Let $I\subset\mathbb{R}$ be an open interval. Consider the following second order IVP \begin{align*} y''+a(x)y'+b(x)y=0,\ y(x_0)=y_0,\ y'(x_0)=y_1,\ x_0\in I, \end{align*} where $a,b:I\rightarrow\mathbb{R}$ are continuous functions and $y\in C^2(I)$. The linear differential operator $L:C^2(I)\rightarrow C(I)$ is defined as $Ly=y''+a(x)y'+b(x)y$.
I know that the solutions of the ODE are in the kernel of $L$, but I need to prove that $\ker(L)$ is isomorphic to $\mathbb{R}^2$ and, therefore, the general solution of the ODE is the linear combination of two linearly independent solutions.
Any suggestions?
Theorem: for every $y_0, y_1$, there exists a unique $y : I \to \mathbb{R}$ solving the initial value problem $L(y) = 0$, $y(x_0) = y_0$, $y'(x_0) = y_1$.
If you can prove this, then you have shown that the linear map $F : \ker(L) \to \mathbb{R}^2$ defined by $F(y) = (y(x_0), y'(x_0))$ is an isomorphism.
Existence and uniqueness follow from the general properties of linear ODEs.