$A$$ is an $$n\times n$$ constant matrix. Since $$A$$ is continuous on $$(-\infty,\infty)$$, Theorem 10.2.1 implies that all solutions of Equation \ref{eq:10.4.1} are defined on $$(-\infty,\infty)$$. Therefore, when we speak of solutions of $${\bf y}'=A{\bf y}$$, we’ll mean solutions on $$(-\infty,\infty)$$.

In this section we assume that all the eigenvalues of $$A$$ are real and that $$A$$ has a set of $$n$$ linearly independent eigenvectors. In the next two sections we consider the cases where some of the eigenvalues of $$A$$ are complex, or where $$A$$ does not have $$n$$ linearly independent eigenvectors.

In Example 10.3.2 we showed that the vector functions

${\bf y}_1=\twocol {-e^{2t}}{2e^{2t}}\quad \text{and} \quad {\bf y}_2=\twocol{-e^{-t}}{e^{-t}}\nonumber$

form a fundamental set of solutions of the system

$\label{eq:10.4.2} {\bf y}'=\left[\begin{array}{cc} {-4}&{-3}\\{6}&{5}\end{array} \right] {\bf y},$

but we did not show how we obtained $${\bf y}_1$$ and $${\bf y}_2$$ in the first place. To see how these solutions can be obtained we write Equation \ref{eq:10.4.2} as

$\label{eq:10.4.3} \begin{array}{ccc} y_1'&=-4y_1-3y_2\\y_2'&=\phantom{-}6y_1+5y_2\end{array}$

and look for solutions of the form

$\label{eq:10.4.4} y_1=x_1e^{\lambda t}\quad \text{and} \quad y_2=x_2e^{\lambda t},$

where $$x_1$$, $$x_2$$, and $$\lambda$$ are constants to be determined. Differentiating Equation \ref{eq:10.4.4} yields

$y_1'=\lambda x_1e^{\lambda t}\quad\mbox{ and }\quad y_2'=\lambda x_2e^{\lambda t}.\nonumber$

Substituting this and Equation \ref{eq:10.4.4} into Equation \ref{eq:10.4.3} and canceling the common factor $$e^{\lambda t}$$ yields

$\begin{array}{ccc}-4x_1-3x_2&=&\lambda x_1 \\ 6 x_1+5x_2&=&\lambda x_2.\end{array}\nonumber$

For a given $$\lambda$$, this is a homogeneous algebraic system, since it can be rewritten as

$\label{eq:10.4.5} \begin{array}{rcl} (-4-\lambda) x_1-3 x_2&=&0\\ 6 x_1+(5-\lambda) x_2&=&0.\end{array}$

The trivial solution $$x_1=x_2=0$$ of this system isn’t useful, since it corresponds to the trivial solution $$y_1\equiv y_2\equiv0$$ of Equation \ref{eq:10.4.3}, which can’t be part of a fundamental set of solutions of Equation \ref{eq:10.4.2}. Therefore we consider only those values of $$\lambda$$ for which Equation \ref{eq:10.4.5} has nontrivial solutions. These are the values of $$\lambda$$ for which the determinant of Equation \ref{eq:10.4.5} is zero; that is,

\begin{aligned} \left|\begin{array}{cc}-4-\lambda&-3\\6&5-\lambda\end{array}\right|&= (-4-\lambda)(5-\lambda)+18\\&=\lambda^2-\lambda-2\\ &=(\lambda-2)(\lambda+1)=0,\end{aligned}\nonumber

which has the solutions $$\lambda_1=2$$ and $$\lambda_2=-1$$.

