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.