for $y''+p(x)y'+q(x)=0$, $-\frac{W'(u',v')(x)}{W(u,v)(x)}=p(x)$ and $q(x)=\frac{W(u',v')(x)}{W(u,v)(x)}$

Question

for $y''+p(x)y'+q(x)=0$, $-\frac{W'(u,v)(x)}{W(u,v)(x)}=p(x)$ and $q(x)=\frac{W(u',v')(x)}{W(u,v)(x)}$. It should be as simple as expanding the Wronskian, but I get a 4th degree derivative.

Answer 1

We recall that the Wronskian of any two differentiable functions $u$ and $v$ is defined as

$W(u, v) = \det \begin{bmatrix} u & v \\ u' & v' \end{bmatrix} = uv' - u'v; \tag 1$

if $u$ and $v$ are twice differentiable we have

$W'(u, v) = (uv' - u'v)' = u'v' + uv'' - u''v - u'v' = uv'' - u''v, \tag 2$

and if $u$ and $v$ are solutions of the ordinary differential equation

$y'' + p(x)y' + q(x)y = 0, \tag 3$

then it is easy to see that

$u'' = -p(x)u' - q(x)u, \tag 4$

$v'' = -p(x)v' - q(x)v. \tag 5$

When $u''$, $v''$ from (4), (5) are substituted into (2) we find

$W'(u, v) = uv'' - u''v = u(-p(x)v' - q(x)v) - (-p(x)u' - q(x)u)v$ $= -p(x)uv' - q(x)uv + p(x)u'v + q(x)uv$ $= -p(x)(uv' - u'v) = -p(x)W(u, v); \tag 6$

when $u$ and $v$ are linearly independent,

$W(u, v) \ne 0, \tag 7$

so (6) may be written

$p(x) = -\dfrac{W'(u, v)}{W(u, v)}. \tag 8$

Continuing in this vein, we have

$W(u', v') = \det \begin{bmatrix} u' & v' \\ u'' & v'' \end{bmatrix}$ $= u'v'' - u''v' = u'(-p(x)v' - q(x)v) - (-p(x)u' - q(x)u)v'$ $= -p(x)u'v' - q(x)u'v + p(x)u'v' + q(x)uv' = q(x)W(u, v), \tag 9$

and again if (7) holds then

$q(x) = \dfrac{W(u', v')}{W(u, v)}. \tag{10}$