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I'm looking at this: https://math.berkeley.edu/~mcivor/math54s11/wronskian.pdf. Which says this: "The fact that the Wronskian is nonzero at $x_0$ means that the square matrix on the left is nonsingular, hence this equation has only the solution $c_1 = c_2 = 0$, so $f$ and $g$ are independent." Where the "square matrix on the left" is the Wronskian and $f$ and $g$ are differentiable functions. I understand how this proves a solution, but I don't understand how this shows that the only solution is $c_1 = c_2 = 0$. Why can't it be any solution as long as $Wc = 0$?

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  • When a matrix $A$ is nonsingular (invertible), the only solution of $A\vec x=\vec 0$ is $\vec x=\vec 0$. – Ted Shifrin May 08 '21 at 00:34
  • @TedShifrin Oh. I did not know that. Thanks!! On further inspection that should have been obvious. – linkedin May 08 '21 at 00:43
  • Note that in $Wc=0$, $c$ is a constant (doesn't depend on $x$) and $W=W(x)$ is a function of $x$. So solving for the one point $x=x_0$ fix the value of $c$ for all $x$. – user10354138 May 08 '21 at 00:44
  • This question, through a little advanced for your needs, has a detail connection between wronskian and differential equations. – Mittens May 08 '21 at 00:50

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