10.5.1: Constant Coefficient Homogeneous Systems II (Exercises)

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Q10.5.1

In Exercises 10.5.1-10.5.12 find the general solution.

1. ${\bf y}'=\left[\begin{array}{cc}{3}&{4}\\[4pt]{-1}&{7}\end{array}\right]{\bf y}$

2. ${\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\[4pt]{1}&{-2}\end{array}\right]{\bf y}$

3. ${\bf y}'=\left[\begin{array}{cc}{-7}&{4}\\[4pt]{-1}&{-11}\end{array}\right]{\bf y}$

4. ${\bf y}'=\left[\begin{array}{cc}{3}&{1}\\[4pt]{-1}&{1}\end{array}\right]{\bf y}$

5. ${\bf y}'=\left[\begin{array}{cc}{4}&{12}\\[4pt]{-3}&{-8}\end{array}\right]{\bf y}$

6. ${\bf y}'=\left[\begin{array}{cc}{-10}&{9}\\[4pt]{-4}&{2}\end{array}\right]{\bf y}$

7. ${\bf y}'=\left[\begin{array}{cc}{-13}&{16}\\[4pt]{-9}&{11}\end{array}\right]{\bf y}$

8. ${\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{1}\\[4pt]{-4}&{6}&{1}\\[4pt]{0}&{4}&{2}\end{array}\right]{\bf y}$

9. ${\bf y}'=\frac{1}{3}\left[\begin{array}{ccc}{1}&{1}&{-3}\\[4pt]{-4}&{-4}&{3}\\[4pt]{-2}&{1}&{0}\end{array}\right]{\bf y}$

10. ${\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{-1}\\[4pt]{-2}&{0}&{2}\\[4pt]{-1}&{3}&{-1}\end{array}\right]{\bf y}$

11. ${\bf y}'=\left[\begin{array}{ccc}{4}&{-2}&{-2}\\[4pt]{-2}&{3}&{-1}\\[4pt]{2}&{-1}&{3}\end{array}\right]{\bf y}$

12. ${\bf y}'=\left[\begin{array}{ccc}{6}&{-5}&{3}\\[4pt]{2}&{-1}&{3}\\[4pt]{2}&{1}&{1}\end{array}\right]{\bf y}$

Q10.5.2

In Exercises 10.5.13-10.5.23 solve the initial value problem.

13. ${\bf y}'=\left[\begin{array}{cc}{-11}&{8}\\[4pt]{-2}&{-3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\[4pt]{2}\end{array}\right]$

14. ${\bf y}'=\left[\begin{array}{cc}{15}&{-9}\\[4pt]{16}&{-9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{5}\\[4pt]{8}\end{array}\right]$

15. ${\bf y}'=\left[\begin{array}{cc}{-3}&{-4}\\[4pt]{1}&{-7}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{2}\\[4pt]{3}\end{array}\right]$

16. ${\bf y}'=\left[\begin{array}{cc}{-7}&{24}\\[4pt]{-6}&{17}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{3}\\[4pt]{1}\end{array}\right]$

17. ${\bf y}'=\left[\begin{array}{cc}{-7}&{3}\\[4pt]{-3}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\[4pt]{2}\end{array}\right]$

18. ${\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{0}\\[4pt]{1}&{-1}&{-2}\\[4pt]{-1}&{-1}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\[4pt]{5}\\[4pt]{-7}\end{array}\right]$

19. ${\bf y}'=\left[\begin{array}{ccc}{-2}&{2}&{1}\\[4pt]{-2}&{2}&{1}\\[4pt]{-3}&{3}&{2}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\[4pt]{-2}\\[4pt]{0}\end{array}\right]$

20. ${\bf y}'=\left[\begin{array}{ccc}{-7}&{-4}&{4}\\[4pt]{-1}&{0}&{1}\\[4pt]{-9}&{-5}&{6}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\[4pt]{9}\\[4pt]{-1}\end{array}\right]$

21. ${\bf y}'=\left[\begin{array}{ccc}{-1}&{-4}&{-1}\\[4pt]{3}&{6}&{1}\\[4pt]{-3}&{-2}&{3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-2}\\[4pt]{1}\\[4pt]{3}\end{array}\right]$

22. ${\bf y}'=\left[\begin{array}{ccc}{4}&{-8}&{-4}\\[4pt]{-3}&{-1}&{-3}\\[4pt]{1}&{-1}&{9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-4}\\[4pt]{1}\\[4pt]{-3}\end{array}\right]$

23. ${\bf y}'=\left[\begin{array}{ccc}{-5}&{-1}&{11}\\[4pt]{-7}&{1}&{13}\\[4pt]{-4}&{0}&{8}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\[4pt]{2}\\[4pt]{2}\end{array}\right]$

Q10.5.3

The coefficient matrices in Exercises 10.5.24-10.5.32 have eigenvalues of multiplicity $3$. Find the general solution.

