Wronskian Calculator

W(f₁,...,fₙ) determinant

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About Wronskian Calculator

A Wronskian calculator for testing linear independence of functions. Computes W(f₁,...,fₙ) = det of the matrix of derivatives. W ≠ 0 implies linear independence. Select from preset function sets. Evaluate at specific points. All calculations are client-side. Key tool for ODE solution verification.

Wronskian Calculator Features

  • Wronskian det
  • Function presets
  • Evaluate at x
  • Independence test
  • Derivative matrix
Wronskian: W(f₁,...,fₙ) = det[fᵢ⁽ʲ⁻¹⁾(x)]. If W(x₀) ≠ 0 for some x₀, functions are linearly independent. For solutions of linear ODEs: W ≡ 0 or W never zero (Abel's theorem). W = ce^(−∫p(x)dx) for y''+p(x)y'+q(x)y=0.

How to Use

Select functions:

  • Functions: Preset sets
  • x: Evaluation point
  • Output: W(x) value

Interpretation

W ≠ 0: linearly independent (conclusive). W ≡ 0: may or may not be dependent (inconclusive for general functions). For ODE solutions: W ≡ 0 ⟺ dependent (stronger result).

Abel's Theorem

For solutions of y''+p(x)y'+q(x)y=0: W(x) = W(x₀)exp(−∫p(t)dt). The Wronskian is either always zero or never zero. This connects the Wronskian to the ODE coefficients.

Step-by-Step Instructions

  1. 1Select function set.
  2. 2Enter x value.
  3. 3View derivative matrix.
  4. 4Get W(x).
  5. 5Test independence.

Wronskian Calculator — Frequently Asked Questions

Can W=0 with independent functions?+

Yes! For general functions (not ODE solutions), W≡0 doesn't imply dependence. Example: f=x², g=x|x| are independent on R, but W=0 everywhere. For analytic functions or ODE solutions, W≡0 does imply dependence.

How is the Wronskian used in solving ODEs?+

1) Verify independence of solutions. 2) Variation of parameters formula uses W in the denominator. 3) Abel's theorem gives W without computing it. 4) Reduction of order uses W to find a second solution from a first.

What is the Wronskian for eˣ, e²ˣ?+

W = |eˣ e²ˣ; eˣ 2e²ˣ| = 2e³ˣ − e³ˣ = e³ˣ ≠ 0. So eˣ and e²ˣ are linearly independent. This confirms they're valid fundamental solutions for y''−3y'+2y=0.

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