Ordinary Differential Equation Consider a system of differential equation
y'(t) = By(t) where B is a 2X2 matrix defined as: B = [α , -β ; β , α]
where α,β ∈ ℝ and (α,β) ≠  (0,0).
A) If the roots of B are purely imaginary verify that the the solution of this system y=( y1 , y2 ) is of the form y1(t) = c1 cos(c2 + βt) and y2(t) = c1 sin(c2 + βt).
B) Characterise the orbit of y, namely y(ℝ+). Graph the orbit in ℝ2.
C) Comment on the stability properties of y when the roots of B are purely imaginary.
D) If the roots are not purely imaginary then verify that the solution of this system y=( y1,y2 ) is of the form y1(t) = c1 exp(αt)cos(c2 + βt) and y2(t) = c1 exp(αt)sin(c2 + βt).
E) Derive and graph the orbit of y when the roots of B are not purely imaginary.
F) Comment on the stability properties of y when the roots of B are not purely imaginary.
G) What determines the direction of rotation of the orbit as t ↑ ∞ ?
 A: Hints (given that this is from an entrance exam).
Part A
We have: $B = \begin{bmatrix}\alpha & -\beta\\\beta & \alpha\end{bmatrix}$
We can find the eigenvalues and eigenvectors of this matrix as:


*

*$\lambda_1 = \alpha - i \beta, ~~v_1 = (-i, 1)$

*$\lambda_2 = \alpha + i \beta, ~~v_2 = (i, 1)$


Since the roots are purely imaginary, that is $\alpha = 0$, we can write the solution $y(t)$ as follows:
$y(t) = \begin{bmatrix} y_1(t) \\ y_2(t)\end{bmatrix} = \begin{bmatrix} c_1  \cos \beta t - c_2 \sin \beta t \\ c_1 \sin \beta t + c_2 \cos \beta t \end{bmatrix} = \begin{bmatrix} c_1 \cos(c_2 + \beta t) \\c_1 \sin(c_2 + \beta t) \end{bmatrix}$
Note: the last part uses the fact that we can shift $\cos$ and $\sin$ terms to combine them. We can easily verify this solution by replacing it in the original equation and verifying it satisfies it.
Part B
The orbits are going to be spirals when the eigenvalues are purely imaginary as can be seen from this phase portrait.
.
Part C
These are stable (but not asymptotically stable); sometimes it is referred to as neutrally stable.
Part D
Form the eigenvalue/eigenvector pairs determined above, we have:
$y(t) = \begin{bmatrix} y_1(t) \\ y_2(t)\end{bmatrix} = \begin{bmatrix} c_1 e^{\alpha t} \cos (\beta t) - c_2 e^{\alpha t} \sin (\beta t) \\ c_1 e^{\alpha t} \sin (\beta t) + c_2 e^{\alpha t} \cos (\beta t) \end{bmatrix} = \begin{bmatrix} c_1 e^{\alpha t} \cos(c_2 + \beta t) \\c_1 e^{\alpha t} \sin(c_2 + \beta t) \end{bmatrix}$
Part E
We can draw a phase portrait for various cases (we already did $\alpha = 0$, so have to see $\alpha > 0$ and $\alpha < 0$).
For $\alpha > 0$, we have:

For $\alpha < 0$, we have:

Part F


*

*$\alpha > 0$, we have an unstable spiral point at the origin.

*$\alpha < 0$, we have stable spiral point at the origin.


Part G
The sign of $\alpha$ as that leads to the time dependent exponential term. Also, the magnitude of $\alpha$ determines the rate.
A: As your eigenvalues are purely imaginary,they have only imaginary parts . That's why the phase portrait looks like circles or ellipses (depends on values of imaginary parts)

When eigenvalues are not purely imaginary, you'll have focus as a phase portrait.
