I am looking at the proof of the following theorem.
Theorem(General solution of the homogeneous differential equation of second order with constant coefficients)
Let the differential equation $y''+a_1y'+a_2y=0, \ a_1, a_2 \in \mathbb{R}$.
We consider the characteristic polynomial of the above differential equation $p(r)=r^2+a_1r+a_2$.
Then, if there are two different solutions of the polynomial $p$, let $r_1, r_2 (r_1, r_2 \in \mathbb{R} \text{ or } r_1,r_2 \in \mathbb{C})$ then the functions $\phi_1(x)=e^{r_1 x}, \phi_2(x)=e^{r_2 x}$ are solutions of the differential equation in $\mathbb{R}$.
If there is a double root of the polynomial $p$, let $r$, then the functions $\phi_1(x)=e^{rx}, \phi_2(x)=xe^{rx}$ are solutions of the differential equation in $\mathbb{R}$.
It holds that : if $\phi$ is a solution of the differential equation in $\mathbb{R}$ then there are $c_1, c_2 \in \mathbb{R} (\text{ or } \mathbb{C})$ so that $\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x) \forall x \in \mathbb{R}$ and obviously for all $c_1, c_2 \in \mathbb{R} (\text{ or } \mathbb{C})$ the function $c_1 \phi_1(x)+c_2 \phi_2(x)$ is a solution of the differential equation.
Proof
We consider the solutions $\phi_1, \phi_2$ of the differential equation as at the formulation of the theorem.
Let $\phi$ be a random solution of the differential equation in $\mathbb{R}$.
We pick a random $x_0 \in \mathbb{R}$.
Then $\phi(x_0)=y_0$ and $\phi'(x_0)=y_1$.
We consider the initial value problem $\left\{\begin{matrix} y''+a_1y'+a_2y=0\\ y(x_0)=y_0\\ y'(x_0)=y_1 \end{matrix}\right.$.
Then we know that there is a solution $\psi$ of the initial value problem of the form $\psi(x)=c_1 \phi_1(x)+c_2 \phi_2(x)$ for appropriate $c_1, c_2$.
Furthermore, $\phi$ is a solution of the same initial value problem.
From the uniqueness of the initial value problem we have $\phi(x)=\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x), \forall x \in \mathbb{R}$.
I haven't understood the following:
We pick a random $x_0$ such that $\phi(x_0)=y_0$ and $\phi'(x_0)=y_1$ and then we consider the initial value problem.
How do we deduce that $c_1 \phi_1(x_0)+ c_2 \phi_2(x_0)=y_0$ and $c_2 \phi_1'(x_0)+c_2 \phi_2'(x)=0$, i.e. that $c_1 \phi_1(x)+c_2 \phi_2(x)$ is a solution of the specific initial value problem?