Is it possible to find general solutions for $n$-th order Euler-Cauchy ODE?

Question

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!

Answer 1

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 $t=e^s$.

Answer 2

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 $x(t)=x(0)\exp [\lambda t]$.