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 January 28th, 2019, 04:28 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 93 Wronskian application What is Wronskian used for? Here is the page of definition, but I need a simple example. https://en.m.wikipedia.org/wiki/Wronskian Last edited by skipjack; January 28th, 2019 at 09:26 PM.
 January 28th, 2019, 08:16 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. There is a section on that wiki page involving ODEs as an example. Thanks from idontknow
 January 29th, 2019, 02:08 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 93 Wronskian shows whether the solutions are linearly independent. If they are linearly independent the page says, $\displaystyle y_1 =xy_2$ . Why not $\displaystyle y_1=(ax+b)y_2$ ?
 January 29th, 2019, 07:36 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 647 Thanks: 412 Math Focus: Dynamical systems, analytic function theory, numerics The only application I know of is an easy proof that some collection of functions are linearly independent. Namely, if $W(f_1,\dots,f_n) \neq 0$ on some domain, $D$, then the set of functions $\{f_1,\dots,f_n\}$ restricted to $D$ form a linearly independent set. Linear independence does not mean $y_1 = xy_2$ nor does it mean $y_1 = (ax + b)y_2$. I don't know where you got either of these formulas. The definition says that if $V$ is a (finite-dimensional) vector space, then $\{v_1,\dots,v_n\} \subset V$ are linearly independent if whenever $a_1v_1 + \dots + a_nv_n = 0$ you must have $a_j = 0$ for every $j$. In English, if a linear combination of linearly independent vectors is zero, then every coefficient is zero. In the case of the Wronskian, the vector space is the space of functions which are $n$ times differentiable so that this determinant makes sense. This is a standard example of a vector space which you should have seen in a first linear algebra class.

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