Consider the following differential equation $$f(x)\partial_ty(x,t)+\hat{M}y(x,t)=0\,,$$ where $\hat{M}$ is a differential operator with respect to $x$ and assume $y(x,t)=T(t)v(x)$. Then, $$f(x)T'(t)v(x)+\hat{M}v(x)T(t)=0$$ and hence $$\frac{T'(t)}{T(t)}=-\frac{\hat{M}v(x)}{f(x)v(x)}=\mathrm{const.}\equiv -\lambda\,.$$ We thus find $$\hat{M}v(x)=\lambda f(x) v(x)$$ and $$T'(t)+\lambda T(t)=0\,.$$ For $f\equiv\mathrm{const.}$, $\hat{M}$ is self-adjoint with the corresponding boundary conditions. In the literature (Rao, Singiresu S. Vibration of continuous systems. John Wiley & Sons, 2007.), I find the following conditions for self-adjointness (for arbitrary (multiplication) operators $f(x)$) $$\int\mathrm{d}x\,u_1(x)\hat{M}u_2(x)=\int\mathrm{d}x\,u_2(x)\hat{M}u_1(x)$$ and simultaneously $$\int\mathrm{d}x\,u_1(x)f(x)u_2(x)=\int\mathrm{d}x\,u_2(x)f(x)u_1(x)\,.$$ The question now is, how can this be true, where I can write the eigenvalue equation in the following way $$\frac{1}{f(x)}\hat{M}v(x)=\lambda v(x)$$ and so define a new operator $\hat{\tilde{M}}\equiv \frac{1}{f(x)}\hat{M}$, which is no more neccessarily self-adjoint with the corresponding boundary conditions as I will get derivatives of $f$ during partial integration?
I appreciate any help, thanks.