Answers to Exercises for Chapter 4

Second-Order Linear Systems

As two first-order differential equations, $ \ddot x + 4\dot x + 5 x=0 $ is $$ \dot x= y, \ \dot y = -4y -5x $$
As a second-order differential equation for $x$, $$\ddot x =2\dot x -3(6x-y)=\dot x-16x. $$ Similarly for $ y$, $$\ddot y = 6(2x-3y)-\dot y= \dot y -16y,$$ so we get the same second-order equation.
For the system $$\dot x = x + 3 y, \ \dot y = 3x -7y,$$ the matrix form is $$M = \begin{pmatrix} 1 & 3\\ 3 & -7 \end{pmatrix}$$ with eigenvalues 2 and -8, eigenvectors $ (3, 1),\ (1,-3)$. These are orthogonal as expected because the matrix is symmetric. The general solution is $$x=3a e^{2t}+ be^{-8t},\ y=a e^{2t}-3be^{-8t}.$$ The specific solution for the initial conditions $x(0)=10, \ y(0)=0$ is $$x=9 e^{2t} + e^{-8t},\ y=3 e^{2t}-3e^{-8t}.$$
1. $$M=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$$ $ T = 5, \ D=-2$, saddle.
2. $$M=\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$$ $T=6,\ D=5$, unstable node.
3. $$M=\begin{pmatrix} 2 & 2 \\ -3 & -2 \end{pmatrix}$$ $T=0,\ D= 2$, centre.
4. $$M=\begin{pmatrix} 2 & 2 \\ -2 & -1 \end{pmatrix}$$ $T=1, \ D=2$, unstable spiral.

1. $$ M = \begin{pmatrix} 2 & -3 \\ 4 & -4\end{pmatrix} $$ Eigenvalues are $-1\pm i \sqrt 3 $, no need for eigenvectors, stable spiral, solutions oscillate and approach the origin.
2. $$ M = \begin{pmatrix} -1 & 1 \\ 6 & -2\end{pmatrix} $$ Eigenvalues 1, -4, so it's a saddle. Eigenvectors $ (1,2), \ (1,-3) $. Solutions tend to infinity in the direction of the vector (1,2).
3. $$ M = \begin{pmatrix} -1 & 2\\ 2 & -4\end{pmatrix} $$ Eigenvalues are 0, with vector $ (2,1) $, and $ -5$ with vector $ (1,-2) $. There are fixed points all along the line $ y=x/2 $, and solutions tend to a point on this line, depending on the initial condition.

For the third-order system $$\dot x = x + y - 4z, \ \dot y = - y, \ \dot z = 2x+3y -3z , $$ the $ y$ variable is uncoupled and decays to zero. That leaves a 2D system for $ x, \ z$ which is a stable spiral, so solutions spiral in to the origin in that plane.
If $M$ is idempotent, $M^2=M$. Hence the exponential of the matrix is $$\exp(M)=I +M +M/2! + M/3! + \dots$$ which can be written (after adding and subtracting $ M$) as $$\exp(M)= e M +I-M . $$ So for the matrix given in the question, first check that it is idempotent, then $$\exp(M)=\begin{pmatrix} e & 3e-3\\0 & 1 \end{pmatrix}.$$