Educational tool; results are estimates for learning, verify with your textbook or instructor.
This page is a Wronskian calculator for the homework search that asks for the Wronskian of 2 functions, or for a way to check linear independence of solutions. You type two or three formulas in the variable x. The page differentiates them, writes the Wronskian matrix, simplifies the determinant when it can, and always evaluates that determinant at a test point x₀ you choose.
The boxes open on e^x and e^(2x), with x₀ = 0. Those two exponentials are the usual first pair from a constant-coefficient equation with two different real roots. Their Wronskian is e^(3x), which is never 0, so the pair is linearly independent on the whole real line. The second worked example, further down, is sin x and cos x. Leave the third box hidden unless you really have three functions.
A nonzero Wronskian at even one point proves linear independence on every interval that contains that point. The converse is false, and it is the mistake this page is built to catch: a Wronskian of 0 at one point does not prove linear dependence. If you already have candidate solutions of a y'' + b y' + c y = 0, build them on the second order differential equation calculator, then bring the two formulas back here. The map around both tools is differential equations.
What the Wronskian is
Take differentiable functions y₁, y₂, and, when you have a third, y₃. The Wronskian matrix puts the functions in the first row and their derivatives down the columns. For two functions the matrix is 2 by 2. For three functions it is 3 by 3, and the last row is the second derivatives. The Wronskian W is the determinant of that matrix.
For two functions the determinant is the single expression W(y₁, y₂) = y₁ y₂′ − y₁′ y₂. The order matters for the sign. The opposite order, y₁′ y₂ − y₁ y₂′, is −W. A test that only asks whether W is zero does not care about that sign. This page keeps the standard order, functions on the top row, so the sign matches the textbook determinant.
The determinant is a function of x, not a single number, until you pick a point. That is why the page prints a formula and then a value W(x₀). The formula is the symbolic derivative when the expression is one this page knows how to differentiate: polynomials, exponentials, sine, cosine, tangent, natural log, square roots, and products and quotients of those. If a formula uses the absolute value, which has a corner, the page says so and switches that run to central differences at x₀. Those numbers are estimates. The symbolic runs are not.
How to read the result
The large line is the simplified determinant. Under it, a sentence says what that determinant does, and does not, imply about linear independence. The table is the matrix itself, so you can check each derivative before you trust the determinant. The steps repeat the same arithmetic in plain text.
Three readings cover the cases.
If W(x₀) is a nonzero number, the functions are linearly independent on every interval that contains x₀. One nonzero value is enough. You do not need the formula to be a nonzero constant. You need one point where it misses zero.
If the simplified formula is the constant 0, the Wronskian is the zero function wherever the derivatives exist. Polynomials, exponentials, sine, cosine, and logarithms are analytic on each interval where they are defined. For those functions, a Wronskian that is identically zero means linear dependence on that interval. A constant multiple is the usual picture: e^x and 2e^x have Wronskian 0 because the second is twice the first.
If W(x₀) = 0 but the formula is not the zero function, stop. That one root does not prove dependence. The pair x and x² has Wronskian x². At x₀ = 0 the value is 0, and the two functions are still linearly independent, because c₁ x + c₂ x² = 0 as a function only when both constants are 0. The preset "Zero at one point" opens on that pair.
Abel's identity, in one paragraph
The subtle reading changes when y₁ and y₂ are not arbitrary formulas but solutions of one second order linear equation y'' + p(x) y' + q(x) y = 0. Abel's identity says
W(x) = W(x₀) e^(−∫ p(s) ds),
with the integral running from x₀ to x. On an interval where p is continuous, the exponential factor is never zero. So W is either zero at every point of that interval or zero at none of them. For genuine solutions of that equation, a single zero really does prove dependence, and a single nonzero value really does prove independence. The identity is a statement about solutions, not about every pair of functions you can type. If you have not yet shown that your formulas solve the equation, a zero at one point is still not a proof of dependence.
