Wronskian

Wronskian

Let L(y)=0 be a homogeneous equation of degree n, and let $y_{1},y_{2},...,y_{n}$ be linearly independent solutions of this equation.

The general solution is then $y=C_{1}y_{1}+C_{2}y_{2}+...+C_{n}y_{n}$.

Suppose instead that the solutions are not linearly independent. Then there exists values for $C_{1},C_{2},...,C_{n}$ not all zero such that $C_{1}y_{1}+C_{2}y_{2}+...+C_{n}y_{n}$=0.

Then the following equations also hold true:

$C_{1}y_{1}'+C_{2}y_{2}'+...+C_{n}y_{n}'=0$
$C_{1}y_{1}''+C_{2}y_{2}''+...+C_{n}y_{n}''=0$
...
$C_{1}y_{1}^{(n-1)}+C_{2}y_{2}^{(n-1)}+...+C_{n}y_{n}^{(n-1)}=0$

When there is to be a non-trivial solution to this system of homogeneous linear equations, the column vectors corresponding to each coefficient are linearly dependent. This is equivalent to saying that the following determinant, called the Wronskian is 0:

${\begin{Vmatrix}y_{1}&y_{2}&...&y_{n}\\y_{1}'&y_{2}'&...&y_{n}'\\y_{1}''&y_{2}''&...&y_{n}''\\...&...&...&...\\y_{1}^{(n-1)}&y_{2}^{(n-1)}&...&y_{n}^{(n-1)}\\\end{Vmatrix}}=0$

Thus, if the solutions are dependent, then their Wronskian is equal to 0, and conversely, and conversely if the Wronskian is equal to 0, then the columns of that matrix must be linearly dependent, indicating that the solutions are linearly dependent.

Resources

• Notes on the Wronskian. Produced by Paul Dawkins, Lamar University

Licensing

Content obtained and/or adapted from:

• Linear Equations, Wikibooks: Ordinary Differential Equations under a CC BY-SA license