10.4 Constant Coefficient Homogeneous Systems I

We’ll now begin our study of the homogeneous system

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 (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

form a fundamental set of solutions of the system

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 (10.4.2) as

and look for solutions of the form

where \(x_1\), \(x_2\), and \(\lambda\) are constants to be determined. Differentiating (10.4.4) yields

Substituting this and (10.4.4) into (10.4.3) and canceling the common factor \(e^{\lambda t}\) yields

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

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 (10.4.3), which can’t be part of a fundamental set of solutions of (10.4.2). Therefore we consider only those values of \(\lambda\) for which (10.4.5) has nontrivial solutions. These are the values of \(\lambda\) for which the determinant of (10.4.5) is zero; that is,

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

Taking \(\lambda=2\) in (10.4.5) yields

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 (10.4.3). We can write this solution in vector form as

Taking \(\lambda=-1\) in (10.4.5) yields the system

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

In (10.4.6) and (10.4.7) the constant coefficients in the arguments of the exponential functions are the eigenvalues of the coefficient matrix in (10.4.2), and the vector coefficients of the exponential functions are associated eigenvectors. This illustrates the next theorem.

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

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

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 \({\bf y}'=A{\bf y}\).

Solution (a) The characteristic polynomial of the coefficient matrix \(A\) in (10.4.8) is

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

with \(\lambda=6\) and \(\lambda=-2\). Setting \(\lambda=6\) in (10.4.10) yields

which implies that \(x_1=x_2\). Taking \(x_2=1\) yields the eigenvector

so

is a solution of (10.4.8). Setting \(\lambda=-2\) in (10.4.10) yields

which implies that \(x_1=-x_2\). Taking \(x_2=1\) yields the eigenvector

so

is a solution of (10.4.8). From Theorem 10.4.1, the general solution of (10.4.8) is

Solution (b) To satisfy the initial condition in (10.4.9), we must choose \(c_1\) and \(c_2\) in (10.4.11) so that

This is equivalent to the system

so \(c_1=2, c_2=-3\). Therefore the solution of (10.4.9) is

or, in terms of components,

Solution (a) The characteristic polynomial of the coefficient matrix \(A\) in (10.4.12) is

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

with \(\lambda=2\), \(3\), \(-1\). With \(\lambda=2\), the augmented matrix of (10.4.14) is

which is row equivalent to

Hence, \(x_1=x_3\) and \(x_2=0\). Taking \(x_3=1\) yields

as a solution of (10.4.12). With \(\lambda=3\), the augmented matrix of (10.4.14) is

which is row equivalent to

Hence, \(x_1=x_3\) and \(x_2=-x_3\). Taking \(x_3=1\) yields

as a solution of (10.4.12). With \(\lambda=-1\), the augmented matrix of (10.4.14) is

which is row equivalent to

Hence, \(x_1=x_3/7\) and \(x_2=-3x_3/7\). Taking \(x_3=7\) yields

as a solution of (10.4.12). By Theorem 10.4.1, the general solution of (10.4.12) is

which can also be written as

Solution (b) To satisfy the initial condition in (10.4.13) we must choose \(c_1\), \(c_2\), \(c_3\) in (10.4.15) so that

Solving this system yields \(c_1=3\), \(c_2=-2\), \(c_3=1\). Hence, the solution of (10.4.13) is

Solution The characteristic polynomial of the coefficient matrix \(A\) in (10.4.16) is

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

which is row equivalent to

Hence, \(x_1=x_2=x_3\), and we choose \(x_3=1\) to obtain the solution

of (10.4.16). Eigenvectors associated with \(\lambda_2=-5\) are solutions of the system with augmented matrix

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

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

are linearly independent eigenvectors associated with \(\lambda_2= -5\), and the corresponding solutions of (10.4.16) are

Because of this and (10.4.17), Theorem 10.4.1 implies that the general solution of (10.4.16) is

Geometric Properties of Solutions when \(n=2\)

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

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 \({\bf y}=\dst{\twocol{y_1}{y_2}}\) is a non-constant solution of (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 (10.4.18).) I’s important to note that \(C\) is the trajectory of infinitely many solutions of (10.4.18), since if \(\tau\) is any real number, then \({\bf y}(t-\tau)\) is a solution of (10.4.18) (Exercise 28(b)), and \((y_1(t-\tau),y_2(t-\tau))\) also moves along \(C\) as \(t\) varies from \(-\infty\) to \(\infty\). Moreover, Exercise 28(c) implies that distinct trajectories of (10.4.18) can’t intersect, and that two solutions \({\bf y}_1\) and \({\bf y}_2\) of (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 28(a), a trajectory of a nontrivial solution of (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 \({\bf y}=\dst{\twocol{k_1}{k_2}}\ne{\bf 0}\) is a constant solution of (10.4.18) (which could occur if zero is an eigenvalue of the matrix of (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

of (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 (10.4.18) is

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

We leave it to you (Exercise 35) to classify the trajectories of (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 (10.4.19) becomes

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 (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 (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 10.4.1), toward it if \(\lambda_1<0\) (Figure 10.4.2). (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 (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 (10.4.19) are both nonzero. In this case, the trajectory of (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 (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

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 10.4.3.

