Recipe

Check two solutions for linear independence with the Wronskian

Use when: you have two solutions y_1,y_2 of the same equation y''+p(x)y'+q(x)y=0 and need to know whether they're independent

Goal: a verdict on independence, valid across the whole interval, from one computation

Taught in The Wronskian

1. 1

   Check that y_1 and y_2 both actually satisfy the same linear homogeneous equation y''+p(x)y'+q(x)y=0 on a common interval where p,q are continuous — that's what makes the "one point settles it" shortcut below valid at all.

   Example

   Take y_1=e^x, y_2=e^{-x}, and check them against y''-y=0: y_1''-y_1=e^x-e^x=0 and y_2''-y_2=e^{-x}-e^{-x}=0. Both check out.

   Why: the Wronskian theorem's all-or-nothing guarantee is proved specifically for two solutions of a shared linear homogeneous equation — it says nothing about the Wronskian of two arbitrary, unrelated functions.

2. 2

   Compute y_1' and y_2'.

   Example

   y_1=e^x \Rightarrow y_1'=e^x. y_2=e^{-x}\Rightarrow y_2'=-e^{-x}.

   Why: the Wronskian's formula needs both derivatives before it can be formed.

3. 3

   Plug the functions and derivatives into the determinant formula, then simplify to as clean a form as possible.

   Example

   W=e^x(-e^{-x})-e^x(e^{-x})=-1-1=-2

   Why: this determinant is exactly the coefficient matrix from the dependence-relation proof; it vanishes precisely when a nontrivial dependence relation exists.

4. 4

   Pick any convenient x_0 in the interval — an already-simplified constant like the one above needs no substitution at all — then read the verdict off its value.

   Example

   W=-2 is already a constant, nonzero for every x — no substitution needed. Since it's nonzero, e^x and e^{-x} are independent for every x.

   • W(x_0)\ne0 at even one point — y_1,y_2 are linearly independent on the whole interval.
   • W(x_0)=0 at even one point — W\equiv0 on the whole interval, and y_1,y_2 are linearly dependent.

   ⚡ Shortcut

   If you can already see that y_2 is a constant multiple of y_1 (or vice versa) just by inspection — say y_2=3y_1 — skip the Wronskian entirely: two functions in a constant ratio are dependent by definition, and the determinant would only confirm W\equiv0 the long way.

   Why: this is the all-or-nothing half of the Wronskian theorem — for solutions of the same linear homogeneous equation, one nonzero value forces the Wronskian nonzero (and the solutions independent) everywhere, and one zero value forces it zero (and the solutions dependent) everywhere.