One can show that \begin{align} G(x;\xi)= \begin{cases} & \bigg(\frac{\xi^4-16}{60\xi^4}\bigg)\bigg(x^3-\frac{1}{x}\bigg),\qquad 1\le x<\xi\\ & \bigg(\frac{\xi^4-1}{60\xi^4}\bigg)\bigg(x^3-\frac{16}{x}\bigg),\qquad \xi< x\le 2 \end{cases} \end{align} is the Green's function for the operator $$L[y]=x^2y''-xy'-3y$$ with boundary conditions $y(1)=y(2)=0$. Hence, one wants to find the solution $y(x)$ to the differential equation $L[y]=x-3$ subject to the above conditions.
The Green's function can be written as $$G(x;\xi)=H(\xi-x)\bigg(\frac{\xi^4-16}{60\xi^4}\bigg)\bigg(x^3-\frac{1}{x}\bigg)+H(x-\xi)\bigg(\frac{\xi^4-1}{60\xi^4}\bigg)\bigg(x^3-\frac{16}{x}\bigg)$$ and the solution of the boundary value problem can be written as $$y(x)=\int_{1}^{2}G(x;\xi)f(\xi)d\xi=\int_{1}^{2}G(x;\xi)(\xi-3)d\xi.$$
But I struggle to evaluate correctly the integral since I do not understand what the functions $H(x-\xi)$ and $H(\xi-x)$ exactly represent and how to write them down precisely.
I would appreciate any help or suggestion. Thank you.
If this is indeed your Green's function, then the solution is
$$y(x) = \left (x^3-\frac{16}{x} \right ) \int_1^x d\xi \, \left (\frac{\xi^4-1}{60 \xi^4} \right ) (\xi-3) + \left (x^3-\frac{1}{x} \right ) \int_x^2 d\xi \, \left (\frac{\xi^4-16}{60 \xi^4} \right ) (\xi-3)$$
As is usually the case, the evaluation of these integrals is more messy than difficult. The solution I get is
$$y(x) = 1-\frac{11}{15 x} - \frac{x}{4} - \frac{x^3}{60} $$