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