# Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent

## Theorem

Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:

$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$

on a closed interval $\closedint a b$.

Then:

$y_1$ and $y_2$ are linearly dependent
the Wronskian $\map W {y_1, y_2}$ of $y_1$ and $y_2$ is zero everywhere on $\closedint a b$.

## Proof

### Sufficient Condition

Let $y_1$ and $y_2$ are linearly dependent.

Suppose either $y_1$ or $y_2$ is zero everywhere on $\closedint a b$.

Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\closedint a b$.

Thus:

$y_1 {y_2}' - y_2 {y_1}' = 0$

and so by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.

Without loss of generality, it follows by definition of linearly dependent that:

$y_2 = C y_1$

for some $C \in \R$.

Thus by Derivative of Constant Multiple:

${y_2}' = C {y_1}'$

Hence:

 $\ds C$ $=$ $\ds \frac {y_2} {y_1}$ $\ds \leadsto \ \$ $\ds {y_2}'$ $=$ $\ds \frac {y_2} {y_1} {y_1}'$ $\ds \leadsto \ \$ $\ds y_1 {y_2}'$ $=$ $\ds y_2 {y_1}'$ $\ds \leadsto \ \$ $\ds y_1 {y_2}' - y_2 {y_1}'$ $=$ $\ds 0$

Hence by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

$\Box$

### Necessary Condition

Let $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.

One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\closedint a b$.

Without loss of generality, suppose $y_1$ is zero everywhere on $\closedint a b$.

Then by Real Function is Linearly Dependent with Zero Function, $y_1$ and $y_2$ are linearly dependent.

Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.

As $y_1$ is continuous, there is some closed interval:

$\closedint c d \subseteq \closedint a b$

on which:

$\forall x \in \closedint c d: \map {y_1} x \ne 0$

Since $\map W {y_1, y_2} = 0$:

$\dfrac {y_1 {y_2}' - y_2 {y_1}'} { {y_1}^2} = 0$

on $\closedint c d$.

By Quotient Rule for Derivatives, this is the same as:

$\paren {\dfrac {y_2} {y_1} }' = 0$

Integrating with respect to $x$, this gives:

$\dfrac {y_2} {y_1} = k$

Explicitly:

$\forall x \in \closedint c d: \map {y_2} x = k \, \map {y_1} x$

As $\map {y_2} x$ and $k \, \map {y_1} x$ have equal values on $\closedint c d$, their derivatives are equal there as well.

$\forall x \in \closedint a b: \map {y_2} x = k \, \map {y_1} x$.

$\blacksquare$