10.6.1: Constant Coefficient Homogeneous Systems III (Exercises)

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Q10.6.1

In Exercises 10.6.1-10.6.16 find the general solution.

1. ${\bf y}'=\left[\begin{array}{cc}{-1}&{2}\\{-5}&{5}\end{array}\right]{\bf y}$

2. ${\bf y}'=\left[\begin{array}{cc}{-11}&{4}\\{-26}&{9}\end{array}\right]{\bf y}$

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

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

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

6. ${\bf y}'=\left[\begin{array}{ccc}{-3}&{3}&{1}\\{1}&{-5}&{-3}\\{-3}&{7}&{3}\end{array}\right]{\bf y}$

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

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

9. ${\bf y}'=\left[\begin{array}{cc}{5}&{-4}\\{10}&{1}\end{array}\right]{\bf y}$

10. ${\bf y}'=\frac{1}{3}\left[\begin{array}{cc}{7}&{-5}\\{2}&{5}\end{array}\right]{\bf y}$

11. ${\bf y}'=\left[\begin{array}{cc}{3}&{2}\\{-5}&{1}\end{array}\right]{\bf y}$

12. ${\bf y}'=\left[\begin{array}{cc}{34}&{52}\\{-20}&{-30}\end{array}\right]{\bf y}$

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

14. ${\bf y}'=\left[\begin{array}{ccc}{3}&{-4}&{-2}\\{-5}&{7}&{-8}\\{-10}&{13}&{-8}\end{array}\right]{\bf y}$

15. ${\bf y}'=\left[\begin{array}{ccc}{6}&{0}&{-3}\\{-3}&{3}&{3}\\{1}&{-2}&{6}\end{array}\right]{\bf y}$

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

Q10.6.2

In Exercises 10.6.17-10.6.24 solve the initial value problem.

17. ${\bf y}'=\left[\begin{array}{cc}{4}&{-6}\\{3}&{-2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{5}\\{2}\end{array}\right]$

18. ${\bf y}'=\left[\begin{array}{cc}{7}&{15}\\{-3}&{1}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{5}\\{1}\end{array}\right]$

19. ${\bf y}'=\left[\begin{array}{cc}{7}&{-15}\\{3}&{-5}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{17}\\{7}\end{array}\right]$

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

21. ${\bf y}'=\left[\begin{array}{ccc}{5}&{2}&{-1}\\{-3}&{2}&{2}\\{1}&{3}&{2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{4}\\{0}\\{6}\end{array}\right]$

22. ${\bf y}'=\left[\begin{array}{ccc}{4}&{4}&{0}\\{8}&{10}&{-20}\\{2}&{3}&{-2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{8}\\{6}\\{5}\end{array}\right]$

23. ${\bf y}'=\left[\begin{array}{ccc}{1}&{15}&{-15}\\{-6}&{18}&{-22}\\{-3}&{11}&{-15}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{15}\\{17}\\{10}\end{array}\right]$

24. ${\bf y}'=\left[\begin{array}{ccc}{4}&{-4}&{4}\\{-10}&{3}&{15}\\{2}&{-3}&{1}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{16}\\{14}\\{6}\end{array}\right]$

Q10.6.3

25. Suppose an $n\times n$ matrix $A$ with real entries has a complex eigenvalue $\lambda=\alpha+i\beta$ ($\beta\ne0$) with associated eigenvector ${\bf x}={\bf u}+i{\bf v}$, where ${\bf u}$ and ${\bf v}$ have real components. Show that ${\bf u}$ and ${\bf v}$ are both nonzero.

26. Verify that

\[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \quad \text{and}\quad \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),\nonumber\]

are the real and imaginary parts of

\[e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}).\nonumber\]

27. Show that if the vectors ${\bf u}$ and ${\bf v}$ are not both ${\bf 0}$ and $\beta\ne0$ then the vector functions

\[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad \mbox{ and }\quad \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)\nonumber\]

are linearly independent on every interval.

28. Suppose ${\bf u}=\left[\begin{array}{c}{u_{1}}\\{u_{2}}\end{array}\right]$ and ${\bf v}=\left[\begin{array}{c}{v_{1}}\\{v_{2}}\end{array}\right]$ are not orthogonal; that is, $({\bf u},{\bf v})\ne0$.

Show that the quadratic equation \[({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0\nonumber\] has a positive root $k_1$ and a negative root $k_2=-1/k_1$.
Let ${\bf u}_1^{(1)}={\bf u}-k_1{\bf v}$, ${\bf v}_1^{(1)}={\bf v}+k_1{\bf u}$, ${\bf u}_1^{(2)}={\bf u}-k_2{\bf v}$, and ${\bf v}_1^{(2)}={\bf v}+k_2{\bf u}$, so that $({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0$, from the discussion given above. Show that \[{\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \quad \text{and} \quad {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.\nonumber\]
Let ${\bf U}_1$, ${\bf V}_1$, ${\bf U}_2$, and ${\bf V}_2$ be unit vectors in the directions of ${\bf u}_1^{(1)}$, ${\bf v}_1^{(1)}$, ${\bf u}_1^{(2)}$, and ${\bf v}_1^{(2)}$, respectively. Conclude from (a) that ${\bf U}_2={\bf V}_1$ and ${\bf V}_2=-{\bf U}_1$, and that therefore the counterclockwise angles from ${\bf U}_1$ to ${\bf V}_1$ and from ${\bf U}_2$ to ${\bf V}_2$ are both $\pi/2$ or both $-\pi/2$.

Q10.6.4

In Exercises 10.6.29-10.6.32 find vectors ${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the trajectories, and plot some typical trajectories.

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

30. ${\bf y}'=\left[\begin{array}{cc}{-15}&{10}\\{-25}&{15}\end{array}\right]{\bf y}$

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

32. ${\bf y}'=\left[\begin{array}{cc}{-3}&{-15}\\{3}&{3}\end{array}\right]{\bf y}$

Q10.6.5

In Exercises 10.6.33-10.6.40 find vectors ${\bf U}$ and ${\bf V}$ parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory.

33. ${\bf y}'=\left[\begin{array}{cc}{-5}&{6}\\{-12}&{7}\end{array}\right]{\bf y}$

34. ${\bf y}'=\left[\begin{array}{cc}{5}&{-12}\\{6}&{-7}\end{array}\right]{\bf y}$

35. ${\bf y}'=\left[\begin{array}{cc}{4}&{-5}\\{9}&{-2}\end{array}\right]{\bf y}$

36. ${\bf y}'=\left[\begin{array}{cc}{-4}&{9}\\{-5}&{2}\end{array}\right]{\bf y}$

37. ${\bf y}'=\left[\begin{array}{cc}{-1}&{10}\\{-10}&{-1}\end{array}\right]{\bf y}$

38. ${\bf y}'=\left[\begin{array}{cc}{-1}&{-5}\\{20}&{-1}\end{array}\right]{\bf y}$

39. ${\bf y}'=\left[\begin{array}{cc}{-7}&{10}\\{-10}&{9}\end{array}\right]{\bf y}$

40. ${\bf y}'=\left[\begin{array}{cc}{-7}&{6}\\{-12}&{5}\end{array}\right]{\bf y}$