Taking $$\lambda=2$$ in Equation \ref{eq:10.4.5} yields

\begin{aligned} -6 x_1-3 x_2&=0\\ 6 x_1+3 x_2&=0,\end{aligned}\nonumber

which implies that $$x_1=-x_2/2$$, where $$x_2$$ can be chosen arbitrarily. Choosing $$x_2=2$$ yields the solution $$y_1=-e^{2t}$$, $$y_2=2e^{2t}$$ of Equation \ref{eq:10.4.3}. We can write this solution in vector form as

$\label{eq:10.4.6} {\bf y}_1=\twocol {-1}{\phantom{-}2} e^{2t}.$

Taking $$\lambda=-1$$ in Equation \ref{eq:10.4.5} yields the system

\begin{aligned} -3 x_1-3 x_2&=0\\ \phantom{-}6 x_1+6 x_2&=0,\end{aligned}\nonumber

so $$x_1=-x_2$$. Taking $$x_2=1$$ here yields the solution $$y_1=-e^{-t}$$, $$y_2=e^{-t}$$ of Equation \ref{eq:10.4.3}. We can write this solution in vector form as

$\label{eq:10.4.7} {\bf y}_2=\twocol{-1}{\phantom{-}1}e^{-t}.$

In Equation \ref{eq:10.4.6} and Equation \ref{eq:10.4.7} the constant coefficients in the arguments of the exponential functions are the eigenvalues of the coefficient matrix in Equation \ref{eq:10.4.2}, and the vector coefficients of the exponential functions are associated eigenvectors. This illustrates the next theorem.

Theorem $$\PageIndex{1}$$

Suppose the $$n\times n$$ constant matrix $$A$$ has $$n$$ real eigenvalues $$\lambda_1,\lambda_2,\ldots,\lambda_n$$ (which need not be distinct) with associated linearly independent eigenvectors $${\bf x}_1,{\bf x}_2,\ldots,{\bf x}_n$$. Then the functions

${\bf y}_1={\bf x}_1e^{\lambda_1 t},\, {\bf y}_2={\bf x}_2e^{\lambda_2 t},\, \dots,\, {\bf y}_n={\bf x}_ne^{\lambda_n t} \nonumber$

form a fundamental set of solutions of $${\bf y}'=A{\bf y};$$ that is$$,$$ the general solution of this system is

${\bf y}=c_1{\bf x}_1e^{\lambda_1 t}+c_2{\bf x}_2e^{\lambda_2 t} +\cdots+c_n{\bf x}_ne^{\lambda_n t}. \nonumber$

Proof

Differentiating $${\bf y}_i={\bf x}_ie^{\lambda_it}$$ and recalling that $$A{\bf x}_i=\lambda_i{\bf x}_i$$ yields

${\bf y}_i'=\lambda_i{\bf x}_ie^{\lambda_it}=A{\bf x}_ie^{\lambda_it} =A{\bf y}_i. \nonumber$

This shows that $${\bf y}_i$$ is a solution of $${\bf y}'=A{\bf y}$$.

The Wronskian of $$\{{\bf y}_1,{\bf y}_2,\ldots,{\bf y}_n\}$$ is

$\left|\begin{array}{cccc} x_{11}e^{\lambda_1 t}& x_{12}e^{\lambda_2 t}&\cdots& x_{1n}e^{\lambda_n t}\\ x_{21}e^{\lambda_1 t}& x_{22}e^{\lambda_2 t}&\cdots& x_{2n}e^{\lambda_n t}\\\vdots&\vdots&\ddots&\vdots\\ x_{n1}e^{\lambda_1 t}& x_{n2}e^{\lambda_2 t}&\cdots& x_{nn}e^{\lambda x_n t}\end{array}\right| =e^{\lambda_1 t}e^{\lambda_2 t}\cdots e^{\lambda_n t} \left|\begin{array}{cccc} x_{11}&x_{12}&\cdots&x_{1n}\cr x_{21}&x_{22}&\cdots&x_{2n}\cr \vdots&\vdots&\ddots&\vdots\cr x_{n1}&x_{n2}&\cdots&x_{nn}\cr \end{array}\right|.\nonumber$

Since the columns of the determinant on the right are $${\bf x}_1$$, $${\bf x}_2$$, …, $${\bf x}_n$$, which are assumed to be linearly independent, the determinant is nonzero. Therefore Theorem 10.3.3 implies that $$\{{\bf y}_1,{\bf y}_2,\ldots,{\bf y}_n\}$$ is a fundamental set of solutions of $${\bf y}'=A{\bf y}$$.

