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The Wronskian of the general solution $$y(x)=ay_1(x)+by_2(x)+cy_3(x)~,$$ to a third order differential equation, is given by $$ W = \begin{vmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \\ \end{vmatrix} $$ But if you have written $y(x)$ in a different order, such as $$by_2(x)+ay_1(x)+cy_3(x)~,$$ then you get a negative value because you've swapped rows of a matrix. So doesn't that mean the Wronskian should be a $\pm$ really?

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The importance of the Wronskian is the fact that its non-zero can tell you about linear independence (if it is zero, you need some additional conditions to say something about linear dependence).

The "sign" doesn't matter -- multiplying any of the functions you're testing by $-1$ would also change the sign of the determinant but lead you to the same logical conclusion (sign is in quotes, since the Wronskian can also be used for complex valued functions). That is, if the set of vectors {x,y,z} are linearly independent, so are {-x,y,z}, {-x,-y,z},{-x,-y,-z}, etc.

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The wronskian is a function, not a number, so you don't can't say it's lower or higher than $0(x)$. You may get either $ g(x) $ or $-g(x)$ depending on row placement but it matters little. You only care about whether or not said $ g(x) $ is 0 for all x.

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    $\begingroup$ Actually, since only one statement is true for Wronskian ("it is constantly zero", "it's never zero") you can compare it with $0(x)$. $\endgroup$ – Evgeny May 14 '15 at 17:00

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