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Consider the following differential equation. For $ y(t)\in C^\infty([0,1]) $, $$y''(t) - 5y'(t) + 6y(t) =0 $$

Here the solutions are $y(t) =Ae^{2t}$ or $A e^{3t}$ where A is a constant.

I want to know the number of independent solutions of the above differential equation. In fact we already know the number. The number is two.

This question can be reduced to the question about dimension of kernel of differential operator :

$$ \frac{d^2}{dt^2} : C^\infty([0,1]) \rightarrow C^\infty([0,1])$$

has the kernel of dimension 2. It is clear since $t^n$ is sended to $nt^{n-2}$. From the above argument we already know that the dimension of kernel of the following operator is at least 2 : $$ \frac{d^2}{dt^2} -5 \frac{d}{dt} + 6 I: C^\infty([0,1]) \rightarrow C^\infty([0,1])$$ where $I$ is an identity map. But how can we show that the dimension of kernel of the operator is two ?

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Providing a basis that has exactly two elements. –  Artem Nov 12 '12 at 2:00
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up vote 2 down vote accepted

If you can show that $\dim\big(\ker\left(\displaystyle\frac d{dt}-c\cdot I\right) \big) =1$, then the following decomposition may help: $\left(\displaystyle\frac d{dt}-2\cdot I\right)\left(\displaystyle\frac d{dt}-3\cdot I\right)$.

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I have posted a similar question about the kerner of a fractional differential operator. math.stackexchange.com/questions/354863/… I don't know if your argument could be adapted to my case. –  Ambesh Apr 9 '13 at 11:46
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