# constant coefficient of an ODE, I did a proof but something it´s wrong

Let an homogeneous ODE with constant coefficient $\sum\limits_{k = 0}^n {a_k y^{\left( k \right)} } = 0$ with clearly $a_k$ denotes the constants, and $y^{\left( k \right)} = \frac{{d^k y}} {{dx^k }}$. I have a question about a proof that I did, it´s obviously bad, it´s a known fact that if $\lambda$ is a solution of the associated polynomial $\sum\limits_{k = 0}^n {a_k x^k }$ then $e^{\lambda x}$ it´s a solution. And if $\lambda$ has algebraic multiplicity m>1 (clearly $m \leqslant n$) then other LI solutions, are $$\lambda e^{\lambda x} ,\lambda ^2 e^{\lambda x} ,...\lambda ^{m - 1} e^{\lambda x}$$ My question it´s about this fact, I want to prove it , But look something. Let´s take a function of the form $f\left( x \right) = \lambda ^j e^{\lambda x}$ let´s check if it´s a solution. Then we replace $$\sum\limits_{k = 0}^n {a_k \left( {f\left( x \right)} \right)^{\left( k \right)} } = \sum\limits_{k = 0}^n {a_k \left( {\lambda ^j e^{\lambda x} } \right)^{\left( k \right)} } = \sum\limits_{k = 0}^n {a_k \lambda ^j \left( {e^{\lambda x} } \right)^{\left( k \right)} } = \sum\limits_{k = 0}^n {a_k \lambda ^j \lambda ^k e^{\lambda x} } = \lambda ^j e^{\lambda x} \sum\limits_{k = 0}^n {a_k \lambda ^k }$$ the last step it was because that terms are constant in the sum. But $$\sum\limits_{k = 0}^n {a_k \lambda ^k }=0$$ thus $\lambda ^j e^{\lambda x}$ it´s a solution , the problem is that this is valid for every $j \in R$ but the functions of the form $$\lambda ^j e^{\lambda x}$$ are always LI if the j are distinct, so we have more than n LI solutions , this it´s false, so my proof it´s false, But I don´t know where is the mistake addition to showing what is wrong, I could help with this event please, Thanks!

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The solutions should be $$e^{\lambda x}, xe^{\lambda x},\dots, x^{m-1}e^{\lambda x}$$ instead of $$e^{\lambda x} ,\lambda e^{\lambda x} ,...\lambda ^{m - 1} e^{\lambda x}.$$