Linear PDE with Variable Coefficients Help I would like to find the closed-form solution (for $x > 0$) to
$$\frac{\partial u}{\partial t} = a \frac{\partial^2 u}{\partial x^2} +b \frac{\partial u}{\partial x} + (cx+d) u,$$
    $$u(0,\,x) = 1,\;\; \frac{\partial u(t,\,0)}{\partial x} = 0,$$
where $a,\,b,\,c,\,d \in \mathbb{R}\backslash\{0\}$ are arbitrary constants. Though I looked in several books, I was not able to find a suitable approach with the given boundary/initial conditions. Any help would be appreciated!
 A: Let 
\begin{align}
u(x,t) = e^{- \lambda t} f(x)
\end{align}
to obtain the differential equation, in $x$, 
\begin{align}
a f^{''} + b f^{'} + (c x + d + \lambda) f = 0.
\end{align}
Now let 
\begin{align}
f(x) = e^{-(b x)/(2a)} g(x)
\end{align}
to obtain the differential equation
\begin{align}
g^{''} + \left( \frac{c}{a} \ x + \frac{d + \lambda}{a} - \frac{b^{2}}{4 a^{2}} \right) g = 0.
\end{align}
This is Airy's differential equation in a slightly different form and has the solution
\begin{align}
g(x) = c_{1} Ai\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) + 
c_{2} Bi\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right), 
\end{align}
where $\alpha = c/a$ and $\beta = (d+\lambda)/a - b^{2}/(4 a^{2})$.
The general solution is then 
\begin{align}
u(x,t) = e^{- \lambda t} \ e^{- bx/2a} \left[ c_{1} Ai\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) + 
c_{2} Bi\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) \right]
\end{align}
where $\alpha = c/a$ and $\beta = (d+\lambda)/a - b^{2}/(4 a^{2})$.
The remainder of the problem is using the initial conditions to find the constants $c_{1}$, $c_{2}$ and the separation constant $\lambda$. 
A: The Laplace transform with respect to $t$ leads to the differential equation
$$sY \left( x \right) -1=a\;{\frac {{\rm d}^{2}}{
{\rm d}{x}^{2}}}Y \left( x \right) +b\;{\frac {\rm d}{{\rm d}x}}Y
 \left( x \right) + (  c x+d)\; Y \left( x \right) 
$$
which has solutions involving Airy functions and integrals of Airy functions times exponentials (not surprising since the Airy differential equation is $w'' = x w$).  I would be surprised if the inverse
Laplace transform could be expressed in closed form.
A: Sorry for the late response. This is what the Polyanin book suggests:
Consider
$$u(t,\,x) = w(t,\,z) \exp\left( ctx-\frac{b}{2a}x+\frac{1}{3}ac^2t^3 + \left(d-\frac{b^2}{4a}\right)t \right),\;\;\;z=x+act^2$$.
Then
$$\partial_t w = a \partial_{zz} w$$.
However, the book makes no mention of boundary conditions. I would appreciate any tips!
Thanks,
Bryon
