Answer to Question #200249 in Functional Analysis for Arun

Question

Show that all solution of a linear, homogeneous and n order ordinary differential equation constitute an n-dimensional linear vector space?

Answer 1

We know that the there exists a unique solution to an initial value problem for a linear homogeneous ordinary differential equation :

$\exists ! f \; \text{such that} \; \begin{cases}g(f,f',...,f^{(n)})=0 \\ f(t_0)=a_0 \\ ... \\ f^{(n-1)}(t_0)=a_{n-1} \end{cases}$

As the solution is uniquely determined by the values of $f(t_0),...,f^{(n-1)}(t_0)$. Let us take $n$ lineraly independent functions as solutions to initial value problems

$f_i - \text{the unique solution of } \begin{cases} g(f,f',...,f^{(n)})=0 \\ f(t_0)=0 \\ ... \\ f^{(i)}(t_0)=1 \\ ... \\ f^{(n-1)}(t_0)=0 \end{cases}$ , i.e. a solution of the equation with all derivatives, excpet for the i-th, vanishing at $t_0$. Therefore, any solution of the initial value problem can be uniquely written as (by linearity of i-th derivative) :

$f= \sum_{i=0}^{n-1} a_i f_i$ and so the space of all solutions is an n-dimensional vector space.