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Prove some properties of the Wronskian

The Wronskian W(x) is defined by

    \[ W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x) \]

for given functions u_1 and u_2.

  1. If the Wronskian W(x) of two functions u_1(x) and u_2(x) is zero for all x in an open interval I, prove that \frac{u_2(x)}{u_1(x)} is constant for all x \in I. Equivalently, if \frac{u_2}{u_1} is not constant on I then there is some c \in I such that W(c) \neq 0.
  2. Prove that the derivative of the Wronskian is given by

        \[ W'(x) = u_1(x) u_2''(x) - u_2(x)u_1''(x). \]

  1. Proof. (Note: I think we need the additional assumption that u_1(x) \neq 0 for any x \in I.) With our additional assumption we have,

        \[ d \left( \frac{u_2}{u_1} \right) = \frac{u_1 u_2' - u_1' u_2}{(u_1)^2} = 0 \]

    since W = u_1 u_2' - u_1' u_2 = 0 by assumption. Thus, \frac{u_2}{u_1} is constant by the zero derivative theorem. \qquad \blacksquare

  2. Proof. This is a direct computation of the derivative of the Wronskian,

        \begin{align*}  &&W &= u_1(x) u_2'(x) - u_1'(x) u_2(x) \\  \implies && W' &= u_1(x)u_2''(x) + u_1'(x) u_2'(x) - u_1''(x) u_2(x) - u_1'(x) u_2'(x) \\  \implies && W' &= u_1(x)u_2''(x) - u_1''(x) u_2(x). \qquad \blacksquare \end{align*}

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