Example $$\PageIndex{1}$$

1. Find the general solution of $\label{eq:10.4.8} {\bf y}'=\left[\begin{array}{cc}{2}&{4}\\{4}&{2}\end{array} \right] {\bf y}.$
2. Solve the initial value problem $\label{eq:10.4.9} {\bf y}'=\left[\begin{array}{cc}{2}&{4}\\{4}&{2}\end{array} \right] {\bf y},\quad{\bf y}(0)=\left[\begin{array}{r}5 \\-1 \end{array}\right].$

Solution a

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.4.8} is

\begin{aligned} \left|\begin{array}{cc} 2-\lambda&4\\4&2-\lambda\end{array}\right| &= (\lambda-2)^2-16\\ &= (\lambda-2-4)(\lambda-2+4)\\ &= (\lambda-6)(\lambda+2).\end{aligned}\nonumber

Hence, $$\lambda_1=6$$ and $$\lambda_2 =-2$$ are eigenvalues of $$A$$. To obtain the eigenvectors, we must solve the system

$\label{eq:10.4.10} \left[\begin{array}{cc} 2-\lambda&4\\4&2-\lambda\end{array}\right] \left[\begin{array}{c} x_1\\x_2\end{array}\right]= \left[\begin{array}{c} 0\\0\end{array}\right]$

with $$\lambda=6$$ and $$\lambda=-2$$. Setting $$\lambda=6$$ in Equation \ref{eq:10.4.10} yields

$\left[\begin{array}{rr}-4&4\\4&-4 \end{array}\right]\left[\begin{array}{c} x_1\\x_2\end{array}\right]=\left[\begin{array}{c} 0\\0\end{array} \right],\nonumber$

which implies that $$x_1=x_2$$. Taking $$x_2=1$$ yields the eigenvector

${\bf x}_1=\left[\begin{array}{c} 1\\1\end{array}\right],\nonumber$

so

${\bf y}_1=\left[\begin{array}{c} 1\\1\end{array}\right]e^{6t}\nonumber$

is a solution of Equation \ref{eq:10.4.8}. Setting $$\lambda=-2$$ in Equation \ref{eq:10.4.10} yields

$\left[\begin{array}{rr} 4&4\\4&4\end{array}\right] \left[\begin{array}{c} x_1\\x_2 \end{array}\right]=\left[\begin{array}{c} 0\\0\end{array}\right],\nonumber$

which implies that $$x_1=-x_2$$. Taking $$x_2=1$$ yields the eigenvector

${\bf x}_2=\left[\begin{array}{r}-1\\1\end{array}\right],\nonumber$

so

${\bf y}_2=\left[\begin{array}{r}-1\\1\end{array} \right]e^{-2t}\nonumber$

is a solution of Equation \ref{eq:10.4.8}. From Theorem $$\PageIndex{1}$$, the general solution of Equation \ref{eq:10.4.8} is

$\label{eq:10.4.11} {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2=c_1\left[\begin{array}{r}1\\1 \end{array}\right]e^{6t}+c_2\left[\begin{array}{r}-1\\1 \end{array}\right]e^{-2t}.$

Solution b

To satisfy the initial condition in Equation \ref{eq:10.4.9}, we must choose $$c_1$$ and $$c_2$$ in Equation \ref{eq:10.4.11} so that

$c_1\left[\begin{array}{r}1\\1\end{array}\right]+c_2\left[ \begin{array}{r}-1\\ 1\end{array}\right]=\left[\begin{array}{r}5\\-1 \end{array}\right].\nonumber$

