Differential equations: Connection between repeated roots of characteristic equation and generalized eigenvectors My question is about homogeneous linear equations with constant coefficients: $ay''+by'+cy=0$.
When you solve this equation via a characteristic equation (see e.g. here) problems arise when you encounter repeated roots in it. What you would basically do is to combine two chained solutions (principle of superposition and reduction of order).
This equation can also be converted to a system of first order equations (see e.g. here). You then have to diagonalize the resulting matrix. Problems arise when you find that it is not diagonizable because it has to few eigenvectors with repeated eigenvalues. Here again you can remedy the situation by using generalized eigenvectors to find solutions.
My questions:


*

*Are both problems and solutions equivalent?

*Could you please provide a simple example where you show that both methods lead to the same solution.

 A: You forgot the second $y$ in your equation.
The characteristic equation is the characteristic polynomial of the matrix of the associated first-order system -as you can check. It tells you what the eigenvalues are.
If the eigenvalues are complex they are conjugate because their product is a real number ($c/a$ of your equation) and their sum is $-b$, so your matrix has distinct eigenvalues and is diagonalizable -it has no strictly generalized eigenvector, they are all true eigenvectors.
In case of real eigenvalues (2 real roots of the characteristic equation) if they are different your matrix is diagonalizable. If they are equal then the matrix may be symmetric or not, equivalently, diagonalizable or not. If it is not you do not have 2 linearly independent eigenvectors, you have 1 and you will have strictly generalized eigenvectors.
An example of the last case:
\begin{pmatrix}
1 & 0 \\
\lambda & 1 \\
\end{pmatrix}
is an improper node for $\lambda\ne 0$. One solution of it is $x=e^{\lambda t}$, $y=e^{\lambda t}+te^{\lambda t}$.
In the (phase) plane the solution curves spiral toward ($\lambda<0$) or outward ($\lambda>0$) the origin.
EDIT: I corrected the remark on complex eigenvalues being conjugate: I replaced $a$ by $c/a$ and added the fact that the sum of the roots also needs to be a real number to deduce they are conjugate. The latter implies that they have opposite imaginary parts and then their real part must be equal, by the multiplication formula for complex numbers ($ad+bc=0$ and $b=-d$ imply $a-c=0$).
Sorry.
EDIT 2: I should not have used the term spiral as the trajectories do not even make a full turn around the origin. Depending on $\lambda$ and on initial conditions they will bend around it more or less. See pictures here. The solutions for various initial conditions are obtained adding appropriate real coefficients to the exponential summands above, i.e. $x=C_1e^{\lambda t}$, $y=C_2e^{\lambda t}+C_1te^{\lambda t}$. As Gerry Myerson comments the solutions do not cross the eigenline, which is the vertical line here, $x_0=0$ for initial conditions, so $C_1=0$.
