Is it possible to find general solutions for $n$-th order Euler-Cauchy ODE? Consider $n$-th order Euler-Cauchy equation:
\begin{equation}
  a_nt^n\frac{d^nx}{dt^n}+a_{n-1}t^{n-1}\frac{d^{n-1}x}{dt^{n-1}}+\cdots+a_{1}t\frac{dx}{dt}+a_0x=0,\quad a_0\ne0,a_{1},\cdots a_{n}\in\Bbb R
\end{equation}
I'm asked to find solutions (not sure "general" solutions or just "special" ones) for this system. The hint is: try solutions of the form $x=t^\lambda$. So I let $x=t^\lambda$ and find that
$$
  t^\lambda\left[\left(a_nn!\binom{\lambda}{n}+a_{n-1}(n-1)!\binom{\lambda}{n-1}+\cdots+a_1\binom{\lambda}{1}+a_0\right)\right]=0
$$
Now it is clear that each $x=t^{\lambda_i}$ for which $\lambda_i$ fits the above equation is a solution for the ODE. Also, all the solutions for the ODE form a vector space.
If I am only required to find special solutions then I'm done. But I don't want to just stop here. I want to find general solutions, if possible. What troubles me is the case of repeated roots, say, $\lambda_i$ with multiplicity $n_i$. I wonder if I could extend the result of homogeneous linear ODE analogously into this case: would 
$$(C_1+C_2t+\cdots+C_{n_i}t^{n_i-1})t^{\lambda_i}$$
also be a solution for the ODE? At first I tried the case of a second order ODE with repeated $\lambda=-1$ and it worked well, so I was encouraged. I found out that to prove my conjecture (if correct) I only needed to show, due to linearity of the solution space, that
$$t^m\cdot t^{\lambda_i},\quad 0\le m\le n_i-1$$
is a solution. But when I plug it back into the ODE the resulting equation looks horrible. So I kinda doubt whether I'm on the right track now.
It's very likely that I have made an incorrect analog here. But anyway, I would like someone to tell me how to find out the general solution for this ODE, if possible.
Best regards!
 A: It may be convenient to rewrite the problem as a a set of coupled first
order equations.
\begin{equation*}
a_{n}\frac{d^{n}x}{dt^{n}}+a_{n-1}\frac{d^{n-1}x}{dt^{n-1}}+\cdots +a_{1}
\frac{dx}{dt}+a_{0}x=0
\end{equation*}
Put
\begin{equation*}
X_{0}=x,\;X_{1}=\frac{dx}{dt},\cdots X_{n}=\frac{d^{n}x}{dt^{n}}
\end{equation*}
Then
\begin{equation*}
X_{n}=-\frac{1}{a_{n}}\{a_{n-1}X_{n-1}+\cdots a_{0}X_{0}\}
\end{equation*}
and
\begin{eqnarray*}
\partial _{t}X_{0} &=&X_{1} \\
\partial _{t}X_{1} &=&X_{2} \\
&&\cdots  \\
\partial _{t}X_{n-1} &=&-\frac{1}{a_{n}}\{a_{n-1}X_{n-1}+\cdots a_{0}X_{0}\}
\end{eqnarray*}
Or, with
\begin{equation*}
\mathbf{X}=\left(
\begin{array}{c}
X_{0} \\
X_{1} \\
\cdots  \\
X_{n-1}
\end{array}
\right)
\end{equation*}
we have
\begin{equation*}
\partial _{t}\mathbf{X}=\mathsf{A}\cdot \mathbf{X}
\end{equation*}
and
\begin{equation*}
\mathbf{X}(t)=\exp [\mathsf{A}t]\mathbf{X}(0)
\end{equation*}
The eigenvalues of the matrix $\mathsf{A}$ correspond to the roots $\lambda
_{j}$ of the equation
\begin{equation*}
a_{n}\lambda ^{n}+a_{n-1}\lambda ^{n-1}+\cdots +a_{0}=0
\end{equation*}
obtained by substituting $x(t)=x(0)\exp [\lambda t]$.
A: If $\lambda$ is a root of multiplicity $k$, then
$$
t^\lambda,\ (\log t)\,t^\lambda,\dots,(\log t)^{k-1}\,t^\lambda
$$
are $k$ linearly independent solutions.
Another way to solve Euler's equation is to transform it into an equation with constant coefficients via de change of variable $t=e^s$.
