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Can someone explain from where the aforementioned fact is derived? By the way $W = \text { Wronskian determinant}.$

David C. Ullrich
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user
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  • Thiis might help!, also observe that linear dependence implies zero Wronskian but not the converse, so the two sided implication is not correct, it should be a one sided $\implies$ implication. – BAYMAX Nov 11 '19 at 20:23
  • @BAYMAX OP has stated in the title "solutions to the differential equation...". The backwards direction becomes correct if you assume this additionally, more exactly, they have to solve a linear ODE, as written in the answer you have linked. – Jan Nov 11 '19 at 21:16
  • The notation $W(x_1(t_0),x_2(t_0))$ doesn't make much sense - probably you meant $W(x_1,x_2)(t_0)$. – David C. Ullrich Nov 13 '19 at 13:49

1 Answers1

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For the non-trivial direction, suppose $x_1$ and $x_2$ are solutions to $$x''+bx'+cx=0,$$define $$W=x_1x_2'-x_1'x_2,$$and assume $$W(t_0)=0.$$Define $$y(t)=x_2'(t_0)x_1(t)-x_1'(t_0)x_2(t).$$Of course $y$ is a solution to the original DE, being a linear combination of $x_1$ and $x_2$. It's clear that $y'(t_0)=0$, and $W(t_0)=0$ shows that $y(t_0)=0$. So uniqueness shows that $$y(t)=0,$$which says that $x_1$ and $x_2$ are dependent.

David C. Ullrich
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