Does this ODE question have closed form solution? These days, I am struggling with following ODE problem when I build up my research model:
$1/2f''(x)+a(b - x) f'(x) -(c+ e^{A+Bx})f(x)=0$
where f(x) is a smooth function, and $a,b,c, A,B$ are all constants.
How to get the closed form of f(x)?
I tried the Laplace transform to work on it, say $F(s) = L(f(x)) $, but because of $e^{A+Bx}$, there will be a term $F(s-B)$ in the transformed equation. How to deal with this term?
I also tried the power series method, but got some very complicate coefficients, which stops me going further. 
I think the term $e^{A+Bx}$ is the difficult part.
Could anyone here tell me how to deal with this kind of problem? Does the solution exit? I tried several ODE books but cannot find similar examples. Or could any one can suggest some relevant books?
Thank you very much. 
 A: Here's a nifty trick - you can transform any linear $n$th-order (in)homogeneous ordinary differential equation into a linear $1$st-order (in)homogeneous $n$-dimensional system of ODEs, which can be solved with the power of matrix exponentials. Behold:
Write $v=y'$, and reinterpret the differential equation as a first-order system:
$$\begin{pmatrix} y \\ v \end{pmatrix}' = \begin{pmatrix}0 & 1\\ -2(c+e^{A+Bx}) & 2a(b-x)\end{pmatrix} \begin{pmatrix} y \\ v \end{pmatrix}.$$
If we denote the column vector $(y,v)^T$ as $\vec{y}$, we can write this system as $\vec{y}'=P(x)\vec{y}$. The solution is actually the same as in the $1$-dimensional case, but you need the matrix exponential,
$$\vec{y} = \exp\left(\int_{x_0}^x P(\tau)d\tau\right) \vec{y}_0 $$
$$= \exp \begin{pmatrix}0 & x-x_0\\ \frac{2}{B}e^A(e^{Bx_0}-e^{Bx})+2c(x_0-x) & 2ab(x-x_0)+a(x_0^2-x^2)\end{pmatrix} \vec{y}_0. $$
Before we actually attempt to synthesize the incoming monstrosity, let's prepare ourselves by writing this matrix using variables: $\begin{pmatrix} 0 & \alpha \\ \beta & \gamma\end{pmatrix}$. Because, let's face it, the full analytical solution is going to be insanely complicated. Also, we are going to do it using the Putzer Algorithm (just keep in mind that we need to distinguish between $x$ and $t$; while using this method we must fix $x$). The eigenvalues are $\lambda_{1,2} = (\gamma\pm\sqrt{\gamma^2+4\alpha\beta})/2$. Solving for $p_1$ and $p_2$ using generic methods, and plugging them into the formula with $t=1$, we get
$$\vec{y}=\left[ \begin{pmatrix} e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2}\end{pmatrix} +\frac{e^{\lambda_1}-e^{\lambda_2}}{\lambda_1-\lambda_2} \begin{pmatrix} -\lambda_1 & \alpha \\ \beta & \gamma-\lambda_1 \end{pmatrix}\right]\vec{y}_0.$$
Just remember the $y$ you want is the first component of $\vec{y}$, and that $\lambda_{1,2}$ are functions of $\alpha,\beta,\gamma$, and that $\alpha,\beta,\gamma$ are all functions of $x$, so this technically contains your full analytical solution, just with a lot of substitutions. Have fun with your research OP. :)
A: Maple doesn't find a closed-form solution, so I suspect there isn't any (other than the trivial $f=0$).  Here are the first few terms of some series solutions: with $f(0)=1$, $f'(0)=0$, 
$f(x) = 1+ \left( c+{{\rm e}^{A}} \right) {x}^{2}+ \left( -2/3\,abc-2/3\,ab{{\rm e}^{A}}+1/3\,{{\rm e}^{A}}B \right) {x}
^{3}$ 
$~~~~~~~~~~~+(~1/6\,{c}^{2}+1/3\,c{{\rm e}^{A}}+1/6\, \left( {{\rm e}^{A
}} \right) ^{2}+1/3\,{a}^{2}{b}^{2}c+1/3\,{a}^{2}{b}^{2}{{\rm e}^{A}}-
1/6\,ab{{\rm e}^{A}}B+1/3\,ac+1/3\,a{{\rm e}^{A}}$
$~~~~~~~~~~~+1/12\,{{\rm e}^{A}}{
B}^{2}~) {x}^{4}+ ( 2/15\,{{\rm e}^{A}}Bc+2/15\, ( {
{\rm e}^{A}} ) ^{2}B+{\frac {1}{60}}\,{{\rm e}^{A}}{B}^{3}-2/15
\,ab{c}^{2}-{\frac {4}{15}}\,cab{{\rm e}^{A}}$
$~~~~~~~~~~~-2/15\,ab ( {{\rm e}
^{A}} ) ^{2}-2/15\,{a}^{3}{b}^{3}c-2/15\,{a}^{3}{b}^{3}{{\rm e}^
{A}}+1/15\,{a}^{2}{b}^{2}{{\rm e}^{A}}B-1/3\,{a}^{2}bc-1/3\,{a}^{2}b{
{\rm e}^{A}}$
$~~~~~~~~~~~-1/30\,ab{{\rm e}^{A}}{B}^{2}+1/10\,a{{\rm e}^{A}}B
 ) {x}^{5}+O( {x}^{6}) $
With $f(0)=0$, $f'(0)=1$:
$f(x) = x-ab{x}^{2}+ \left( 2/3\,{a}^{2}{b}^{2}+1/3\,a+1/3\,c+1/3\,{{\rm e}^{A}} \right) {x}^{3}$
$+\left( -1/3\,abc-1/3\,ab{
{\rm e}^{A}}-1/3\,{a}^{3}{b}^{3}-1/2\,{a}^{2}b+1/6\,{{\rm e}^{A}}B
 \right) {x}^{4}$
$+(~1/5\,{a}^{2}{b}^{2}c+1/5\,{a}^{2}{b}^{2}{
{\rm e}^{A}}+2/15\,{a}^{4}{b}^{4}+2/5\,{a}^{3}{b}^{2}-1/6\,ab{{\rm e}^
{A}}B+1/10\,{a}^{2}+2/15\,ac+2/15\,a{{\rm e}^{A}}$
$~~~~+1/30\,{c}^{2}+1/15\,
c{{\rm e}^{A}}+1/30\, ({{\rm e}^{A}}) ^{2}+1/20\,{{\rm e}
^{A}}{B}^{2}) {x}^{5}+O(x^{6})~) $
A: I am not sure if I have the right idea of how to go about solving this. With the knowledge that $a,~b,~c,~A,\text{ and }B$ are all constants, selecting nice choices for these will result the ODE equation to be of the form:
$$\dfrac{d^{2}y}{dx^{2}}-2x\dfrac{dy}{dx}+\lambda y=0$$
Thinking along these lines would result in using a series method to solve this ODE. This looks to me to be a Hermite Differential Equation problem. I would pursue along the lines in that direction, just hoping that making the assumption of the constants was not a terrible idea. 
I assumed that if $a=1,~b=0,~c=0,~A=1,\text{ and }B=0$.
Hope this helps out any and sorry for confusion if this is not applicable to your desired problem at hand.
Good~Luck.
A: Have you looked at Variation of constants?
http://en.wikipedia.org/wiki/Variation_of_parameters
A: One thing you can do is use a change of variables $x -> p x + q$ to reduce the number of parameters by two: you could, for example, take $A=0$ and $B=1$.
