Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution

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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.

Let $\map {y_2} x$ be the real function defined as:

$\map {y_2} x = \map v x \map {y_1} x$

where:

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


Then $y_2$ and $y_1$ are linearly independent.


Proof

This will be demonstrated by calculating the Wronskian of $y_1$ and $y_2$ and demonstrating that it is non-zero everywhere.

First we take the derivative of $v$ with respect to $x$:

$v' = \dfrac 1 { {y_1}^2} e^{- \int P \rd x}$
\(\ds \map W {y_1, y_2}\) \(=\) \(\ds y_1 {y_2}' - y_2 {y_1}'\) Definition of Wronskian
\(\ds \) \(=\) \(\ds y_1 \paren {v y_1}' - v y_1 {y_1}'\) Definition of $y_2$
\(\ds \) \(=\) \(\ds y_1 \paren {v' y_1 + v {y_1}'} - v y_1 {y_1}'\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds {y_1}^2 v'\) simplifying
\(\ds \) \(=\) \(\ds {y_1}^2 \dfrac 1 { {y_1}^2} e^{-\int P \rd x}\) substituting for $v'$
\(\ds \) \(=\) \(\ds e^{-\int P \rd x}\) simplifying

As $\ds -\int P \rd x$ is a real function, $e^{-\int P \rd x}$ is non-zero wherever it is defined.

Hence from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent, $y_1$ and $y_2$ are linearly independent.

$\blacksquare$


Sources