An nth-order ODE has n linearly independent solutions MathWorld states:

"In general, an $n^\text{th}$-order ODE has $n$ linearly independent solutions".

Are they referring to linear ODEs? I only know why it should be true for ODEs with constant coefficients, by the following observations:
The solutions to the differential equation $a_0f+\dots +a_nf^{(n)}=0$, where $a_n\ne 0$ form a vector space $V$ (check).
Let $f\in C^n(\mathbb{R})$ s.t. $a_0f+\dots +a_nf^{(n)}=0$, where $a_n\ne 0$.
Let $\vec{a}=(a_0,a_1,\dots,a_{n-1})$ and $\vec{f}=(f,f^{(1)},\dots,f^{(n-1)})$.
$f^{(n)}=-a_n^{-1}(a_0f+\dots +a_{n-1}f^{(n-1)})=-a_n^{-1}\vec{a}\cdot\vec{f}$ is differentiable, and the $m^\text{th}$ derivative of $\vec{b}\cdot\vec{f}$ is:
$$\vec b\left(\matrix{\vec{e_2}\\\vdots\\\vec{e_n}\\-a_n^{-1}\vec{a}}\right)^m\cdot \vec{f}$$
Hence $f$ is infinitely differentiable. Moreover, the coefficients above are bounded above by an exponential in $m$. For any closed interval $[-d,d]$, $\vec{f}$ is continuous and therefore bounded. This means that the Taylor series for $f$ converges to $f$ in the interval by Taylor's theorem for the expansion about $x=0$ (using the Lagrange form of the remainder on the whole interval). Hence the Taylor series for $f$ about $x=0$ converges to $f$ for all $\mathbb{R}$, i.e. $f$ is analytic.
Now consider the linear transformation $L:V\to\mathbb{R}^n,f\mapsto (f(0),f^{(1)}(0),\dots,f^{(n-1)}(0))$. To prove surjectivity, use the differential equation to produce a Taylor series and show that it is a solution. Injectivity is proven by the below:

If $L(f)=L(g)$ for some solutions $f,g$, then $\forall k=0,1,\dots,n-1, f^{(k)}(0)=g^{(k)}(0)$, and by the differential equation, this also holds for all $k\in\mathbb{N}$. $f$ and $g$ are analytic and since the Taylor series is unique, $f=g$.

Hence, $V$ has dimension $n$.
Is my proof correct?
Is the theorem for general linear ODEs true, and how do I prove it?
 A: Your proof is now good.  For the homogeneous non-constant coefficient system $$a_n(t)x^{(n)}+\cdots +a_0(t)x=0$$ the solutions again form a vector space, there is nothing different in this respect. How to find its dimension (in particular, prove that the dimension is not zero)?
One cannonball way to proceed is to rewrite it as a first order vector valued equation by introducing variables $x_1=x',\cdots$.  This furnishes the first order equation $$X'(t)=A(t)X(t).$$  Here the function $A$ is assumed nicely behaved from some interval $(a,b)$ to the Banach space $\mathbb{R}^n$, e.g. take the entries to be Lipschitz and $a_n(t)\neq 0$ on $(a,b)$ to avoid the degenerate locus (in fact, assume $a_i/a_n(t)$ are also Lipschitz on the interval).  Then we can form any IVP we want $$X'=A(t)X(t),\quad X(0)=X_0$$ and the Picard-Lindelof theorem furnishes a unique solution with the initial condition $X_0$ (if one tries to simplify the proof of P-L for linear systems one still ends up having to perform the crucial approximation process and prove the uniform convergence, which is the big deal in P-L itself). 
But now the fact that you can find solutions corresponding to $n$ linearly independent choices of $X_0$ implies that the vector space of solutions is exactly $n$-dimensional by the theorem on the Wronskian (the Wronskian of $n$ solutions is either identically zero or never zero; if the initial vectors $X_0$ are linearly independent, the Wronskian is nonzero somewhere, so it is never zero on $(a,b)$ and thus the solutions are independent).