Constant-coefficient equations are the course's main source of those solutions. The characteristic equation decides whether you get two real exponentials, a repeated root with an extra factor of x, or a sine and cosine pair. That algebra is on the second order page. The glossary line for Wronskian is the short version of this paragraph.
Two worked examples
Both examples are written out here, so you can read them with scripts turned off. The form opens on the first one. The "Sine and cosine" button loads the second.
Two exponentials. y₁ = e^x, y₂ = e^(2x).
Derivatives
y₁′ = e^x
y₂′ = 2 e^(2x)
Wronskian matrix
y₁ y₂
function e^x e^(2x)
y′ e^x 2 e^(2x)
Determinant
W = y₁ y₂′ − y₁′ y₂
= (e^x)(2 e^(2x)) − (e^x)(e^(2x))
= 2 e^(3x) − e^(3x)
= e^(3x)
At x₀ = 0
W(0) = e^0 = 1, which is not 0.
Conclusion
e^(3x) is never 0 for real x. The two functions are linearly independent on the whole real line. A nonzero value at one point is enough to prove independence. A Wronskian of 0 at one point would not have proved dependence.
Sine and cosine. y₁ = sin x, y₂ = cos x.
Derivatives
y₁′ = cos x
y₂′ = −sin x
Wronskian matrix
y₁ y₂
function sin x cos x
y′ cos x −sin x
Determinant
W = (sin x)(−sin x) − (cos x)(cos x)
= −sin(x)^2 − cos(x)^2
= −1
At x₀ = 0
W(0) = −1, which is not 0.
Conclusion
The determinant is the constant −1, so it is never 0. Sine and cosine are linearly independent on the whole real line. The same constant is what Abel's identity predicts for y'' + y = 0, where p(x) = 0 and the exponential factor is 1. A Wronskian of 0 at one point does not prove linear dependence. This example never meets that case.
Common mistakes
The first mistake is treating W(x₀) = 0 as a finished proof that the functions are dependent. It is not. Dependence means some constants, not all zero, make c₁ y₁ + c₂ y₂ + c₃ y₃ the zero function, everywhere on the interval, not at one sample. The pair x and x² is the counterexample to keep: the Wronskian vanishes at 0, and no such constants exist.
The second mistake is a sign slip in the 2 by 2 determinant, computing y₁′ y₂ − y₁ y₂′ and then comparing the formula with a solution key that used the other order. Both determinants are zero together. If the key says e^(3x) and you got −e^(3x), check the order of subtraction before you recompute the derivatives.
The third mistake is evaluating outside the domain. ln x and 1/x are not defined at 0, and tan x blows up at π/2. A blank or an infinite entry is not a zero Wronskian. Move x₀ to a point where every function and every derivative in the matrix is a real number. For a natural log, any positive test point works. At x₀ = 1, the Wronskian of ln x and x is ln(x) − 1, which equals −1.
The fourth mistake, with three functions, is to stop after a 2 by 2 minor. Independence of a triple is the full 3 by 3 determinant, with second derivatives in the last row. The preset "Three polynomials" uses 1, x, and x². The determinant is the constant 2, so the three monomials are linearly independent. They are the usual basis for quadratics.
The fifth mistake is quoting an identically zero Wronskian as a proof for functions that are not analytic. There are textbook pairs, built from absolute values, whose Wronskian is zero everywhere on an interval even though the functions are linearly independent. This page will not silently differentiate an absolute value. When the formula it did differentiate simplifies to 0, the analytic case applies, and the functions are dependent on each interval where they are all defined.
Related reading
Place the equation on the differential equations map. The short definitions are Wronskian, linear independence, differential equation, and determinant.
Build the candidate solutions on the second order differential equation calculator. A first order linear equation is a different object: the integrating factor calculator solves y' + P(x) y = Q(x). The same initial value problem can start from a transform instead. The Laplace transform calculator names the table row and inverts a proper rational function of s. If you want the determinant arithmetic alone, without derivatives, use the 2 by 2 matrix calculator. After you have an explicit formula, the numerical derivative calculator can audit a single slope.
Last reviewed: 25 October 2026.
