Prove that a second order diff. eq. has only two linearly independent solutions. Let $p(t)$ and $q(t)$ be two continuous functions. Prove that the second order linear equation $$y'' + p(t)y' + q(t)y = 0$$ has two, and only two linearly independent solutions.
$\textbf{Sketch of Proof:}$
D$^2$y + pDy + qy = 0
(D$^2$ + pD + q)y = 0
Let L = D$^2$ + pD + q
L is a linear operator, Ly = 0
Claim: {c$_1$y$_1$ + c$_2$y$_2$} are all the solutions
Because L is a linear operator $\rightarrow$ L(c$_1$y$_1$ + c$_2$y$_2$) = L(c$_1$y$_1$) + L(c$_2$y$_2$)
= c$_1$L(y$_1$) + c$_2$L(y$_2$)
Since c$_1$y$_1$ + c$_2$y$_2$ is a solution and L(c$_1$y$_1$ + c$_2$y$_2$) = 0 $\rightarrow$ 
c$_1$L(y$_1$) = 0 and c$_2$L(y$_2$) = 0
I know I need to prove that c$_1$ and c$_2$ are 0, but I'm not sure how to prove that this is the only case.
 A: Assume that this refers to an open $t$-interval $\Omega$ with $0\in\Omega$. According to the standard existence theorem (extended to second-order ODEs) there are two solutions $y_1$, $y_2:\>\Omega\to{\mathbb R}$ realizing the particular initial values 
$$y_1(0)=1, \quad y_1'(0)=0,\qquad{\rm resp.}\qquad y_2(0)=0,\quad y_2'(0)=1\ .$$
The existence theorem guarantees this first for some neighborhood of $0$, but in the case of a linear ODE the two solutions can actually be extended to all of $\Omega$. The two functions $y_1$ and $y_2$ are obviously linearly independent. It follows that the solution space ${\cal L}$, which is a vector space in our case, has dimension $\geq 2$.
Now let $t\mapsto y(t)$ be an arbitrary solution. Then the function
$$y_*(t):=y(t)- y(0)y_1(t)-y'(0)y_2(t)$$
is also a solution, and in addition satisfies $y_*(0)=y_*'(0)=0$. But the zero solution satisfies these two conditions as well, and by the uniqueness part of the main theorem about ODE's it follows that in fact $y_*(t)\equiv0$. This proves that $y_1$, $y_2$ generate all of ${\cal L}$, whence ${\rm dim}({\cal L})\leq2$ as well.
