I am trying to find the Green's function for the following ODE, $$[(1+x)u']'+(\sin{x})u=0$$ with the boundary conditions $u(0)=u(1)=0$ and $x\in[0,1]$.
What I have done is tried to integrate the equation after simplifying (i.e differentiating the first term) and put the given conditions, but got stuck in mid way. How can I do this?
My work: $$(1+x)u''+u'+(\sin{x})u=0$$ $$\int_{0}^{x}(1+x)u''dx+u(x)-u(0)+\int_{0}^{x}\sin{x}~u~dx=0$$ $$(1+x)u'(x)-u'(0)+u(x)+\int_{0}^{x}\sin{x}~u~dx=0$$ (using integration by parts and $u(0)=0$). Repeating the same step again for last equality I couldn't simplify it to find a neat result.
Second thought: If I can find solution to this ODE $$u''+\dfrac{1}{1+x}u'+\dfrac{\sin{x}}{1+x}u=0$$