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Considering the following ODE : find $f(x)$ such that $$\frac{\sigma^{2}}{2}\frac{d^2}{dx^2}f(x)+a(b-x)\frac{d}{dx}f(x)-(\rho+\lambda)f(x)=-\lambda g(x) $$ Where, $a,b,\rho,\lambda,\sigma\in(0,+\infty)$, and $f(x)$ and $g(x)$ are assumed to have enough "good properties" !

Using the Fourier transform, one specific solution could be found, but I am interested in finding a general solution. It would be great if some one could give me some ideas. Thanks for your time and consideration.

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closed as too localized by Victor Protsak, Will Jagy, José Figueroa-O'Farrill, Yemon Choi, Harald Hanche-Olsen Sep 12 '10 at 21:46

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It's linear in $f$ –  Will Jagy Sep 12 '10 at 19:11
    
Hi Will, I am little confused by your comment, it would be great if you could clarify your comment for me –  Nameless Sep 12 '10 at 19:16
    
The difference of two solutions of an inhomogeneous linear equation is a solution of the corresponding homogeneous equation. –  Victor Protsak Sep 12 '10 at 19:41
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Nameless, given the elementary nature of this question and your comments indicating lack of understanding of basic facts about linear ODEs, I suggest (a) reading a textbook on ODEs; (b) asking your teachers (if you are a student); (c) posting follow-up questions on other sites listed at the FAQ. Voting to close. –  Victor Protsak Sep 12 '10 at 20:32

1 Answer 1

Will + Victor noted the general solution is your particular solution plus the general solution of the corresponding homogeneous DE. And, note, that this homogeneous DE does not depend on $g$. According to Maple, the solution of that homogeneous DE is... $$ f_\mathrm{homog} (x) = C_1 \mathrm{KummerM} \left(\frac{\rho + \lambda}{2 a},\frac{1}{2},\frac{a (b - x)^{2}}{\sigma^{2}}\right) + C_2 \mathrm{KummerU} \left(\frac{\rho + \lambda}{2 a},\frac{1}{2},\frac{a (b - x)^{2}}{\sigma^{2}}\right) $$

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Dear Gerald I thank for your comment but it is really sorry to say the solution you gave makes nonsense for me. All I need are the techniques used to solve the problem. Thanks again –  Nameless Sep 12 '10 at 20:20
    

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