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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.16$: Problem $1$