I'm looking to solve the BVP $$(x-1)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + y = 1$$ subject to the conditions $y(0)=0,y(1)=2$, firstly by getting the problem into self adjoint form and then by finding the Greens function.
I think I have the self adjoint form for the problem, namely $$\frac{d}{dx}(\frac{e^{-x}}{x-1}\frac{dy}{dx}) +\frac{e^{-x}}{(x-1)^2}y=\frac{e^{-x}}{(x-1)^2}$$ where the boundary conditions remain unchanged as I have just multiplied through by a factor and then used the chain rule to simplify.
Then, I have noted that $y=x$ and $y=e^x$ are solutions to the homogeneous form of the above which are linearly independent since they have non-zero Wronskian on the domain we are considering.
Here I've also noted that $u=2x$ satisfies the homogeneous problem and the inhomogeneous boundary conditions, so then $\hat{y} = y - u$ is a solution to the inhomogeneous problem with the homogeneous boundary conditions (which we can find by the Greens function and use to get the solution $y$ of the problem we are considering).
It is in finding the Greens function that I'm struggling, as I seem to always get a wrong answer. I wonder if my understanding of the method is wrong. Could someone provide a step by step process of how to get the Greens function in this case, and then how to get a solution from that?
$$(x-1)y''-xy'+y(x)=1~~~(1)$$ let $z(x)=y(x)-1$, then n(1) can be written as $$(x-1)z''(x)-xz'+z=0$$ By inspection $z=x=z_1$ is one so;ution of this equation, then the other solution can be found as $$z_2=z_1\int \frac{1}{z_1^2} e^{\int x(x-1)^{-1}} dx= x\int e^{x}\left(\frac{1}{x}-\frac{1}{x^2}\right) dx= x \frac{e^{x}}{x}=e^x$$ So the general solution of (1) is $$y(x)=z(x)+1=C_1 x +C_2 e^x+1$$