How can you tell that a general solution to a DE is general? At school, or in a first-year course on DEs, we learn (perhaps in less abstract language) that if you have a linear $n$th-order differential equation
$$Ly = f$$
then the general solution is something of the form
$$y = a_1 y_1 + ... + a_n y_n + g$$
where the $y_i$ are independent and satisfy $Ly_i = 0$, and $g$ satisfies $Lg = f$. Then we receive lots of training in how to find the $y_i$ and $g$.
Obviously any choice of the $a_i$ will give us a solution to $Ly=f$. But how do you know that there aren't any more solutions? 
We justify this by making an analogy with systems of linear equations $Ax=b$, saying something along the lines of 'the space of solutions has the same dimension as the kernel of $A$'. But that works in finite dimensions - how do we know that the same is true with linear operators? 
 A: For some specific cases: 
Consider the first order DE: $$\frac{dx}{dt} = ax$$ where a is a constant.
First we show that the general solution of this equation is of the form: $c.{e}^{at}$ 
Let $y(t)$ be a solution of the DE. Consider: $q(t) = \frac{{e}^{at}}{y}$.  We have:
$$q'(t) = \frac{a{e}^{at}y-{e}^{at}y'}{{y}^{2}}=\frac{a{e}^{at}y-a{e}^{at}y}{{y}^{2}} =0$$  So $q(t)$ is a constant. This shows that $y(t)$ is of the form $c{e}^{at}$.
We conclude that this is the general solution of the DE.
Now consider a second order DE: $$ay''+by'+cy=0$$
By introducing the variable: $x = y'$ we can rewrite the DE in: 
$ax'+bx+cy=0$ and $y = x'$. This can be put in matrix form:
$$X' = AX$$ With $A$ a matrix and $X = (x,y)$. A solution of this equation is:
$${c}_{1}.U{e}^{kt}+{c}_{2}.V{e}^{lt} $$  (1). Where the c's are constants and U,V the eigenvectors of A and k and l the corresponding eigenvalues. We can see that the eigenvectors of A, corresponding to the second order DE, are linear independent. So any solution $f(t)$ of $X' =AX$ can be writen in:
$$f(t) = g(t)U+h(t)V$$ By defenition: $f'(t) = Af(t)$. So: 
$$g(t)kU+h(t)lV=g'(t)U+h'(t)V$$ Because the U and V are linearly independent:
$g(t)k = g'(t)$ and $h(t)k = h'(t)$. We already have shown that the g and h are of the form $c{e}^{at}$. So f is of the sugested form (1). So this is the general solution.
This can also be done for higher dimensions.
A: 
Theorem [Existence and Uniqueness]: Let $a_{0}(x),a_{1}(x),\cdots,a_{n-1}(x)$ be continuous functions on $[a,b]$ and let constants $y_{0},y_{1},\cdots,y_{n-1}$ be given. Then there exists one and only one n-times continuously differentiable solution $y(x)$ on $[a,b]$ of the differential equation
  $$
      y^{(n)}(x)+a_{n-1}y^{(n-1)}(x)+\cdots+a_{1}y^{(1)}(x)+a_{0}y(x) = 0
$$
  with initial endpoint values
  $$
       y(a) = y_{0},\; y'(a)=y_{1},\; y''(a)=y_{2},\cdots, y^{(n-1)}(a)=y_{n-1}.
$$

In other words, the solution of the homogenous differential equation $y$ is uniquely determined once you know the values of $y(a),y'(a),y^{(2)}(a),\cdots,y^{(n-1)}(a)$. So the following map is a linear bijection between the solution space of the homogeneous equation and $\mathbb{R}^{n}$:
$$
              y \mapsto \left[\begin{array}{c}y(a)\\y^{(1)}(a)\\ \vdots\\ y^{(n-1)}(a)\end{array}\right]
$$
If you choose $y_{j}$ to be the unique solution whose image under this map is the j-th standard basis vector for $\mathbb{R}^{n}$, then every solution of the homogeneous equation can be written uniquely as $\alpha_{1}y_{1}(x)+\cdots+\alpha_{n}y_{n}(x)$ for real constants $\alpha_{j}$ (why?). The difference of any two solutions solutions of the inhomogeneous equation is a solution of the homogeneous equation and, hence, must be such a linear combination.
