I have this eigenvalue problem:
$$-x^2u_{xx} + 2xu_x - 2u = \lambda x^2 u$$
for $0 < x < 1$, with boundary conditions $u_x(0) = 0, u(1) = u_x(1)$.
I'm trying to find a function $M(x)$, such that under the change of variable $u(x) = M(x)w(x)$, it can be transformed to
$$w_{xx} = - \lambda w$$
for $0 < x < 1$.
The solution says this:
With $u = Mw$, the ODE $-x^2 u'' + 2xu' - 2u = \lambda x^2u$ becomes
$$-x^2 (M'' -aM' - bM) + 2x(M' w + Mw') - 2Mw = \lambda x^2Mw$$
$$-x^2 Mw'' + (2xM - 2x^2 M')w' + (-x^2 M'' + 2xM' - 2M)w = \lambda x^2Mw$$
I was wondering if someone could please explain where the $a$ and $b$ came from? Thanks.
Seems like a typo to me, since $u=Mw$ means $u''=M''w +2M'w' +Mw''$ rather than $M'' -aM' -bM$.