The direction of \({\bf y}(t)\) in (10.4.19) is the same as that of

and of

Since the right side of (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 (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 (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 10.4.4 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

so, although the trajectory is asymptotically parallel to \(L_2\) as \(t\to\infty\), it’s 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 10.4.5 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

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

Case 3: \(\lambda_2>0>\lambda_1\)

Figure 10.4.6 shows some typical trajectories. In this case,

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

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 \(L_2\).

10.4 Exercises

In Exercises 1–15 find the general solution.

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}1\\1\end{array} \right]e^{3t}+c_2\left[\begin{array}{r}1\\-1\end{array}\right] e^{-t}}\)  

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}1\\1\end{array} \right]e^{-t/2}+c_2\left[\begin{array}{r}-1\\1\end{array}\right]e^{-2t}}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}-3\\1\end{array} \right]e^{-t}+c_2\left[\begin{array}{r}-1\\2\end{array} \right]e^{-2t}}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}2\\1\end{array} \right]e^{-3t}+c_2\left[\begin{array}{r}-2\\1\end{array} \right]e^t}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}1\\1 \end{array}\right]e^{-2t}+c_1\left[\begin{array}{r}-4\\1\end{array} \right]e^{3t}}\)  

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}3\\2\end{array} \right]e^{2t}+c_2\left[\begin{array}{r}1\\1\end{array}\right]e^t}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}-3\\1 \end{array} \right]e^{-5t}+c_2\left[\begin{array}{r}-1\\1\end{array} \right]e^{-3t}}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}1\\2\\1 \end{array}\right]e^{-3t}+c_2\left[\begin{array}{r}-1\\-4\\1 \end{array} \right]e^{-t}+c_3\left[\begin{array}{r}-1\\-1\\1\end{array} \right]e^{2t}}\)

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

   Show answer

   \({\bf y}=\dst{c_1\left[\begin{array}{r}2\\1\\2 \end{array}\right]e^{-16t}+c_2\left[\begin{array}{r}-1\\2\\ 0\end{array} \right]e^{2t}+c_3\left[\begin{array}{r}-1\\0\\1 \end{array}\right]e^{2t}}\)

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

    Show answer

    \({\bf y}=\dst{c_1\left[\begin{array}{r}-2\\-4\\3 \end{array}\right]e^t+c_2\left[\begin{array}{r}-1\\1\\ 0\end{array}\right] e^{-2t}+c_3\left[\begin{array}{r}-7\\-5\\4\end{array}\right] e^{2t}}\)

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

    Show answer

    \({\bf y}=\dst{c_1\left[\begin{array}{r}-1\\-1\\1 \end{array}\right]e^{-2t}+c_2\left[\begin{array}{r}-1\\-2\\1 \end{array} \right]e^{-3t}+c_3\left[\begin{array}{r}-2\\-6\\3\end{array} \right]e^{-5t}}\)

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

    Show answer

    \({\bf y}=\dst{c_1\left[\begin{array}{r}11\\\phantom{1}7 \\\phantom{1}1\end{array}\right]e^{3t}+c_2\left[\begin{array}{r}1\\2 \\1\end{array}\right]e^{-2t}+c_3\left[\begin{array}{r}1\\1\\1 \end{array}\right]e^{-t}}\)

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

    Show answer

    \({\bf y}=\dst{c_1\left[\begin{array}{r}4\\-1\\1\end{array} \right]e^{-4t}+c_2\left[\begin{array}{r}-1\\-1\\1 \end{array}\right]e^{6t}+c_3\left[\begin{array}{r}-1\\0\\1 \end{array}\right]e^{4t}}\)

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

    Show answer

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

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

    Show answer

    \({\bf y}=\dst{c_1\left[\begin{array}{r}1\\ -1\\2 \end{array}\right]+c_2\left[\begin{array}{r}-1\\0\\3\end{array} \right]e^{6t}+c_3\left[\begin{array}{r}1\\3\\0 \end{array}\right]e^{6t}}\)

16. \(\dst{{\bf y}'=\twobytwo{-7}4{-6}7{\bf y},\quad{\bf y}(0)=\twocol2{-4}}\)

