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The Wronskian of functions \(f_1, \dots, f_n\) is the determinant of the matrix whose first row is the functions and whose next rows are their first, second, … derivatives. If the Wronskian is not zero at some point, the functions are linearly independent. For example, \(W(e^x, e^{2x}) = e^{3x}\), which is never 0.
The Wronskian Formula
Two functions
\[W(f, g) = \begin{vmatrix} f & g \\ f’ & g’ \end{vmatrix} = f g’ – g f’\]
Three functions
\[W(f, g, h) = \begin{vmatrix} f & g & h \\ f’ & g’ & h’ \\ f” & g” & h” \end{vmatrix}\]
Each new function adds one more row of derivatives. So 4 functions need derivatives up to the third order. The calculator finds every derivative, builds the matrix, expands the determinant, and simplifies the answer.
Example 1: \(e^x\) and \(e^{2x}\)
\[W = \begin{vmatrix} e^x & e^{2x} \\ e^x & 2e^{2x} \end{vmatrix} = e^x \cdot 2e^{2x} – e^{2x}\cdot e^x = 2e^{3x} – e^{3x} = e^{3x}.\]
Because \(e^{3x} > 0\) for every \(x\), the two functions are linearly independent. They form a fundamental set of solutions of \(y” – 3y’ + 2y = 0\). You can solve that equation with the differential equation solver.
Example 2: \(\sin x\) and \(\cos x\)
\[W = \sin x\,(-\sin x) – \cos x\,(\cos x) = -(\sin^2 x + \cos^2 x) = -1.\]
The Wronskian is a nonzero constant, so sine and cosine are independent. They are the two building blocks of every solution of \(y” + y = 0\).
Example 3: \(1, x, x^2\)
\[W = \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & 2x \\ 0 & 0 & 2 \end{vmatrix} = 1\cdot1\cdot2 = 2.\]
The matrix is triangular, so the determinant is just the product of the diagonal.
How to Read Your Result
| Result | What it means |
|---|---|
| \(W \ne 0\) at even one point | The functions are linearly independent. |
| \(W = 0\) everywhere, and the functions solve the same linear differential equation | The functions are linearly dependent. |
| \(W = 0\) everywhere, for any other functions | The test does not decide. Check directly whether one function is a combination of the others. |
For example, \(W(x, 2x) = 0\) because \(2x\) is just 2 times \(x\). And \(W(e^x, e^{-x}, \cosh x) = 0\) because \(\cosh x = \tfrac12 e^x + \tfrac12 e^{-x}\).
Abel’s theorem shortcut. For two solutions of \(y” + p(x)y’ + q(x)y = 0\), the Wronskian is \(W = C e^{-\int p(x)\,dx}\). So it is either zero everywhere or never zero. In Example 1, \(p = -3\), and indeed \(W = e^{3x}\).
Learn the Math Behind the Wronskian Calculator
The Wronskian is a determinant of functions and their derivatives. If it is not zero, the solutions of a linear differential equation are independent.
Frequently Asked Questions
Why is the Wronskian used in differential equations?
A second-order linear equation needs two independent solutions to build the general solution \(y = C_1 y_1 + C_2 y_2\). The Wronskian is the quick test that the two solutions really are independent. It also appears in the variation of parameters formula.
Can the Wronskian be negative?
Yes. Only whether it is zero or not matters for independence. A negative Wronskian, like \(-1\) for sine and cosine, still means the functions are independent.
Does the order of the functions matter?
Swapping two functions swaps two columns of the matrix, which flips the sign of the Wronskian. It does not change whether it is zero.
Can I use t instead of x?
Yes. Type your functions with t, and change the Variable box to t.
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