How many methods are there for solving repeated roots of differential equations? I am studying differential equations and the book states there are many methods of finding the solutions of a differential equation that has repeated roots, but it only gives one method.  D'Alembert method. 
D'Alembert method:is to replace the constant of $cy_1(t)$ with $v(t)$.  
Could anyone inform me of the other methods? 
An example for the smarter users to use as a template. 
$y^{\prime\prime}+4y^{\prime}+4y=0$
I know how to solve this with the D'Alembert method, what are the others? 
 A: The Exponential Shift Theorem.  See e.g. http://www.math.ubc.ca/~israel/m215/coco/coco.html
A: 3 methods come to mind.
First, as you mention, you can take a known solution $y$ (so that $cy$ are solutions for all constants $c$ and replace it with $vy$ for some unknown function $v$ and then get a lower order equation to solve to get a second solution. This is essentially a baby case of variation of parameters.
Next, homogeneous linear differential equations can be recast as a system of first order linear differential equations with constant coefficients. For your example:
$y''+4y'+4y=0$ we can set $y_1=y$ and $y_2=y_1'=y'$ so that $y_2'+4y_2+4y_1=0$. Rewriting we have $y_1'=y_2$ and $y_2'=-4y_1-4y_2$. In matrix form this is...
$$ \begin{bmatrix} y_1' \\ y_2' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -4 & -4 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix}$$
For short write ${\bf y}'=A{\bf y}$. 
The coefficient matrix we get here is called a companion matrix.
It turns out that eigenvectors yield solutions for such matrix systems. If $A$ is diagonalizable we can find a full set of solutions. However, if $A$ isn't diagonalizable (i.e. not enough eigenvectors - as in the case of a repeat root), we can get additional solutions by considering generalized eigenvectors. More succinctly the solution is given by ${\bf y}=e^{At}{\bf c}$ (where $e^{At}$ is the matrix exponential and ${\bf c}$ is some vector of constants). The first row of such vector solutions are solutions of the original equation. So new solutions for the repeated root case come from generalized eigenvectors.
Finally, you can differentiate with respect to the eigenvalue and pop-out new solutions in the repeated root case. This reveals an underlying symmetry of this class of differential equations. Specifically, if we have a repeated root type equation: $y''-2\lambda y'+\lambda^2=0$. Call the differential operator $D=\frac{d}{dt}$. Then the equation becomes $(D^2-2\lambda D+\lambda^2)[y]=0$ which is $(D-\lambda)^2[y]=0$. Essentially because $\partial_\lambda= \frac{\partial}{\partial \lambda}$ and $D$ commute, we get $\partial_\lambda (D-\lambda)^2[y]=\partial_\lambda(0)=0$ and so $2(D-\lambda)[y](-1)+(D-\lambda)^2[\partial_\lambda y]=0$. So if $y$ is a solution of the lower order equation $(D-\lambda)[y]=0$ (i.e. $y'-\lambda y=0$), then $\partial_\lambda y$ solves the repeated root equation. Specifically, if $y=e^{\lambda t}$ is a solution, then so is $\partial_\lambda y = \lambda e^{\lambda t}$. 
