linear ODE with constant coefficients, proof Let $
\sum\limits_{k = 0}^n {y^{\left( k \right)} a_k }  = 0
$ an homogeneous ODE, where $a_k$ are constants. How can I solve the equation when the roots are repeated?  One way, that I saw in wikipedia, is using the fact that if $ e^{cx} $ is a solution, then $ (x^r)(e^{cx}) $ is also too, How can I prove this? It´s difficult to me, to evaluate the sum, because i want to show that $
\sum\limits_{k = 0}^n {\left( {x^r e^{cx} } \right)^{\left( k \right)} a_k }  = 0
$ but I need to evaluate $
{\left( {x^r e^{cx} } \right)^{\left( k \right)} }
$ and I don´t know how to do it Dx

An alternative to looking at the corresponding first order linear system is to consider factoring linear differential operators. 
Given a homogeneous linear differential equation with constant coefficients, $a_ny^{(n)}+\cdots+a_1y'+a_0y=0$, we can define a linear differential operator $L=a_nD^n+\dots+a_1D+a_0$ where $D[y]=y'$, $D^2[y]=D[D[y]]=y''$, etc. and each constant stands for left multiplication by that constant: $a_k[y]=a_ky$.
So our differential equation becomes the operator equation: $L[y]=0$.
Example: $3y''+2y'+5y=0$ $\Longleftrightarrow$ $L=3D^2+2D+5$ and $L[y]=0$
Next, $D$ is a linear operator, so it commutes with constants: $D[ky]=kD[y]$. In addition, powers of $D$ commute with each other. This means we can factor each operator (like a polynomial).
Example: $y''-y=0$ gives us $L=D^2-1=(D-1)(D+1)=(D+1)(D-1)$ these are all equal because everything commutes.
Now to solve our equation we can focus on one factor at a time.
Example: Suppose $L=(D-1)(D-2)$. Then $L[e^t]=(D-1)(D-2)[e^t]=(D-2)(D-1)[e^t]=(D-2)[0]=0$ since $(D-1)[e^t]=e^t-e^t=0$ and $(D-2)[0]=0-2(0)=0$. Next, $L[e^{2t}]=(D-1)(D-2)[e^{2t}]=(D-1)[2e^{2t}-2e^{2t}]=(D-1)[0]=0$. This shows that both $y=e^t$ and $y=e^{2t}$ are solutions and hence $y=c_1e^t+c_2e^{2t}$ is the general solution (accepting some theory which tell us that 2 solutions are linearly independent and the space of solutions is 2-dimensional). 
Let's focus on a single factor: $L=(D-r)^n$.
Consider $y=t^ke^{rt}$. $(D-r)[y]=(kt^{k-1}e^{rt}+t^kre^{rt})-r(t^ke^{rt})=kt^{k-1}e^{rt}$.
Next, $(D-r)^2[y]=(D-r)(D-r)[y]=(D-r)[kt^{k-1}e^{rt}]=k(k-1)t^{k-2}e^{rt}$. And in general $(D-r)^{k+1}[y]=k(k-1)\cdots 2\cdot 1\cdot 0 \cdot e^{rt}=0$.
This means that $(D-r)^n[t^ke^{rt}]=0$ when $n>k$.
Thus if the characteristic polynomial has a factor $(x-r)^n$ so that the corresponding linear differential operator $L$ has a factor $(D-r)^n$, we must have solutions $e^{rt}$, $te^{rt}$, $\dots$, $t^{n-1}e^{rt}$. These can be shown to be linearly independent by using a Wronskian determinant (this is fairly straight forward but difficult for a beginner). Solutions corresponding to distinct factors are independent because they are in distinct spaces of generalized eigenvectors.
Example: Suppose $y^{(5)}-4y^{(4)}+y'''+10y''-4y'-8y=0$. Then $L=D^5-4D^4+D^3+10D^2-4D-8=(D-2)^3(D+1)^2$. Solving $(D-2)^3[y]=0$, we get $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}$. Solving $(D+1)^2[y]=0$, we get $y=c_4e^{-t}+c_5te^{-t}$. The general solution is $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}+c_4e^{-t}+c_5te^{-t}$.
Similar methods allow one to deal with (repeated) irreducible quadratic factors.