    Show answer

    \({\bf y}=\dst{-\twocol26e^{5t}+\twocol42e^{-5t}}\)

17. \(\dst{{\bf y}'={1\over6}\twobytwo72{-2}2{\bf y},\quad{\bf y}(0)=\twocol0{-3}}\)

    Show answer

    \({\bf y}=\dst{\twocol2{-4}e^{t/2}+\twocol{-2}1e^{t}}\)

18. \(\dst{{\bf y}'=\twobytwo{21}{-12}{24}{-15}{\bf y},\quad{\bf y}(0)=\twocol53}\)

    Show answer

    \({\bf y}=\dst{\twocol77e^{9t}-\twocol24e^{-3t}}\)

19. \(\dst{{\bf y}'=\twobytwo{-7}4{-6}7{\bf y},\quad{\bf y}(0)=\twocol{-1}7}\)

    Show answer

    \({\bf y}=\dst{\twocol39e^{5t}-\twocol42e^{-5t}}\)

20. \(\dst{{\bf y}'={1\over6}\threebythree1204{-1}0003{\bf y},\quad{\bf y}(0)=\threecol471}\)

    Show answer

    \({\bf y}=\dst{\threecol550e^{t/2}+\threecol001e^{t/2}+ \threecol{-1}20e^{-t/2}}\)  

21. \(\dst{{\bf y}'={1\over3}\threebythree2{-2}3{-4}43210{\bf y},\quad{\bf y}(0)=\threecol115}\)

    Show answer

    \({\bf y}=\dst{\threecol333e^t+\threecol{-2}{-2}2e^{-t}}\)

22. \(\dst{{\bf y}'=\threebythree6{-3}{-8}21{-2}3{-3}{-5}{\bf y},\quad{\bf y}(0)=\threecol0{-1}{-1}}\)

    Show answer

    \({\bf y}=\dst{\threecol2{-2}2e^t-\threecol303e^{-2t}+\threecol110e^{3t}}\)

23. \(\dst{{\bf y}'={1\over3}\threebythree24{-7}15{-5}{-4}4{-1}{\bf y},\quad{\bf y}(0)=\threecol413}\)

    Show answer

    \({\bf y}=\dst{-\threecol121e^t+\threecol424e^{-t}+\threecol110e^{2t}}\)

24. \(\dst{ {\bf y}'=\threebythree301{11}{-2}7103{\bf y},\quad {\bf y}(0)=\threecol276}\)

    Show answer

    \({\bf y}=\dst{ \threecol{-2}{-2}2e^{2t}-\threecol030e^{-2t}+ \threecol4{12}4e^{4t}}\)

25. \(\dst{ {\bf y}'=\threebythree{-2}{-5}{-1}{-4}{-1}145{3}{\bf y},\quad {\bf y}(0)=\threecol8{-10}{-4}}\)

    Show answer

    \({\bf y}=\dst{ \threecol{-1}{-1}1e^{-6t}+\threecol2{-2}2e^{2t}+ \threecol7{-7}{-7}e^{4t}}\)

26. \(\dst{ {\bf y}'=\threebythree3{-1}04{-2}04{-4}2{\bf y},\quad {\bf y}(0)=\threecol7{10}2}\)

    Show answer

    \({\bf y}=\dst{ \threecol144e^{-t}+\threecol66{-2}e^{2t}}\)

27. \(\dst{{\bf y}'= \left[\begin{array}{rrr}-2&2&6\\2&6&2\\-2&-2& 2\end{array}\right]{\bf y}},\quad{\bf y}(0)=\threecol6{-10}7\)

    Show answer

    \({\bf y}=\dst{ \threecol4{-2}2+\threecol3{-9}6e^{4t}+\threecol{-1}1{-1}e^{2t}}\)

28. Let \(A\) be an \(n\times n\) constant matrix. Then Theorem 10.2.1 implies that the solutions of

    \[ {\bf y}'=A{\bf y} \tag*{\rm(A)} \]

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

    1. Use Theorem 10.2.1 to show that the only solution of (A) that can ever equal the zero vector is \({\bf y}\equiv{\bf0}\).

    2. Suppose \({\bf y}_1\) is a solution of (A) 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 (A).

    3. Suppose \({\bf y}_1\) and \({\bf y}_2\) are solutions of (A) 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 (A), and apply the uniqueness assertion of Theorem 10.2.1.

29. C/G \({\bf y}'=\dst{\twobytwo111{-1}}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=y_1\) and \(L_2:y_2=-y_1\) are trajectories other trajectories

    are asymptotically tangent to \(L_1\) as \(t\to-\infty\) and asymptotically tangent to \(L_2\) as \(t\to\infty\).

