 Solution

Expand the two operators before comparing coefficients:Equality $L=w\tilde{L}$ requires $a=\tilde{a}$, $c=\tilde{c}$, and $(aw{)}^{\prime }=w(a^{\prime }+\tilde{b})$. Thus $a{w}^{\prime }=\tilde{b}w$, giving the positive integrating factorWith $p=aw>0$, the result is the Sturm-Liouville form $L=-(pd/dx{)}^{\prime }-wc$. The arbitrary positive multiplier $C$ rescales both sides of the eigenvalue problem and does not change its eigenvalues. The coefficient ${\tilde{a}}^{\prime }$ is already present in the divergence term of $\tilde{L}$; it must not be counted a second time in the exponential.