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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.

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Because of the $e^{A+Bx}$ coefficient, I would be surprised if there was a closed form solution. An approximate solution generated by power series would at least give you an idea about the general behavior of the solution. If you have some particular application in mind, perhaps more information about the parameters (their magnitudes, e.g.) can be used to simplify further? – cch Apr 21 '11 at 15:14
I tried the power series method. But it is too complicate for my model... Thank you all the same. – William Apr 22 '11 at 3:23

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. :)

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Nice answer, very usefull, I think. – leo Aug 2 '11 at 14:05
Wrong. When the matrices $P(\tau)$ don't commute, you need a "time-ordered exponential", not an ordinary matrix exponential. – Robert Israel Feb 21 at 5:29

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})~) $

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:how do you derive the equation f(x) listed above. – qaz Jun 12 '12 at 19:25
@liu : I used Maple, but in principle what you can do is substitute $f(x) = 1 + c_2 x^2 + c_3 x^3 + \ldots$ or $x + c_2 x^2 + c_3 x^3 + \ldots$ and the Maclaurin series for $e^{A+Bx}$ into the differential equation, expand it out and look at terms in each power of $x$. – Robert Israel Jun 12 '12 at 20:11

Have you looked at Variation of constants?

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Variation of parameters can help solve the inhomogeneous equation, but you need the solutions to the homogeneous equation first. – cch Apr 21 '11 at 15:15
Thank you~~~. I will check this out. I think this may help. – William Apr 21 '11 at 16:05
Yes, cch is right. I just looked in to the methods. But the first thing is I need to get the fundamental solutions to the homogeneous equation. – William Apr 22 '11 at 20:11

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.


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Thank you for your answer. The parameters are non-zeros. So, it is kind of hard to handle. – William Apr 21 '11 at 15:52
@William: Sorry that didn't help. I will take some further looking into it. – night owl Apr 22 '11 at 3:16

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 $B=1$.

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