This is equivalent to the system

\begin{aligned} c_1-c_2&=\phantom{-}5\\ c_1+c_2&=-1,\end{aligned}\nonumber

so $$c_1=2, c_2=-3$$. Therefore the solution of Equation \ref{eq:10.4.9} is

${\bf y}=2\left[\begin{array}{r}1\\1\end{array}\right]e^{6t}-3 \left[\begin{array}{r}-1\\1\end{array}\right]e^{-2t},\nonumber$

or, in terms of components,

$y_1=2e^{6t}+3e^{-2t},\quad y_2=2e^{6t}-3e^{-2t}.\nonumber$

Example $$\PageIndex{2}$$

1. Find the general solution of $\label{eq:10.4.12} {\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\-2& 3& 2\\4&-1&-2\end{array}\right]{\bf y}.$
2. Solve the initial value problem $\label{eq:10.4.13} {\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\-2&3& 2\\4&-1&-2\end{array} \right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{r}2\\ -1\\8\end{array}\right].$

Solution a

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.4.12} is

$\left|\begin{array}{ccc}3-\lambda&-1&-1\\-2&3-\lambda& 2\\4 &-1&-2-\lambda\end{array}\right|=-(\lambda-2)(\lambda-3)(\lambda+1).\nonumber$

Hence, the eigenvalues of $$A$$ are $$\lambda_1=2$$, $$\lambda_2=3$$, and $$\lambda_3=-1$$. To find the eigenvectors, we must solve the system

$\label{eq:10.4.14} \left[\begin{array}{ccc}3-\lambda&-1&-1\\-2&3-\lambda& 2\\4&-1& -2-\lambda\end{array}\right]\left[\begin{array}{c} x_1\\x_2\\x_3 \end{array} \right]=\left[\begin{array}{r}0\\0\\0\end{array}\right]$

with $$\lambda=2$$, $$3$$, $$-1$$. With $$\lambda=2$$, the augmented matrix of Equation \ref{eq:10.4.14} is

$\left[\begin{array}{rrrcr} 1&-1&-1&\vdots&0\\-2& 1&2&\vdots&0\\4&-1&-4&\vdots&0 \end{array}\right],\nonumber$

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 $$x_2=0$$. Taking $$x_3=1$$ yields

${\bf y}_1=\left[\begin{array}{rrr}1\\0\\1\end{array}\right]e^{2t}\nonumber$

as a solution of Equation \ref{eq:10.4.12}. With $$\lambda=3$$, the augmented matrix of Equation \ref{eq:10.4.14} is

$\left[\begin{array}{rrrcr}0&-1&-1&\vdots&0\\-2& 0& 2&\vdots&0\\4&-1&-5&\vdots&0 \end{array}\right],\nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&-1&\vdots&0\\0&1&1& \vdots&0\\0&0&0&\vdots&0\end{array}\right].\nonumber$

Hence, $$x_1=x_3$$ and $$x_2=-x_3$$. Taking $$x_3=1$$ yields

${\bf y}_2=\left[\begin{array}{r}1\\-1\\1\end{array} \right]e^{3t}\nonumber$

as a solution of Equation \ref{eq:10.4.12}. With $$\lambda=-1$$, the augmented matrix of Equation \ref{eq:10.4.14} is

$\left[\begin{array}{rrrcr} 4&-1&-1&\vdots&0\\-2&4& 2&\vdots&0\\4&-1&-1&\vdots&0 \end{array}\right],\nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&-{1\over 7}&\vdots&0\\0&1& {3\over 7}&\vdots&0\\0&0&0&\vdots&0\end{array}\right].\nonumber$

Hence, $$x_1=x_3/7$$ and $$x_2=-3x_3/7$$. Taking $$x_3=7$$ yields

${\bf y}_3=\left[\begin{array}{r}1\\-3\\7\end{array} \right]e^{-t}\nonumber$

as a solution of Equation \ref{eq:10.4.12}. By Theorem $$\PageIndex{1}$$, the general solution of Equation \ref{eq:10.4.12} is

