Wronskian test for linear independence of solutions

Statement

Let $I$ be an open interval, let $a_0, \dots, a_{n-1} : I \to \mathbb{R}$ be continuous, and let $y_1, \dots, y_n$ be solutions on $I$ of the homogeneous linear equation $y^{(n)} + a_{n-1}(x) y^{(n-1)} + \dots + a_0(x) y = 0$. Let $W = W(y_1, \dots, y_n)$ be their Wronskian. Then these three conditions are equivalent. (a) The functions $y_1, \dots, y_n$ are linearly independent on $I$. (b) There is a point $x_0 \in I$ with $W(x_0) \ne 0$. (c) For each $x \in I$, $W(x) \ne 0$.

Remarks

So the Wronskian of $n$ solutions is either zero at each point of $I$ or zero at no point of $I$. Abel's identity gives the same result for $n = 2$. For arbitrary functions, $W(x_0) \ne 0$ at one point still implies linear independence. But the converse fails: $x^2$ and $x |x|$ are linearly independent on $\mathbb{R}$, and their Wronskian is $0$ at each point. Example: $e^{x}$ and $e^{-x}$ solve $y'' - y = 0$, and their Wronskian is $-2$, so they are independent.

Explanations

L1 human

To check that $n$ solutions are independent, compute one determinant at one point. If it is not zero, the solutions are independent and form a basis of all solutions. For solutions of a linear equation, the value at one point already decides the question.