$Y''(x)-\frac{Y'(x)}{x}=-a$, for $x \geq c$,
(1) $a>0$ and $c>0$ are constants.
(2) Boundary conditions are: $Y(c)=0$ and $Y'(\infty)=0$
I used $x=\tan(u)$ to change the variables into $u$. Hoping I didn't make any mistake in this process, it didn't lead to a reasonable solution.
Thanks for your help and consideration.
the solution of the homogeneous problem is $$y = Ax^2 + B \tag1 $$ we will try the variation of parameters by making $(1)$ a solution $$ y'' - \frac1x y' = -a \tag 2$$ by making constraints on $A, B$ which are now functions of $x.$
$$y' = A'x^2 + 2Ax + B' = 2Ax \text{ and } A' x^2 + B' = 0 \text{ so }y'' = 2A'x + 2A. \tag 3$$
putting these in $(2),$ we get another equation$$-a = 2A'x + 2A-\frac1x2Ax = 2A'x, A = -\frac a2 \ln x + C $$ and solving for $B'$ gives $$B' = \frac a 2 x, B = \frac a4+Dx^2 $$ finally we have for $$y = x^2\left(-\frac a2 \ln x + C \right) + \frac a4+Dx^2, y'=2Ax = -ax\ln x + 2Cx$$
the boundary condition $y'(\infty) = 0$ cannot be satisfied for any $C$. so i would say that your initial value problem has no solution.