24. ${\bf y}'=\left[\begin{array}{ccc}{5}&{-1}&{1}\\[4pt]{-1}&{9}&{-3}\\[4pt]{-2}&{2}&{4}\end{array}\right]{\bf y}$

25. ${\bf y}'=\left[\begin{array}{ccc}{1}&{10}&{-12}\\[4pt]{2}&{2}&{3}\\[4pt]{2}&{-1}&{6}\end{array}\right]{\bf y}$

26. ${\bf y}'=\left[\begin{array}{ccc}{-6}&{-4}&{-4}\\[4pt]{2}&{-1}&{1}\\[4pt]{2}&{3}&{1}\end{array}\right]{\bf y}$

27. ${\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{-2}\\[4pt]{-1}&{5}&{-3}\\[4pt]{1}&{1}&{1}\end{array}\right]{\bf y}$

28. ${\bf y}'=\left[\begin{array}{ccc}{-2}&{-12}&{10}\\[4pt]{2}&{-24}&{11}\\[4pt]{2}&{-24}&{8}\end{array}\right]{\bf y}$

29. ${\bf y}'=\left[\begin{array}{ccc}{-1}&{-12}&{8}\\[4pt]{1}&{-9}&{4}\\[4pt]{1}&{-6}&{1}\end{array}\right]{\bf y}$

30. ${\bf y}'=\left[\begin{array}{ccc}{-4}&{0}&{-1}\\[4pt]{-1}&{-3}&{-1}\\[4pt]{1}&{0}&{-2}\end{array}\right]{\bf y}$

31. ${\bf y}'=\left[\begin{array}{ccc}{-3}&{-3}&{4}\\[4pt]{4}&{5}&{-8}\\[4pt]{2}&{3}&{-5}\end{array}\right]{\bf y}$

32. ${\bf y}'=\left[\begin{array}{ccc}{-3}&{-1}&{0}\\[4pt]{1}&{-1}&{0}\\[4pt]{-1}&{-1}&{-2}\end{array}\right]{\bf y}$

Q10.5.4

33. Under the assumptions of Theorem 10.5.1, suppose ${\bf u}$ and $\hat{\bf u}$ are vectors such that

\[(A-\lambda_1I){\bf u}={\bf x}\quad\mbox{and }\quad (A-\lambda_1I)\hat{\bf u}={\bf x},\nonumber \]

and let

\[{\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t} \quad\mbox{and }\quad \hat{\bf y}_2=\hat{\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}.\nonumber \]

Show that ${\bf y}_2-\hat{\bf y}_2$ is a scalar multiple of ${\bf y}_1={\bf x}e^{\lambda_1t}$.

34. Under the assumptions of Theorem 10.5.2, let

\[\begin{aligned} {\bf y}_1 &={\bf x} e^{\lambda_1t},\\[4pt] {\bf y}_2&={\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\[4pt] {\bf y}_3&={\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}.\end{aligned}\nonumber \]

Complete the proof of Theorem 10.5.2 by showing that ${\bf y}_3$ is a solution of ${\bf y}'=A{\bf y}$ and that $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is linearly independent.

35. Suppose the matrix \[A=\left[\begin{array}{cc}{a_{11}}&{a_{12}}\\[4pt]{a_{21}}&{a_{22}}\end{array}\right]\nonumber \] has a repeated eigenvalue $\lambda_1$ and the associated eigenspace is one-dimensional. Let ${\bf x}$ be a $\lambda_1$-eigenvector of $A$. Show that if $(A-\lambda_1I){\bf u}_1={\bf x}$ and $(A-\lambda_1I){\bf u}_2={\bf x}$, then ${\bf u}_2-{\bf u}_1$ is parallel to ${\bf x}$. Conclude from this that all vectors ${\bf u}$ such that $(A-\lambda_1I){\bf u}={\bf x}$ define the same positive and negative half-planes with respect to the line $L$ through the origin parallel to ${\bf x}$.

Q10.5.5

In Exercises 10.5.36-10.5.45 plot trajectories of the given system.

36. ${\bf y}'=\left[\begin{array}{cc}{-3}&{-1}\\[4pt]{4}&{1}\end{array}\right]{\bf y}$

37. ${\bf y}'=\left[\begin{array}{cc}{2}&{-1}\\[4pt]{1}&{0}\end{array}\right]{\bf y}$

38. ${\bf y}'=\left[\begin{array}{cc}{-1}&{-3}\\[4pt]{3}&{5}\end{array}\right]{\bf y}$

39. ${\bf y}'=\left[\begin{array}{cc}{-5}&{3}\\[4pt]{-3}&{1}\end{array}\right]{\bf y}$

40. ${\bf y}'=\left[\begin{array}{cc}{-2}&{-3}\\[4pt]{3}&{4}\end{array}\right]{\bf y}$

41. ${\bf y}'=\left[\begin{array}{cc}{-4}&{-3}\\[4pt]{3}&{2}\end{array}\right]{\bf y}$

42. ${\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\[4pt]{1}&{-2}\end{array}\right]{\bf y}$

43. ${\bf y}'=\left[\begin{array}{cc}{0}&{1}\\[4pt]{-1}&{2}\end{array}\right]{\bf y}$

44. ${\bf y}'=\left[\begin{array}{cc}{-2}&{1}\\[4pt]{-1}&{0}\end{array}\right]{\bf y}$

45. ${\bf y}'=\left[\begin{array}{cc}{0}&{-4}\\[4pt]{1}&{-4}\end{array}\right]{\bf y}$

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