Inhomogeneous Boundary Value Problem I am trying to solve the following BVP
$a V^{\prime \prime}(x) + b[c - x]V^{\prime}(x) + x V(x) = 0$ 
with the boundary conditions
$V(-\infty) = 0 \quad \mbox{and} \quad V(0) = 1$
I tried some standard approaches but nothing worked. 
However - Plugging into mathematica I get something in terms of hermite polynomials and hypergeometric functions which I've verified numerically for some parameters $a,b,c$ .
Can someone suggest whether titis is actually solvable ?
 A: To maintain the meaning of this question, $a\neq0$ and $b\neq0$ should be restricted
Hint:
Let $V(x)=\int_Ce^{xs}K(s)~ds$ ,
Then $a(\int_Ce^{xs}K(s)~ds)''+b(c-x)(\int_Ce^{xs}K(s)~ds)'+x\int_Ce^{xs}K(s)~ds=0$
$\int_C(as^2+bcs)e^{xs}K(s)~ds-\int_C(bs-1)e^{xs}K(s)~d(xs)=0$
$\int_C(as^2+bcs)e^{xs}K(s)~ds-\int_C(bs-1)K(s)~d(e^{xs})=0$
$\int_C(as^2+bcs)e^{xs}K(s)~ds-[(bs-1)e^{xs}K(s)]_C+\int_Ce^{xs}~d((bs-1)K(s))=0$
$\int_C(as^2+bcs)e^{xs}K(s)~ds-[(bs-1)e^{xs}K(s)]_C+\int_C((bs-1)K'(s)+bK(s))e^{xs}~ds=0$
$\int_C((bs-1)K'(s)+(as^2+bcs+b)K(s))e^{xs}~ds-[(bs-1)e^{xs}K(s)]_C=0$
$\therefore(bs-1)K'(s)+(as^2+bcs+b)K(s)=0$
$(bs-1)K'(s)=-((as+bc)s+b)K(s)$
$\dfrac{K'(s)}{K(s)}=-\dfrac{\left(a\left(s-\dfrac{1}{b}\right)+\dfrac{a}{b}+bc\right)s+b}{bs-1}$
$\int\dfrac{K'(s)}{K(s)}ds=-\int\dfrac{as\left(s-\dfrac{1}{b}\right)+\dfrac{(a+b^2c)s}{b}+b}{bs-1}ds$
$\int\dfrac{K'(s)}{K(s)}ds=-\int\dfrac{as\left(s-\dfrac{1}{b}\right)+\dfrac{a+b^2c}{b}\left(s-\dfrac{1}{b}\right)+\dfrac{a+b^2c}{b^2}+b}{bs-1}ds$
$\int\dfrac{K'(s)}{K(s)}ds=-\int\left(\dfrac{as}{b}+\dfrac{a+b^2c}{b^2}+\dfrac{a+b^3+b^2c}{b^2(bs-1)}\right)ds$
$\ln K(s)=-\dfrac{as^2}{2b}-\dfrac{(a+b^2c)s}{b^2}-\dfrac{a+b^3+b^2c}{b^3}\ln\left(s-\dfrac{1}{b}\right)+c_1$
$K(s)=C\left(s-\dfrac{1}{b}\right)^{-\frac{a+b^3+b^2c}{b^3}}e^{-\frac{as^2}{2b}-\frac{(a+b^2c)s}{b^2}}$
