Second order equation. (i)Show that the ODE $$y''+[b'(x)/b(x)]y'-[a^2/b^2(x)]
y=0$$ has a pair of linearly independant solutions that are reciprocals, where $a$ is a constant and $b(x)$ is a function of x. Find them in terms of $a$ and $b(x)$.
(ii)If the ODE $$y''+p(x)y'+2y=0$$ has solutions $y$ and $y^2$, find $y$ and find $p(x)$. Find both the possibilities.
 A: This is a seriously weird question. For the first one, first set $y=e^w$, so
$$ y' = w'y, \qquad y'' = (w''+w'^2)y, $$
and then the equation becomes
$$ w''+w'^2 + \frac{b'}{b}w' - \frac{a^2}{b^2} = 0 $$
Now, if both $y$ and $1/y$ satisfy the original equation, both $w$ and $-w$ satisfy this equation. Substituting $-w$ in,
$$ -w'' + w'^2 - \frac{b'}{b}w' - \frac{a^2}{b^2} = 0, $$
and adding these two equations gives
$$ w'^2 - \frac{a^2}{b^2} = 0, $$
and hence $w' = \pm a/b$. Then
$$ y = \exp{\left( \pm a \int \frac{dx}{b(x)} \right)}, $$
and we can check that this works:
$$ y' = \pm\frac{a}{b} y \qquad y'' = \left( \mp\frac{ab'}{b^2} + \frac{a^2}{b^2} \right)y, $$
which you can easily stick into the original equation.

For the second one, the same $y=e^w$ trick works, and substituting $w$ and $2w$ in and eliminating the second derivative gives
$$ w'^2 - 1 = 0, $$
so we find that $w = \pm x$. Substituting this into the $w''$ equation, we find that
$$ 0+1 \pm p(x) + 2 = 0, $$
so $p(x)=\mp 3$ when $x=\pm 1$.
