# What is the Wronskian?

Finds the Wronskian determinant of 2 to 4 functions of one variable and whether it shows linear independence, checked numerically.

- Page: https://www.acalculator.org/math/wronskian-calculator
- JSON spec: https://www.acalculator.org/math/wronskian-calculator.json
- Version: 30ab5c2616a2

## Default answer

Example with the default inputs (Functions e^x, e^(-x)): The Wronskian of e^x, e^(-x) is -2.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| f | Functions | 2 to 4 functions of one variable, separated by commas. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| w | Wronskian W | The determinant of the functions and their derivatives. |
| verdict | Linear independence | What W says about linear independence. |

## Method

W is the determinant whose rows are the functions and their 1st to (n − 1)th derivatives, each derivative checked against difference quotients.

## Assumptions

- Radians. An answer that fails its check is not shown.

## Worked examples

1. f = e^x, e^(-x) gives w = -2. Source: Trench (2013), Elementary Differential Equations, 5.1, Example 5.1.5 (a).
2. f = x^2, 1/x^2 gives w = -4/x. Source: Trench (2013), Elementary Differential Equations, 5.1, Example 5.1.5 (c).

## FAQ

### What is the Wronskian?

For functions y₁, …, yₙ of x, the Wronskian W is the determinant of the n × n matrix whose first row is y₁, …, yₙ, whose second row is their first derivatives, and so on up to the (n − 1)th derivatives. For two functions, W = y₁y₂′ − y₁′y₂.

### How does the Wronskian test linear independence?

If W is not 0 at some point of an interval, the functions are linearly independent on that interval. For e^x and e^(−x), W = −2, so they are independent. The converse needs care: W = 0 everywhere does not by itself prove dependence (x² and x|x| have W = 0 but are independent), but for solutions of one linear differential equation it does.

### Why is the Wronskian used with differential equations?

Solutions y₁, y₂ of a second-order linear equation y″ + p(x)y′ + q(x)y = 0 form a fundamental set, so every solution is c₁y₁ + c₂y₂, exactly when their Wronskian is not 0. Abel’s formula shows that such a Wronskian is either never 0 or always 0 on an interval.

### How is the answer checked?

Each derivative comes from a computer algebra system and must match a numeric difference quotient at 15 test points. The determinant is built from those checked derivatives, and each simplified form of it must equal that determinant at the same points. An answer that fails is not shown.

### How does the page decide that W is 0?

W counts as 0 when, at every one of 15 test points where it is defined, its value is within 10⁻⁹ of the sum of the sizes of the terms of the determinant. The page then says that the test does not show independence. W itself shows as 0 when the algebra reduces it to 0, and otherwise as the expression, such as sin(x)² + cos(x)² − 1.

### Which functions can I type?

Any functions of one letter (x, t, or another letter except e), separated by commas: polynomials, e^(kx), sin, cos, ln, sqrt and so on. A constant such as 1 is allowed. Two different letters give a message.

## Sources

- William F. Trench, Elementary Differential Equations with Boundary Value Problems, section 5.1 Homogeneous Linear Equations (LibreTexts): https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/05%3A_Linear_Second_Order_Equations/5.01%3A_Homogeneous_Linear_Equations
- OpenStax, Calculus Volume 3, section 7.1 Second-Order Linear Equations: https://openstax.org/books/calculus-volume-3/pages/7-1-second-order-linear-equations