I have the boundary value problem $$xy''=ky$$ $$y(0)=y'(1)=0$$ and want to find the adjoint, so I have
$L=x\frac{d^2}{dx^2}-k I$ $$(Lu,v)=(x\frac{d^2u}{dx^2}-k u,v)=\int_0^1 v(xu''-k u)dx=$$ $$\int_0^1 vxu''dx-\int_0^1 k uv dx=$$ $$[u'xv]_0^1-\int_0^1 u'(v+v'x)dx-\int_0^1 k uv dx=$$ $$-\int_0^1 u'vdx -\int_0^1 u'v'x -\int_0^1 k uv dx =$$ $$-[uv]_0^1 +\int_0^1 uv'dx -[uv'x]_0^1 +\int_0^1 u(v''x+v')dx-\int_0^1 k uv =$$ $$-u(1)v(1)-u(1)v'(1)+\int_0^1 u(xv''+2v'-k v)dx =-u(1)v(1)-u(1)v'(1)+(u,L^*v)$$
Now I demand $v(1)=0$ and $v'(1)=0$ so that the boundary terms vanish, so the adjoint operator is $$L^*=x \frac{d^2}{dx^2}+2\frac{d}{dx} +kI$$.
Is this correct?