How do you put
$u'' +c u' +d =0$
into regular SL-form? Can not see how it's an eigenvalue problem without a first order term. But the theorem states EVERY second order operator can be put into SL form, right?
How do you put
$u'' +c u' +d =0$
into regular SL-form? Can not see how it's an eigenvalue problem without a first order term. But the theorem states EVERY second order operator can be put into SL form, right?
To put the ODE into a SLE form, we need to find $p$ and $q$ such that
$$ -(p u')' + q u = f \iff u'' + cu' +d = 0 $$
If we expand the SLE, we see
$$ - p u'' -p'u' + qu = f \iff u '' + \frac{ p'}{p}u' - \frac{q}{p} u = -\frac{f}{p} $$
By simple comparison we see
$$q \equiv 0, \quad \& \quad \frac{p'}{p} = c, \quad \& \quad \frac{f}{p} = d $$