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Prove the Wronskian satisfies a first-order differential equation

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.

Let W be the Wronskian of two solutions u_1 and u_2 of the differential equations

    \[ y'' + ay' + by = 0, \]

where a and b are constants.

  1. Prove that W satisfies the first-order linear differential equation

        \[ W' + aW = 0 \]

    and hence,

        \[ W(x) = W(0) e^{-ax}. \]

    By this formula we can see that if W(0) \neq 0 then W(x) \neq 0 for all x.

  2. Assume u_1 is not identically zero and prove that W(0) = 0 if and only if \frac{u_2}{u_1} is constant.

  1. First, we evaluate W'(x) + aW(x) where W(x) = u_1(x) u_2'(x) - u_1'(x) u_2(x) is the Wronskian of the two functions u_1(x) and u_2(x).

        \begin{align*}  W'(x) + aW(x) &= u_1 (x) u_2''(x) - u_1''(x) u_2(x) + au_1(x) u_2'(x) - au_1'(x) u_2(x) \\  &= u_1 (x) \left( u_2''(x) + au_2'(x) \right) - u_2(x) \left(u_1''(x) + au_1'(x) \right) \\  &= u_1(x) \left( -bu_2(x)\right) - u_2(x) \left( -bu_1(x) \right) \\  &= 0. \qquad \blacksquare \end{align*}

    Furthermore, by Theorem 8.3 (page 310 of Apostol), since W(x) is a solution of W'(x) + aW(x) = 0 we know

        \[ W(x) = ce^{-ax} = W(0)e^{-ax} \]

    since W(0) = c. Hence, W(x) \neq 0 if W(0) \neq 0.

  2. Assume W(0) = 0. Then W(x) = 0 for all x. By part (a) of the previous exercise (Section 8.14, Exercise #21) we know \frac{u_2}{u_1} is constant.

    Conversely, assume \frac{u_2}{u_1} is constant. Then, again by the previous exericse, we have W(x) = 0 for all x. Hence, W(0) = 0. \qquad \blacksquare

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