General solution to linear second-order homogeneous ODE question I understand that the solution to a linear second-order homogeneous ODE is in the form $y=c_1y_1+c_2y_2$ where $y_1$ and $y_2$ are solutions to the ODE. I also understand why a linear combination of two unique solutions is also a solution. I do not understand why $y=c_1y_1+c_2y_2$ will give the general solution which will take into account all particular solutions. Could anyone please clear up my misunderstanding? Thank you. 
 A: A linear second-order homogeneous ODE is has the form 
$y''+a(x)y'+b(x)y=0$,
where $a$ and $b$ are continuous functions on an intervall $I$.
Let $L$ be the set of all functions in $C^2(I)$, which are solutions of this ODE.
Then $L$ is a vector space and $ \dim L=2$. 
If $\{y_1,y_2\}$ is a basis of $L$ , then each solution of the ODE has the form $c_1y_1+c_2y_2$.
A: This is part of the general theory on first-order linear homogeneous systems of ODEs.
Namely, if you have a system in $\mathbb{R}^n$ of the form $x' = A(t) x$, with $A\colon I \to M_n$ continuous on some interval $I$ (here $M_n$ denotes the vector space of $n\times n$ matrices with real elements), then the set of all maximal solutions is a vector subspace of $C^1(I, \mathbb{R}^n)$ of dimension $n$.
Given this result, you have only to recast your second-order homogeneous linear equation to a first-order linear system in $\mathbb{R}^2$.
Indeed, if your equation is of the form
$$
y'' + a(t) y' + b(t) y = 0
$$
you can define $x_1 = y$, $x_2 = y'$, obtaining the system
$$
\begin{cases}
x_1' = x_2, \\
x_2' = - a(t) x_2 - b(t) x_1,
\end{cases}
$$
i.e. $x' = A x$ with
$$
A = \begin{pmatrix}
0 & 1\\
-b(t) & -a(t)
\end{pmatrix}\,.
$$
A: Let $y_0$ and $y_1$ be solutions with $$(y_0(0),y'_0(0))=(1,0)\text{ and }(y_1(0),y'_1(0))=(0,1).$$ These exist by the existence theorem.
For any other solution $y$ form the function $$u(x)=y(0)y_0(x)+y'(0)y_1(x).$$ Then this function is also a solution as a linear combination of solutions and it has the same initial values as $y$ in $x=0$. By the uniqueness theorem $u=y$ everywhere.
This proves that the dimension of the solution space is $2$, as $(y_0,y_1)$ is a basis.