${\bf y}=c_1\left[\begin{array}{r}1\\0\\1\end{array}\right]e^{2t} +c_2\left[\begin{array}{r}1\\-1\\1\end{array}\right] e^{3t}+c_3 \left[\begin{array}{r}1\\-3\\7\end{array}\right]e^{-t},\nonumber$

which can also be written as

$\label{eq:10.4.15} {\bf y}=\left[\begin{array}{crc}e^{2t}&e^{3t}&e^{-t} \\0&-e^{3t}& -3e^{-t}\\e^{2t}&e^{3t}&\phantom{-}7e^{-t}\end{array} \right]\left[\begin{array}{c} c_1\\c_2\\c_3\end{array}\right].$

Solution b

To satisfy the initial condition in Equation \ref{eq:10.4.13} we must choose $$c_1$$, $$c_2$$, $$c_3$$ in Equation \ref{eq:10.4.15} so that

$\left[\begin{array}{rrr}1&1&1\\0&-1&-3\\ 1&1&7\end{array}\right] \left[\begin{array}{c} c_1\\c_2\\c_3\end{array}\right]= \left[\begin{array}{r}2\\-1\\8\end{array}\right].\nonumber$

Solving this system yields $$c_1=3$$, $$c_2=-2$$, $$c_3=1$$. Hence, the solution of Equation \ref{eq:10.4.13} is

\begin{aligned} {\bf y}&=\left[\begin{array}{ccc}e^{2t}&e^{3t}& e^{-t}\\0&-e^{3t} &-3e^{-t}\\e^{2t}&e^{3t}&7e^{-t}\end{array} \right] \left[\begin{array}{r}3\\-2\\1\end{array}\right]\\ &=3\left[\begin{array}{r}1\\0\\1\end{array}\right]e^{2t}-2 \left[\begin{array}{r}1\\-1\\1\end{array}\right] e^{3t}+\left[\begin{array}{r}1\\-3\\7\end{array} \right]e^{-t}.\end{aligned}\nonumber

Example $$\PageIndex{3}$$

Find the general solution of

$\label{eq:10.4.16} {\bf y}'=\left[\begin{array}{rrr}-3&2&2\\ 2&-3&2\\2&2&-3 \end{array}\right]{\bf y}.$

Solution

The characteristic polynomial of the coefficient matrix $$A$$ in Equation \ref{eq:10.4.16} is

$\left|\begin{array}{ccc}-3-\lambda&2&2\\2&-3-\lambda&2\\2&2 &-3-\lambda\end{array}\right|=-(\lambda-1)(\lambda+5)^2.\nonumber$

Hence, $$\lambda_1=1$$ is an eigenvalue of multiplicity $$1$$, while $$\lambda_2=-5$$ is an eigenvalue of multiplicity $$2$$. Eigenvectors associated with $$\lambda_1=1$$ are solutions of the system with augmented matrix

$\left[\begin{array}{rrrcr}-4&2&2&\vdots&0\\ 2 &-4&2&\vdots&0\\2&2&-4& \vdots&0\end{array}\right],\nonumber$

which is row equivalent to

$\left[\begin{array}{rrrcr} 1&0&-1&\vdots& 0\\0&1&-1 &\vdots& 0 \\0&0&0&\vdots&0\end{array}\right].\nonumber$

Hence, $$x_1=x_2=x_3$$, and we choose $$x_3=1$$ to obtain the solution

$\label{eq:10.4.17} {\bf y}_1=\left[\begin{array}{r}1\\1\\1\end{array}\right]e^t$

of Equation \ref{eq:10.4.16}. Eigenvectors associated with $$\lambda_2=-5$$ are solutions of the system with augmented matrix

$\left[\begin{array}{rrrcr} 2&2&2&\vdots&0\\2&2&2&\vdots&0 \\2&2&2&\vdots&0\end{array}\right].\nonumber$

Hence, the components of these eigenvectors need only satisfy the single condition

