Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another

Theorem
Let $\map {y_1} x$ be a particular solution to the homogeneous linear second order ODE:
 * $(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

such that $y_1$ is not the trivial solution.

Then there exists a standard procedure to determine another particular solution $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are linearly independent.

Proof
Let $\map {y_1} x$ be a non-trivial particular solution to $(1)$.

Thus, for all $C \in \R$, $C y_1$ is also a non-trivial particular solution to $(1)$.

Let $\map v x$ be a function of $x$ such that:
 * $(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$

is a particular solution to $(1)$ such that $y_1$ and $y_2$ are linearly independent.

Thus:
 * $(3): \quad {y_2}'' + \map P x {y_2}' + \map Q x y = 0$

Differentiating $(2)$ twice $x$:
 * ${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$

and:
 * ${y_2} = \map {v} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}''} x$

Substituting these into $(3)$ and rearranging:
 * $v paren { {y_1} + P {y_1}' + Q y_1} + v y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$

Since $y_1$ is a particular solution of $(1)$, this reduces to:
 * $v y_1 + v' \paren {2 {y_1} + P y_1} = 0$

which can be expressed as:
 * $\dfrac {v''} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$

Integration gives:
 * $\ds \ln v' = - 2 \ln y_1 - \int P \rd x$

This leads to:
 * $v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$

and so:
 * $\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

From Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution, $y_1$ and $v y_1$ are linearly independent.