2013-01-13T14:06:45ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=18808Thank you, this remark: "Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent." opened my eyes ]]>http://www.mathisfunforum.com/profile.php?id=1910882013-01-13T14:06:45Zhttp://www.mathisfunforum.com/viewtopic.php?pid=248631#p248631 I see. Well, if are differentiable, then linear independence implies for all . If they are not both differentiable, then it is possible that they are linearly independent yet . See http://en.wikipedia.org/wiki/Wronskian# … dependence for an example. ]]>http://www.mathisfunforum.com/profile.php?id=502822013-01-12T14:39:21Zhttp://www.mathisfunforum.com/viewtopic.php?pid=248483#p248483Of course but what i meant was: can I avoid to include W(t0)!=0 for some t0 in my hypotesis? In other words, if I have two linearly independent solutions u and v, can I automatically say W(u,v)!=0 for all t?]]>http://www.mathisfunforum.com/profile.php?id=1910882013-01-12T14:06:52Zhttp://www.mathisfunforum.com/viewtopic.php?pid=248478#p248478 Yes, if you can show that for some then for all .]]>http://www.mathisfunforum.com/profile.php?id=502822013-01-12T12:35:41Zhttp://www.mathisfunforum.com/viewtopic.php?pid=248476#p248476Hi guys, I'm studying some demonstrations about 2nd order differential equations of the form: y''+2by'+ay=f(t) where a,b are constants. Suppose that u,v are linearly independent solutions. Now, in several demonstrations, it's needed that the Wronksian determinant of u,v it's different from zero. I see from Abel's identity (http://en.wikipedia.org/wiki/Abel's_identity) that if this is true for some t0 value, then it's true for all t. Provided this, can I always say that the Wronksian of u,v is always non-zero??]]>http://www.mathisfunforum.com/profile.php?id=1910882013-01-12T10:06:44Zhttp://www.mathisfunforum.com/viewtopic.php?pid=248469#p248469