$x_1+x_2+x_3=0.\nonumber$

Since there’s only one equation here, we can choose $$x_2$$ and $$x_3$$ arbitrarily. We obtain one eigenvector by choosing $$x_2=0$$ and $$x_3=1$$, and another by choosing $$x_2=1$$ and $$x_3=0$$. In both cases $$x_1=-1$$. Therefore

$\left[\begin{array}{r}-1\\0\\1\end{array}\right]\quad \mbox{ and }\quad\left[\begin{array}{r}-1\\1\\0 \end{array}\right]\nonumber$

are linearly independent eigenvectors associated with $$\lambda_2= -5$$, and the corresponding solutions of Equation \ref{eq:10.4.16} are

${\bf y}_2=\left[\begin{array}{r}-1\\0\\1\end{array} \right]e^{-5t}\quad \mbox{ and }\quad{\bf y}_3=\left[\begin{array}{r}-1\\1\\ 0\end{array}\right]e^{-5t}.\nonumber$

Because of this and Equation \ref{eq:10.4.17}, Theorem $$\PageIndex{1}$$ implies that the general solution of Equation \ref{eq:10.4.16} is

${\bf y}=c_1\left[\begin{array}{r}1\\1\\ 1\end{array}\right]e^t+c_2 \left[\begin{array}{r}-1\\0\\1\end{array}\right] e^{-5t}+c_3\left[\begin{array}{r}-1\\1\\0\end{array} \right]e^{-5t}.\nonumber$

## Geometric Properties of Solutions when $$n=2$$

We’ll now consider the geometric properties of solutions of a $$2\times2$$ constant coefficient system

$\label{eq:10.4.18} \twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]\twocol{y_1}{y_2}.$

It is convenient to think of a “$$y_1$$-$$y_2$$ plane," where a point is identified by rectangular coordinates $$(y_1,y_2)$$. If  $${\bf y}=\left[\begin{array}{c}{y_{1}}\\{y_{2}}\end{array} \right]$$ is a non-constant solution of Equation \ref{eq:10.4.18}, then the point $$(y_1(t),y_2(t))$$ moves along a curve $$C$$ in the $$y_1$$-$$y_2$$ plane as $$t$$ varies from $$-\infty$$ to $$\infty$$. We call $$C$$ the trajectory of $${\bf y}$$. (We also say that $$C$$ is a trajectory of the system Equation \ref{eq:10.4.18}.) I’s important to note that $$C$$ is the trajectory of infinitely many solutions of Equation \ref{eq:10.4.18}, since if $$\tau$$ is any real number, then $${\bf y}(t-\tau)$$ is a solution of Equation \ref{eq:10.4.18} (Exercise 10.4.28b), and $$(y_1(t-\tau),y_2(t-\tau))$$ also moves along $$C$$ as $$t$$ varies from $$-\infty$$ to $$\infty$$. Moreover, Exercise 10.4.28c implies that distinct trajectories of Equation \ref{eq:10.4.18} can’t intersect, and that two solutions $${\bf y}_1$$ and $${\bf y}_2$$ of Equation \ref{eq:10.4.18} have the same trajectory if and only if $${\bf y}_2(t)={\bf y}_1(t-\tau)$$ for some $$\tau$$.

From Exercise 10.4.28a, a trajectory of a nontrivial solution of Equation \ref{eq:10.4.18} can’t contain $$(0,0)$$, which we define to be the trajectory of the trivial solution $${\bf y}\equiv0$$. More generally, if $${\bf y}=\left[\begin{array}{c}{k_{1}}\\{k_{2}}\end{array} \right] \ne{\bf 0}$$ is a constant solution of Equation \ref{eq:10.4.18} (which could occur if zero is an eigenvalue of the matrix of Equation \ref{eq:10.4.18}), we define the trajectory of $${\bf y}$$ to be the single point $$(k_1,k_2)$$.

