Why does linear independence of solutions to a second order ODE imply generality? First of all, I've searched this site for an answer and found this question, but the answers did not tell what I want to know.
If we have a second order homogeneous differential equation 
$$ p(t) y'' + q(t) y' + r(t) y = 0 $$
and two solutions $y_1, y_2$, I know that:


*

*Their linear combinations will be solutions as well

*For such a solution to be general, the two solutions $y_1$ and $y_2$ must be linearly independent (general solution of this form implies linear independence)


However, none of the answers tell why does linear independence imply generality of the solution obtained as a linear combination of the two functions. How do we know there is no other function that is a solution and that is outside the span of the two solutions $y_1, y_2$?
The course I've been following for a while gives a proof that is unclear to me. It says that for a solution to be general, it must satisfy the general initial conditions $y(t_0) = y_0, y'(t_0) = y'_0$, so I guess this means, for all $t_0, y_0$ and $y'_0$. But how do we know that, even if our solution $y$ does in fact, satisfy any initial conditions, there is no other solution that satisfies some set of initial conditions for some $(t_0, y_0, y'_0)$, but does not fall in the form of $y$?
 A: Consider a general $n$-th order linear homogeneous ODE:
$$\sum_{i=0}^na_i(t)y^{(i)}(t)=0\tag{1}$$
where $a_n(t)\neq 0$ on a certain interval $J$ around a point $t_0$.
First, we should emphasize that solutions of the ODE $(1)$ are only defined on some small intervals. So what we want to prove is:

If $t_0$ is a point on which all $a_i$ are defined and $C^1$ on some interval around $t_0$, and $a_n(t_0)\neq 0$, then there exists an interval $I$ around $t_0$ such that the solution set of $(1)$ has dimension $n$.
Conversely, if $K$ is any interval on which $a_n\neq 0$, then there exist $n$ linearly independent solutions $y_1,\ldots,y_n$ of $(1)$, then these solutions generate the whole solution set of $(1)$ on $K$.

The easiest way is to use the Picard-Lindelof Theorem, which works in $\mathbb{R}^n$. Here is a restricted version of the general statement:

Theorem [Picard-Lindelof]: Consider a Cauchy problem
$$y'(t)=F(t,y(t)),\qquad y(t_0)=(f_1,\ldots,f_n)\tag{2}$$
where $F=F(t,x_1,\ldots,x_n)$ is a function, $C^1$ from some neighbourhood of $(t_0,f_1,\ldots,f_n)$ to $\mathbb{R}^n$. Then

*

*There exists an interval $I$ around $t_0$ and a fucntion $y=y(t)$ which is a solution of $(2)$. (This is the existence of solutions.)

*Any two solutions $y_1,y_2$ of $(2)$, defined on intervals $I$ and $K$, coincide on $I\cap K$. (This is the "uniqueness" of solutions.)



Solving the ODE $(1)$ is equivalent to solving the following ODE on $n$ dimensions: Again, recall that we assume $a(t_0)\neq 0$ on some interval $J$ around $t_0$. Let $z(t)=(z_0(t),\ldots,z_{n-1}(t))$. Then consider the problem
$$z'(t)=F(t,z(t))\tag{3}$$
where $F(t,x_0,\ldots,x_{n-1})=\left(x_1,\ldots,x_{n-1},-\sum_{i=0}^{n-1}\frac{a_i(t)}{a_n(t)}x_i\right)$.
The equation $(3)$ means that
$$z_0'=z_1,\qquad z_1'=z_2,\ldots\qquad\text{or more generally }z_i=z_0^{(i)},$$
and $z_{n-1}'=-\sum_{i=0}^{n-1}\frac{a_i(t)}{a_n(t)}z_i$.
Therefore, $z$ is a solution of $(3)$ iff $y:=z_0$ is a solution of $(1)$. Moreover, the map $z\mapsto z_0=y$ is a linear bijection from the solution space of $(3)$ to the solution space of $(1)$, so they have the same dimensions.
The function $F$ is defined and $C^1$ on $J\times\mathbb{R}^n$. Take a basis $e_1,\ldots,e_n$ of $\mathbb{R}^n$ and solutions $z^1,\ldots,z^n$ of $(3)$, defined on some interval $I$ around $t_0$, and satisfying $z^i(t_0)=e_i$. These $z^i$ are clearly linearly independent.
If $z$ is any other solution of $(3)$ on $I$, write $z(t_0)=\sum\lambda_i e_i$. By the uniqueness part of Picard-Lindelof, $z=\sum\lambda_i z^i$ on $I$.
