Help with matrix differential equation y'''-3y''+3y'-y=0: repeated eigenvalues!

Keatenclarno

New member
Joined
Mar 19, 2019
Messages
1
Hey, I'm trying to solve the differential equation

y'''-3y''+3y'-y=0

I have to solve it by separating into a system of linear first order differential equations. By doing that and converting it into matrix form, I have:

|0 1 0|
X' = |0 0 1| X
|1 -3 3|

I just need help finding the eigenvectors from the repeated eigenvalues of the matrix:

Top: 0 1 0
Middle: 0 0 1
Bottom: 1 -3 3


So essentially, I need help finding the eigenvalues and eigenvectors of the matrix. It has repeated eigenvalues, and I'm not sure how to tackle that and find the eigenvectors. Thank you!
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,877
Your given matrix is $\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}$. The eigenvalue equation is $\left|\begin{array}{ccc} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 1 & -3 & 3- \lambda \end{array}\right|= -\lambda \left|\begin{array}{cc} -\lambda & 1 \\ -2 & -3- \lambda \end{array}\right|- \left|\begin{array}{cc}0 & 1 \\ 1 & 3- \lambda\end{array}\right|=-(\lambda- 1)^3= 0$. Yes, $\lambda= 1$ is an eigenvalue with "algebraic multiplicity" 3.

An "eigenvector", v, corresponding to eigenvalue $\lambda$, of a linear transformation, A, satisfies $Av= \lambda v$. Here, that means that an eigenvector, $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$, corresponding to eigenvalue 1 satisfies $\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix} y \\ z \\ x- 3y+ 3z\end{bmatrix}= \begin{bmatrix x \\ y \\ z\end{bmatrix}$. So we have the three equations y= x, z= y, and x- 3y+ 3z= z. From the first three equations, x= y= z so the third equation can be written z- 3z+ 3z= z which is true for all z. Any eigenvector can be written in the form $\begin{bmatrix} z \\ z \\ z \end{bmatrix}= z\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix}$.​
Since any eigenvector is a multiple of that single vector, the "geometric multiplicity" of eigenvalue 1 is only 1.​
Personally, I would not have written this problem in terms of matrices at all! The original differential equation is y'''- 3y''+ 3y'- y= 0 which has "characteristic equation" $r^3- 3r^2+ 3r- 1= (r- 1)^3= 0$ (Yes, that is the same as the eigenvalue equation). That means that the general solution to the differential equation is $y(x)= Ae^x+ Bxe^x+ Cx^2e^x= (Cx^2+ Bx+ A)e^x$.​
 
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