30. C/G \({\bf y}'=\dst{\twobytwo{-4}3{-2}{-11}}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=-2y_1\) and \(L_2:y_2=-y_1/3\) are trajectories

    other trajectories are asymptotically parallel to \(L_1\) as \(t\to-\infty\) and asymptotically tangent to \(L_2\) as \(t\to\infty\).

31. C/G \({\bf y}'=\dst{\twobytwo9{-3}{-1}{11}}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=y_1/3\) and \(L_2:y_2=-y_1\) are trajectories other trajectories

    are asymptotically tangent to \(L_1\) as \(t\to-\infty\) and asymptotically parallel to \(L_2\) as \(t\to\infty\).

32. C/G \({\bf y}'=\dst{\twobytwo{-1}{-10}{-5}4}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=y_1/2\) and \(L_2:y_2=-y_1\) are trajectories other trajectories

    are asymptotically tangent to \(L_1\) as \(t\to-\infty\) and asymptotically tangent to \(L_2\) as \(t\to\infty\).

33. C/G \({\bf y}'=\dst{\twobytwo5{-4}1{10}}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=-y_1/4\) and \(L_2:y_2=-y_1\) are trajectories other trajectories

    are asymptotically tangent to \(L_1\) as \(t\to-\infty\) and asymptotically parallel to \(L_2\) as \(t\to\infty\).

34. C/G \({\bf y}'=\dst{\twobytwo{-7}13{-5}}{\bf y}\)

    Show answer

    Half lines of \(L_1:y_2=-y_1\) and \(L_2:y_2=3y_1\) are trajectories other trajectories

    are asymptotically parallel to \(L_1\) as \(t\to-\infty\) and asymptotically tangent to \(L_2\) as \(t\to\infty\).

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 \({\bf x}_1\).

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

    2. 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 \(\mu<0\).

36. C/G \({\bf y}'=\dst{\twobytwo{-1}11{-1}}{\bf y}\)

    Show answer

    Points on \(L_2:y_2=y_1\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_1\), parallel to \(\left[\begin{array}{r} 1\\{-1} \end{array}\right]\), traversed toward \(L_1\).

37. C/G \({\bf y}'=\dst{\twobytwo{-1}{-3}26}{\bf y}\)

    Show answer

    Points on \(L_1:y_2=-y_1/3\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_1\), parallel to \(\left[\begin{array}{r} -1\\2 \end{array}\right]\), traversed away from \(L_1\).

38. C/G \({\bf y}'=\dst{\twobytwo1{-3}{-1}3}{\bf y}\)

    Show answer

    Points on \(L_1:y_2=y_1/3\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_1\), parallel to \(\left[\begin{array}{r} 1\\{-1} \end{array}\right]\), \(\dst{\twocol{\phantom-}1{-1}}\), traversed away from \(L_1\).

39. C/G \({\bf y}'=\dst{\twobytwo1{-2}{-1}2}{\bf y}\)

    Show answer

    Points on \(L_1:y_2=y_1/2\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_1\), parallel to \(\left[\begin{array}{r} 1\\-1 \end{array}\right]\), \(L_1\).

40. C/G \({\bf y}'=\dst{\twobytwo{-4}{-4}11}{\bf y}\)

    Show answer

    Points on \(L_2:y_2=-y_1\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_2\), parallel to \(\left[\begin{array}{r} -4\\1 \end{array}\right]\), traversed toward \(L_1\).

41. C/G \({\bf y}'=\dst{\twobytwo3{-1}{-3}1}{\bf y}\)

    Show answer

    Points on \(L_1:y_2=3y_1\) are trajectories of constant solutions. The trajectories

    of nonconstant solutions are half-lines on either side of \(L_1\), parallel to \(\left[\begin{array}{r} 1\\{-1} \end{array}\right]\), traversed away from \(L_1\).

42. L 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,

    \[ P'=aP \mbox{\quad and \quad}Q'=bQ, \tag*{\rm(A)} \]

    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 (A) 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 \(Q(0)=Q_0>0\).

    1. For several choices of \(a\), \(b\), \(\alpha\), and \(\beta\), verify experimentally (by graphing trajectories of (A) in the \(P\)-\(Q\) plane) that there’s a constant \(\rho>0\) (depending upon \(a\), \(b\), \(\alpha\), and \(\beta\)) with the following properties:

       1. If \(Q_0>\rho P_0\), then \(P\) decreases monotonically to zero in finite time, during which \(Q\) remains positive.

       2. If \(Q_0<\rho P_0\), then \(Q\) decreases monotonically to zero in finite time, during which \(P\) remains positive.

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

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

       \[ A=\twobytwo a{-\alpha}{-\beta}b \]

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