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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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1 Answer 1

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

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Ok, But the proof it´s what I need , and I writed that, I don´t understand your answer =S –  August Nov 23 '11 at 19:17

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