To be specific, this is the question: What do the trajectories look like, and how are they traversed? In this section we’ll answer this question, assuming that the matrix

$A=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]\nonumber$

of Equation \ref{eq:10.4.18} has real eigenvalues $$\lambda_1$$ and $$\lambda_2$$ with associated linearly independent eigenvectors $${\bf x}_1$$ and $${\bf x}_2$$. Then the general solution of Equation \ref{eq:10.4.18} is

$\label{eq:10.4.19} {\bf y}= c_1{\bf x}_1 e^{\lambda_1 t}+c_2{\bf x}_2e^{\lambda_2 t}.$

We’ll consider other situations in the next two sections.

We leave it to you (Exercise 10.4.35) to classify the trajectories of Equation \ref{eq:10.4.18} if zero is an eigenvalue of $$A$$. We’ll confine our attention here to the case where both eigenvalues are nonzero. In this case the simplest situation is where $$\lambda_1=\lambda_2\ne0$$, so Equation \ref{eq:10.4.19} becomes

${\bf y}=(c_1{\bf x}_1+c_2{\bf x}_2)e^{\lambda_1 t}.\nonumber$

Since $${\bf x}_1$$ and $${\bf x}_2$$ are linearly independent, an arbitrary vector $${\bf x}$$ can be written as $${\bf x}=c_1{\bf x}_1+c_2{\bf x}_2$$. Therefore the general solution of Equation \ref{eq:10.4.18} can be written as $${\bf y}={\bf x}e^{\lambda_1 t}$$ where $${\bf x}$$ is an arbitrary $$2$$-vector, and the trajectories of nontrivial solutions of Equation \ref{eq:10.4.18} are half-lines through (but not including) the origin. The direction of motion is away from the origin if $$\lambda_1>0$$ (Figure $$\PageIndex{1}$$), toward it if $$\lambda_1<0$$ (Figure $$\PageIndex{2}$$). (In these and the next figures an arrow through a point indicates the direction of motion along the trajectory through the point.)

Now suppose $$\lambda_2>\lambda_1$$, and let $$L_1$$ and $$L_2$$ denote lines through the origin parallel to $${\bf x}_1$$ and $${\bf x}_2$$, respectively. By a half-line of $$L_1$$ (or $$L_2$$), we mean either of the rays obtained by removing the origin from $$L_1$$ (or $$L_2$$).

Letting $$c_2=0$$ in Equation \ref{eq:10.4.19} yields $${\bf y}=c_1{\bf x}_1e^{\lambda_1 t}$$. If $$c_1\ne0$$, the trajectory defined by this solution is a half-line of $$L_1$$. The direction of motion is away from the origin if $$\lambda_1>0$$, toward the origin if $$\lambda_1<0$$. Similarly, the trajectory of $${\bf y}=c_2{\bf x}_2e^{\lambda_2 t}$$ with $$c_2\ne0$$ is a half-line of $$L_2$$.

Henceforth, we assume that $$c_1$$ and $$c_2$$ in Equation \ref{eq:10.4.19} are both nonzero. In this case, the trajectory of Equation \ref{eq:10.4.19} can’t intersect $$L_1$$ or $$L_2$$, since every point on these lines is on the trajectory of a solution for which either $$c_1=0$$ or $$c_2=0$$. (Remember: distinct trajectories can’t intersect!). Therefore the trajectory of Equation \ref{eq:10.4.19} must lie entirely in one of the four open sectors bounded by $$L_1$$ and $$L_2$$, but do not any point on $$L_1$$ or $$L_2$$. Since the initial point $$(y_1(0),y_2(0))$$ defined by

${\bf y}(0)=c_1{\bf x}_1+c_2{\bf x}_2\nonumber$

is on the trajectory, we can determine which sector contains the trajectory from the signs of $$c_1$$ and $$c_2$$, as shown in Figure $$\PageIndex{3}$$.

