# Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE

## 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 $\closedint a b$.

Let $y_1$ and $y_2$ be linearly independent.

Then their Wronskian is either never zero, or zero everywhere on $\closedint a b$.

## Proof

 $\displaystyle \map W {y_1, y_2}$ $=$ $\displaystyle y_1 {y_2}' - y_2 {y_1}'$ $\displaystyle \leadsto \ \$ $\displaystyle \map {W'} {y_1, y_2}$ $=$ $\displaystyle \paren {y_1 {y_2}'' + {y_1}' {y_2}'} - \paren {y_2 {y_1}'' + {y_2}' {y_1}'}$ Product Rule $\displaystyle$ $=$ $\displaystyle y_1 {y_2}'' - y_2 {y_1}''$

Because $y_1$ and $y_2$ are both particular solutions of $(1)$:

 $\text {(2)}: \quad$ $\displaystyle {y_1}'' + \map P x {y_1}' + \map Q x y_1$ $=$ $\displaystyle 0$ $\text {(3)}: \quad$ $\displaystyle {y_2}'' + \map P x {y_2}' + \map Q x y_2$ $=$ $\displaystyle 0$ $\text {(4)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle y_2 {y_1}'' + \map P x y_2 {y_1}' + \map Q x y_2 y_1$ $=$ $\displaystyle 0$ $(2)$ multiplied by $y_2$ $\text {(5)}: \quad$ $\displaystyle y_1 {y_2}'' + \map P x y_1 {y_2}' + \map Q x y_1 y_2$ $=$ $\displaystyle 0$ $(3)$ multiplied by $y_1$ $\text {(6)}: \quad$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {y_1 {y_2}'' - y_2 {y_1}''} + \map P x \paren {y_1 {y_2}' - y_2 {y_1}'}$ $=$ $\displaystyle 0$ $(5)$ subtracted from $(6)$

That is:

$\dfrac {\d P} {\d W} + P W = 0$

This is a linear first order ODE.

$W = C e^{-\int P \rd x}$

The exponential function is never zero:

Therefore:

$W = 0 \iff C = 0$

and the result follows.

$\blacksquare$