Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution

Theorem
Let $y_1 \left({x}\right)$ and $y_2 \left({x}\right)$ be particular solutions to the homogeneous linear second order ODE:
 * $(1): \quad \dfrac {\mathrm d^2 y} {\mathrm d x^2} + P \left({x}\right) \dfrac {\mathrm d y} {\mathrm d x} + Q \left({x}\right) y = 0$

on a closed interval $\left[{a \,.\,.\, b}\right]$.

Let $y_1$ and $y_2$ be linearly independent.

Then the general solution to $(1)$ is:
 * $y = C_1 y_1 \left({x}\right) + C_2 y_2 \left({x}\right)$

where $C_1 \in \R$ and $C_2 \in \R$ are arbitrary constants.

Proof
Let $y \left({x}\right)$ be any particular solution to $(1)$ on $\left[{a \,.\,.\, b}\right]$.

It is to be shown that constants $C_1$ and $C_2$ can be found such that:
 * $y \left({x}\right) = C_1 y_1 \left({x}\right) + C_2 y_2 \left({x}\right)$

for all $x \in \left[{a \,.\,.\, b}\right]$.

By Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions:
 * a particular solution to $(1)$ over $\left[{a \,.\,.\, b}\right]$ is completely determined by:
 * its value
 * and:
 * the value of its derivative

at a single point.

From Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE:
 * $C_1 y_1 \left({x}\right) + C_2 y_2 \left({x}\right)$

is a particular solution to $(1)$ over $\left[{a \,.\,.\, b}\right]$

We also have:
 * $y \left({x}\right)$

is a particular solution to $(1)$ over $\left[{a \,.\,.\, b}\right]$

Thus it is sufficient to prove that:
 * $\exists x_0 \in \left[{a \,.\,.\, b}\right]: \exists C_1, C_2 \in \R$ such that:
 * $ C_1 y_1 \left({x_0}\right) + C_2 y_2 \left({x_0}\right) = y \left({x_0}\right)$
 * and:
 * $ C_1 {y_1}' \left({x_0}\right) + C_2 {y_2}' \left({x_0}\right) = y \left({x_0}\right)$

For this system to be solvable for $C_1$ and $C_2$ it is necessary that:
 * $\begin{vmatrix}

y_1 \left({x}\right) & y_2 \left({x}\right) \\ {y_1}' \left({x}\right) & {y_2}' \left({x}\right) \\ \end{vmatrix} = y_1 \left({x}\right) {y_2}' \left({x}\right) - y_2 \left({x}\right) {y_1}' \left({x}\right) \ne 0$

That is, that the Wronskian $W \left({y_1, y_2}\right) \ne 0$ at $x_0$.

From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE:
 * if $W \left({y_1, y_2}\right) \ne 0$ at $x_0$, then $W \left({y_1, y_2}\right) \ne 0$ for all $x \in \left[{a \,.\,.\, b}\right]$.

Hence it does not matter what point is taken for $x_0$; if the Wronskian is non-zero at one such point, it will be non-zero for all such points.

From Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
 * $W \left({y_1, y_2}\right) = 0$ for all $x \in \left[{a \,.\,.\, b}\right]$ $y_1$ and $y_2$ are linearly dependent.

But we have that $y_1$ and $y_2$ are linearly independent.

Hence:
 * $\forall x \in \left[{a \,.\,.\, b}\right]: W \left({y_1, y_2}\right) \ne 0$

and so:
 * $\exists x_0 \in \left[{a \,.\,.\, b}\right]: \exists C_1, C_2 \in \R$ such that:
 * $ C_1 y_1 \left({x_0}\right) + C_2 y_2 \left({x_0}\right) = y \left({x_0}\right)$
 * and:
 * $ C_1 {y_1}' \left({x_0}\right) + C_2 {y_2}' \left({x_0}\right) = y \left({x_0}\right)$

The result follows.