Wronskian Determinants of Two Functions

# Wronskian Determinants of Two Functions

We are going to look more into second order linear homogenous differential equations, but before we do, we need to first learn about a type of determinant known as a Wronskian Determinant which we define below.

 Definition: Let $f$ and $g$ be two differentiable functions. Then the Wronskian Determinant of $f$ and $g$ is the $2 \times 2$ determinant $W(f, g) = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{vmatrix} = f(x)g'(x) + f'(x)g(x)$.

Sometimes the term "Wronskian" by itself is used to mean the same thing as "Wronskian Determinant". Furthermore, sometimes we can just write "$W$", or "$W(x)$" instead of $W(f, g)$ to represent the Wronskian of $f$ and $g$.

Let's look at some examples of computing the Wronskian determinant of two differentiable functions.

## Example 1

Determine the Wronskian of the functions $f(x) = x^2$ and $g(x) = 3x^2$. For what values of $x$ is the Wronskian equal to zero?

We note that $f$ and $g$ are both differentiable functions and that $f'(x) = 2x$ and $g'(x) = 6x$. Therefore the Wronskian of $f$ and $g$ is:

(1)
\begin{align} \quad \begin{vmatrix} x^2 & 3x^2\\ 2x & 6x \end{vmatrix} = x^2 \cdot 6x - 3x^2 \cdot 2x = 6x^3 - 6x^3 = 0 \end{align}

Therefore the Wronskian of $f$ and $g$ is equal to zero for all $x \in \mathbb{R}$.

## Example 2

Determine the Wronskian of the functions $f(x) = e^x \sin t$ and $g(x) = e^x \cos x$. For what values of $x$ is the Wronskian equal to zero?

We note that $f$ and $g$ are both differentiable functions and that $f'(x) = e^x \sin x + e^x \cos x$ and $g'(x) = e^x \cos x - e^x \sin x$. Therefore the Wronskian of $f$ and $g$ is:

(2)
\begin{align} \quad \begin{vmatrix} e^x \sin x & e^x \cos x \\ e^x \sin x + e^x \cos x & e^x \cos x - e^x \sin x \end{vmatrix} = \left ( e^x\sin x \right ) \cdot \left (e^x \cos x - e^x \sin x \right ) - \left ( e^x \cos x \right ) \cdot \left ( e^x \sin x + e^x \cos x \right ) \\ \quad = e^{2x} \sin x \cos x - e^{2x} \sin^2 x - e^{2x} \sin x \cos x - e^{2x} \cos^2 x = -e^{2x}(\cos^2 x + \sin^2 x) = -e^{2x} \end{align}

Note that $-e^{2x} < 0$ for all $x \in \mathbb{R}$ so the Wronskian of $f$ and $g$ is zero nowhere.