HomeWork FOUR

5.2.1 The eigenvalues are $2$ and $3$, and eigenvectors are $(1, 2)^T$ and $(1,1)^T$. The general solution is then

$$\left(\begin{array}{c}x\\ y \end{array}\right) = c_1 \left(\begi... ...ay}\right)e^{2 t} + c_2 \left(\begin{array}{c}1\\ 1 \end{array}\right)e^{3 t}.$$

The particular solution is then $c_1=c_2=1$, or

$$\left(\begin{array}{cc}x\\ y \end{array}\right) = \left(\begin{a... ...array}\right)e^{2 t} + \left(\begin{array}{cc}2\\ 2 \end{array}\right)e^{3 t}.$$

5.2.2 We have done this procedure during the class, so please consult lecture notes 5.2.6

 
StreamPlot[{-3 x+2 y, x - 2y}, {x,-5,5},{y,-5,5}]

5.2.13

$$\frac{d}{d t}\left(\begin{array}{cc}x\\ y \end{array}\right) = \... ...frac{-b}{m} \end{array}\right) \left(\begin{array}{c}x\\ y \end{array}\right).$$

with the eigenvalue

$$\lambda=\frac{-b\pm \sqrt{b^2 - 4 m k }}{2 m}.$$

If $b^2>4 m k$ then there are two real negative roots and the origin is stable, and the phase portrait will be stable node:

 
StreamPlot[{x+y, -2x - 3y}, {x,-5,5},{y,-5,5}]

Here we take $m=1/2, b = 3/2; k=1.$ This is over damped oscillator.

If $b^2 = 2 m$ then there are two identical roots, and the oscillator becomes critically damped, which corresponds to the stable deffective (degenerate) node:

 
StreamPlot[{x+y, -x - 2 y}, {x,-5,5},{y,-5,5}]

Here we take $m=1, b = 2, k =1$ If $b^2 < 4 m k$ then the origin is assymptotically stable, since there are two complex roots with negative real part. The phase portrait becomes the stable spiral:

 
StreamPlot[{x+y, -x - .2 y}, {x,-5,5},{y,-5,5}]

5.3.3

$$\dot R = a J, \dot J = b R,$$

with

$$\lambda = \pm \sqrt{ a b}.$$

The case $a b<0$ corresponds to the center, and $a b>0$ to the saddle point 5.3.5 Here

$$\lambda = a\pm b.$$

Then

• $a<b$ is a saddle point,
• $a<b<0$ is a stable node, and
• $0<b<a$ is and unstable node