Author Topic: Wronskian for real equal roots  (Read 719 times)

Zhiya Lou

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Wronskian for real equal roots
« on: October 07, 2018, 10:39:37 PM »
For Wronskian determinant, $y_1$ and $y_2$ is the fundamental set of solution if W is not 0 (for all $t$?)
But, if W could be 0 for some t, does the statement still hold?

Particularly in the real equal root example:

$y''$- $2y'$ + $y$ = 0
$y_1$ = $e^t$
$y_2$ = $t$$e^t$

$W = t^2 e^{2t}-t e^{2t}  = 0$ if $t=1$ or $0$    WRONG

How to think about this kind of situation? if there is no restriction on $t$
« Last Edit: October 08, 2018, 12:41:40 AM by Victor Ivrii »

Kathryn Bucci

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Re: Wronskian for real equal roots
« Reply #1 on: October 08, 2018, 01:10:10 PM »
In that example $W=e^{2t}$ (you made a mistake somewhere). The exponential function is never zero so the Wronskian is always nonzero. Also look at Abel's theorem - that equation satisfies the conditions since it's a 2nd order linear homogenous equation - if you look at the identity in the theorem, you'll see that $W$ must either be always zero or never zero.