Consider $n$-th order [Euler-Cauchy equation](https://en.m.wikipedia.org/wiki/Cauchy%E2%80%93Euler_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!