Basis of solution sets I know that the collection of all solutions to $\sum_{i=0}^nA_iy^{(i)}(t)=0$ form a vector space. But in which way can one find out its basis? Of course I already learned what the basis is. But the key question is, how to prove that it is?
 A: Let's look at an example, $$y''+y'-2y=0$$ It seems that you know that $y_1(t)=e^t$ and $y_2(t)=e^{-2t}$ form a basis for the vector space of all solutions, but you want to know why. 
First, let's prove that these two solutions are linearly independent. Suppose $$Ae^t+Be^{-2t}=0$$ identically (that is, for all $t$). Let $t=0$; you get $A+B=0$, so $B=-A$, so $$Ae^t-Ae^{-2t}=0$$ so $A=0$ or else $e^t-e^{-2t}$ is identically zero. The latter is nonsense ($e^t>1>e^{-2t}$ for $t>0$), so $A=0$, so $B=0$, so the two functions are linearly independent. 
Given any solution $f$ of the differential equation, we can find $A,B$ such that if $g=f-Ae^t-Be^{-2t}$ then $g(0)=g'(0)=0$. But $g$ satisfies the differential equation, so $g$ is identically zero, so $f(t)=Ae^t+Be^{-2t}$. 
A: A general answer to this is:
$\text{The Basis theorem}$ (from linear algebra):
Let $A$ be a $p$-dimensional subspace of $R^n$. Any linearly independent set of exactly $p$ elements in $A$ is automatically a basis for $A$. Also, any set of $p$ elements of $A$ that $span$ $A$ is a basis for $A$.
Note that a vector space (which in your case is given by the solutions of your system) is something completely different than the basis of a matrix. A vector space is just a mathematical structure formed by vectors, which can be added together and multiplied by numbers.
