-2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ It's linear in $f$ $\endgroup$
    – Will Jagy
    Sep 12 '10 at 19:11
  • $\begingroup$ Hi Will, I am little confused by your comment, it would be great if you could clarify your comment for me $\endgroup$
    – Nameless
    Sep 12 '10 at 19:16
  • $\begingroup$ The difference of two solutions of an inhomogeneous linear equation is a solution of the corresponding homogeneous equation. $\endgroup$ Sep 12 '10 at 19:41
  • 9
    $\begingroup$ 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. $\endgroup$ Sep 12 '10 at 20:32
4
$\begingroup$

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... $$ f_\mathrm{homog} (x) = C_1 \mathrm{KummerM} \left(\frac{\rho + \lambda}{2 a},\frac{1}{2},\frac{a (b - x)^{2}}{\sigma^{2}}\right) + C_2 \mathrm{KummerU} \left(\frac{\rho + \lambda}{2 a},\frac{1}{2},\frac{a (b - x)^{2}}{\sigma^{2}}\right) $$

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Nameless
    Sep 12 '10 at 20:20
  • $\begingroup$ en.wikipedia.org/wiki/Confluent_hypergeometric_function $\endgroup$
    – Will Jagy
    Sep 12 '10 at 21:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.