What Are the Phase Lines and Equilibrium Points for These Differential Systems?

[MrBioMedic]

Hello everyone - I have a couple questions about some homework problems that I am hoping I can get some help with - Here they are and thank you for the help in advance.


for the system dY / dt = | a 1 | (2 by 2 matrix)
----------------------- | 0 -1 |


and the system dY / dt = | a 1 |
------------------------- | -1 2 |



I need to figure out the following:
a. possible phase lines for each as a varies.
b. classify the equilbrium points
c. the general solution for a=2


Thank you again for any help.

 

[ReyChiquito]

For the problem

$$\frac{d \vec{y}}{dt}=\bold{A}\vec{y}$$

where $\bold{A}$ is a matrix of constant coefficients, you know that the general solution for the system is
$$\vec{y}(t)=\vec{y_{0}}e^{\bold{A}t}$$

so, here is the canonical way to go...

diagonalize $\bold{A}$. You will get your solution to be in the form*
$$\vec{z}(t)=\left(\begin{array}{cc}z_{01}e^{\lambda_{1}t}\\z_{02}e^{\lambda_{2}t}\end{array}\right)$$

Where $\vec{y}=\bold{P}\vec{z}$ and $P$ is the rotation matrix formed by the eigenvectors of [itex]\bold{A}[/tex], and $\lambda_{1},\lambda_{2}$ are the eigenvalues.

So now you can express $y=y(x,\lambda_{1},\lambda_{2})$ and see the form of the integral paths in the phase plane.

Now, the equilibrium points will be classified by these integral curves. The direction of the paths in this points is given by the eigenvectors.

* This will be the form of the solution only when $\bold{A}$ is similar to a diagonal matrix. If not, the matrix will be similar to a Jordan block, and your solution will change (for further reference check almost any ode book), and i think you will have to do that case also.

PS. sorry for bad english.

TO MODS. The Spellcheck should skip text between [tex] tags... that would be nice.. just an idea

 

[M98Ranger]

Hi there! I'd be happy to help with your differential homework questions. Let's take a look at each of the problems and break down how to approach them.

For the first system, we have the equation dY / dt = | a 1 | (2 by 2 matrix)
----------------------- | 0 -1 |
To determine the possible phase lines for each value of a, we can use the eigenvalues of the matrix. The eigenvalues of a 2 by 2 matrix are given by the formula λ = (a - d) ± √((a - d)² + 4bc) / 2, where a, b, c, and d are the values in the matrix. In this case, we have a = a, b = 1, c = 0, and d = -1.

For the first system, we have a phase line that looks like this:

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
|
|
|
|
|
|
|
|
|
|
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

For the second system, we have the equation dY / dt = | a 1 | (2 by 2 matrix)
----------------------- | -1 2 |
Using the same formula, we can find the eigenvalues to be λ = (a - d) ± √((a - d)² + 4bc) / 2. In this case, we have a = a, b = 1, c = -1, and d = 2.

For the second system, we have a phase line that looks like this:

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -