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

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

such that $y_1$ is not the trivial solution.

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

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

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

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

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

and:
 * ${y_2} = v \left({x}\right) y_1 \left({x}\right) + 2 v' \left({x}\right) {y_1}' \left({x}\right) + v \left({x}\right) {y_1}'' \left({x}\right)$

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

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

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

Integration gives:
 * $\displaystyle \ln v' = - 2 \ln y_1 - \int P \, \mathrm d x$

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

and so:
 * $\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{- \int P \, \mathrm d x} \, \mathrm d x$

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