Given a general nth degree linear ODE, what's the easiest way to prove that there are precisely n linearly independent solutions?
closed as too localized by Andrey Rekalo, Will Jagy, Deane Yang, Qiaochu Yuan, Andres Caicedo Nov 5 '10 at 2:39This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


By the unicity of the Cauchy problem, and linearity. 


I don't know of an easy/easiest way to prove it for a general $n$th degree linear ODE, but it is worth pointing out that in the constant coefficient case you can get this from elementary linear algebra. The idea is that if $N$ is a positive integer and you have complex numbers $c_1, \dots, c_N$, then the solutions to the differential equation $$ \sum_{n=0}^N c_k y^{(k)} = 0 $$ (here $y^{(k)}$ denotes the $k$th derivative of $y$, interpreted as $y$ when $k=0$) are precisely the elements of the kernel of the operator $$ T = \sum_{n=0}^N c_k D^k $$ where $D$ is differentation, regarded as an operator on a vector space $V$ of functions (there is some freedom in what particular space you choose here; say the set of all infinitely differentiable functions $\mathbb{R} \to \mathbb{C}$). From the fundamental theorem of algebra, you know there are complex numbers $\omega, \omega_1, \dots, \omega_N$ with the property that the polynomial $\sum_{n=0}^N c_k z^k$ factors as $\omega \prod_{n=1}^n (z  \omega_n)$; it follows that your operator $T$ also factors, in the algebra of operators on $V$, as $$ T = \omega \prod_{n=1}^N (D  \omega_n I), $$ where $I$ denotes the identity operator on $V$. The point is that each of the operators $D  \omega_n I$ has a onedimensional kernel by basic calculus. (For any $k$, the function $f(t) = \exp(kt)$ is a solution to $y' = k y$, and if $g$ is any other, the quotient rule for derivatives shows that $(g/f)' = 0$. So by a standard argument involving the mean value theorem, $g/f$ is constant; so $\{f\}$ is a basis for $D  kI$.) And it is a basic linear algebra fact that a product of $n$ operators with onedimensional kernel, can have kernel of dimension at most $n$. (Follows from the more general assertion that if $S_1: V \to V$ and $S_2: V \to V$ are any operators, the dimension of the kernel of $S_1 S_2$ is at most the dimension of the kernel of $S_1$ plus the dimension of the kernel of $S_2$. This very easy consequence of the ranknullity theorem and does not require $V$ to be finite dimensional.) Why is the kernel of $T$ exactly $n$dimensional? Well, just write down $n$ linearly independent elements in it, as they do in textbooks. (Of course, if you have the better sort of textbook, the entire argument just given is in there.) For nonconstant coefficients, factoring the corresponding differential operator is no longer the way you want to approach this. But for a lot of ODE, you can still get reasonably elementary theorems about the dimension of the kernel of the operator by applying some kind of transform (e.g. the Laplace transform) and getting in a position where it is just algebra again. 