The direction of $${\bf y}(t)$$ in Equation \ref{eq:10.4.19} is the same as that of

$\label{eq:10.4.20} e^{-\lambda_2 t}{\bf y}(t)= c_1{\bf x}_1e^{-(\lambda_2-\lambda_1)t}+c_2{\bf x}_2$

and of

$\label{eq:10.4.21} e^{-\lambda_1 t}{\bf y}(t)=c_1{\bf x}_1+c_2{\bf x}_2e^{(\lambda_2-\lambda_1)t}.$

Since the right side of Equation \ref{eq:10.4.20} approaches $$c_2{\bf x}_2$$ as $$t\to\infty$$, the trajectory is asymptotically parallel to $$L_2$$ as $$t\to\infty$$. Since the right side of Equation \ref{eq:10.4.21} approaches $$c_1{\bf x}_1$$ as $$t\to-\infty$$, the trajectory is asymptotically parallel to $$L_1$$ as $$t\to-\infty$$.

The shape and direction of traversal of the trajectory of Equation \ref{eq:10.4.19} depend upon whether $$\lambda_1$$ and $$\lambda_2$$ are both positive, both negative, or of opposite signs. We’ll now analyze these three cases.

Henceforth $$\|{\bf u}\|$$ denote the length of the vector $${\bf u}$$.

### Case 1: $$\lambda _{2}>\lambda _{1}>0$$

Figure $$\PageIndex{4}$$ shows some typical trajectories. In this case, $$\lim_{t\to-\infty}\|{\bf y}(t)\|=0$$, so the trajectory is not only asymptotically parallel to $$L_1$$ as $$t\to-\infty$$, but is actually asymptotically tangent to $$L_1$$ at the origin. On the other hand, $$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty$$ and

$\lim_{t\to\infty}\left\|{\bf y}(t)-c_2{\bf x}_2e^{\lambda_2 t}\right\|=\lim_{t\to\infty}\|c_1{\bf x_1}e^{\lambda_1t}\|=\infty,\nonumber$

so, although the trajectory is asymptotically parallel to $$L_2$$ as $$t\to\infty$$, it is not asymptotically tangent to $$L_2$$. The direction of motion along each trajectory is away from the origin.

### Case 2: $$0>\lambda _{2}>\lambda _{1}$$

Figure $$\PageIndex{5}$$ shows some typical trajectories. In this case, $$\lim_{t\to\infty}\|{\bf y}(t)\|=0$$, so the trajectory is asymptotically tangent to $$L_2$$ at the origin as $$t\to\infty$$. On the other hand, $$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty$$ and

$\lim_{t\to-\infty}\left\|{\bf y}(t)-c_1{\bf x}_1e^{\lambda_1 t}\right\|=\lim_{t\to-\infty}\|c_2{\bf x}_2e^{\lambda_2t}\|=\infty,\nonumber$

so, although the trajectory is asymptotically parallel to $$L_1$$ as $$t\to-\infty$$, it is not asymptotically tangent to it. The direction of motion along each trajectory is toward the origin.

### Case 3: $$\lambda _{2}>0>\lambda _{1}$$

Figure $$\PageIndex{6}$$ shows some typical trajectories. In this case,

$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to\infty}\left\|{\bf y}(t)-c_2{\bf x}_2e^{\lambda_2 t}\right\|=\lim_{t\to\infty}\|c_1{\bf x}_1e^{\lambda_1t}\|=0,\nonumber$

so the trajectory is asymptotically tangent to $$L_2$$ as $$t\to\infty$$. Similarly,

$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to-\infty}\left\|{\bf y}(t)-c_1{\bf x}_1e^{\lambda_1 t}\right\|=\lim_{t\to-\infty}\|c_2{\bf x}_2e^{\lambda_2t}\|=0,\nonumber$

so the trajectory is asymptotically tangent to $$L_1$$ as $$t\to-\infty$$. The direction of motion is toward the origin on $$L_1$$ and away from the origin on $$L_2$$. The direction of motion along any other trajectory is away from $$L_1$$, toward $$L_2$$.