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!
