Question #200249

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


1
Expert's answer
2021-05-31T17:42:38-0400

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

!f  such that  {g(f,f,...,f(n))=0f(t0)=a0...f(n1)(t0)=an1\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(t0),...,f(n1)(t0)f(t_0),...,f^{(n-1)}(t_0). Let us take nn lineraly independent functions as solutions to initial value problems

fithe unique solution of {g(f,f,...,f(n))=0f(t0)=0...f(i)(t0)=1...f(n1)(t0)=0f_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 t0t_0. Therefore, any solution of the initial value problem can be uniquely written as (by linearity of i-th derivative) :

f=i=0n1aifif= \sum_{i=0}^{n-1} a_i f_i and so the space of all solutions is an n-dimensional vector space.


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